Free Math Worksheets

- Read a normal vector from each plane equation. - The acute angle between two planes equals the acute angle between their normals. - Use the dot-product angle formula. - The requested plane angle is between \(0^\circ\) and \(90^\circ\).

1. Normal vectors are \(\vec{n}_1=\begin{pmatrix}1\\1\\0\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}1\\0\\1\end{pmatrix}\). 2. Their magnitudes are both \(\sqrt{2}\), and their dot product is \(1\). 3. The acute angle between the planes equals the acute angle between their normal vectors, so \(\cos\phi=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\). 4. Therefore, \(\phi=60^\circ\).

\(60^\circ\)

- How does a normal vector relate to every vector that lies in the plane? - Use the given point to form the displacement vector \(\langle x-7, y+2, z-4\rangle\). - Expand the dot product to obtain the standard equation.

1. In point-normal form, a point \(P\) on the plane and a normal vector \(\vec{n}\) give \(\vec{n}\cdot(\vec{x}-\vec{p})=0\). Thus, \(E: \langle 3, 5, -1\rangle\cdot\langle x-7, y+2, z-4\rangle=0\). 2. Expand the dot product: \(3(x-7)+5(y+2)-(z-4)=0\). 3. Simplify: \(3x+5y-z=7\).

Point-normal form: \(E: \langle 3, 5, -1\rangle\cdot\langle x-7, y+2, z-4\rangle=0\) Standard form: \(E: 3x+5y-z=7\)

- Subtract the coordinates of \(A\) from the coordinates of \(B\). - A normal vector is orthogonal to every displacement vector within the plane. - Substitute the coordinates of the given point to determine the constant in the standard equation.

1. Compute the normal vector: \(\overrightarrow{AB}=\langle 1-3, 2-(-1), 5-2\rangle=\langle -2, 3, 3\rangle\). 2. A point \((x, y, z)\) lies in the plane when its displacement from \(P\) is orthogonal to the normal vector: \(\langle -2,3,3\rangle\cdot\langle x-4,y,z-1\rangle=0\). 3. Expand and simplify: \(-2(x-4)+3y+3(z-1)=0\), so \(-2x+3y+3z=-5\).

\(E: -2x+3y+3z=-5\)

- What is the relationship between the direction vector of a line perpendicular to a plane and the plane’s normal vector? - Find a direction vector from the two points on the line. - Use the point on the \(z\)-axis in a point-normal equation.

1. Because the line is perpendicular to the plane, a direction vector of the line is a normal vector to the plane. Compute \(\overrightarrow{LM}=\langle 5-2,2-2,-1-2\rangle=\langle 3,0,-3\rangle\). 2. Use the simpler parallel normal vector \(\vec{n}=\langle 1,0,-1\rangle\). 3. Apply the point-normal condition at \(S\): \(\langle 1,0,-1\rangle\cdot\langle x,y,z-4\rangle=0\). 4. Simplifying gives \(x-z=-4\).

\(F: x-z=-4\)

- Read a normal vector from the coefficients of each equation. - Compare each normal vector with the direction vectors of the coordinate axes. - Check whether the origin satisfies each plane equation. - A zero dot product indicates orthogonal vectors.

1. Normal vectors are \(\vec{n}_1=\langle 4,0,3\rangle\) and \(\vec{n}_2=\langle 0,5,0\rangle\). 2. The direction vector of the \(y\)-axis is \(\langle 0,1,0\rangle\). Since \(\vec{n}_1\cdot\langle 0,1,0\rangle=0\), the \(y\)-axis is parallel to \(E_1\). The equation is satisfied at the origin, so the entire \(y\)-axis lies in \(E_1\). 3. The vector \(\vec{n}_2\) is parallel to the \(y\)-axis, so \(E_2\) is perpendicular to the \(y\)-axis and parallel to the \(xz\)-plane. Because \(5y=0\) is equivalent to \(y=0\), \(E_2\) is exactly the \(xz\)-plane.

\(E_1\) contains the \(y\)-axis. \(E_2\) is the \(xz\)-plane.

- Which coordinate-axis direction is perpendicular to the \(xy\)-plane? - Use the distance and the positive-axis condition to locate a point on the plane. - Combine that point with a normal vector in a point-normal equation.

1. A plane parallel to the \(xy\)-plane has a normal vector parallel to the \(z\)-axis, so use \(\vec{n}=\langle 0,0,1\rangle\). 2. The plane is \(4\) units from the origin and lies on the positive side of the \(z\)-axis, so it passes through \(P(0, 0, 4)\). 3. Point-normal form is \(\langle 0,0,1\rangle\cdot\langle x,y,z-4\rangle=0\). 4. The equivalent standard equation is \(z=4\).

Point-normal form: \(\langle 0,0,1\rangle\cdot\langle x,y,z-4\rangle=0\) Standard form: \(z=4\)

- Which coordinate remains constant on a plane parallel to the \(yz\)-plane? - Use a normal vector parallel to the \(x\)-axis. - For this special plane, the shortest distance is the absolute difference of the relevant coordinates.

1. A plane parallel to the \(yz\)-plane has normal vector \(\langle 1,0,0\rangle\). Because \(Q\) lies on the plane, every point on \(G\) has \(x=-2\). 2. Therefore the standard equation is \(x=-2\), and a point-normal equation is \(\langle 1,0,0\rangle\cdot\langle x+2,y-5,z-1\rangle=0\). 3. The perpendicular distance from \(R\) to this plane is the difference between the \(x\)-coordinates: \(\lvert 3-(-2)\rvert=5\).

a) \(x=-2\) b) \(\langle 1,0,0\rangle\cdot\langle x+2,y-5,z-1\rangle=0\) c) \(5\) units

- How do the components of a normal vector appear in a plane’s standard equation? - A point lies in a plane when its coordinates satisfy the equation. - Substitute \((0, 0, 0)\) into the general equation.

1. The components of the normal vector are the coefficients in the standard equation, so begin with \(4x+7y-z=d\). 2. Substitute the origin: \(4(0)+7(0)-0=d\). Therefore \(d=0\), and the plane is \(4x+7y-z=0\). 3. In general, substituting \((0, 0, 0)\) into \(ax+by+cz=d\) gives \(0=d\). Thus a plane in this form contains the origin exactly when \(d=0\).

a) \(4x+7y-z=0\) b) \(d=0\)

- The shortest segment from a point to a plane is parallel to the plane's normal vector. - Find the magnitude of the normal vector. - Use the point-to-plane distance formula, which is a scalar projection onto the normal.

1. A normal vector to the plane is \(\mathbf{n}=\langle 6,3,-2\rangle\), with magnitude \(\lVert\mathbf{n}\rVert=\sqrt{6^2+3^2+(-2)^2}=7\). 2. The signed plane expression at \(S\) is \(6(7)+3(9)-2(10)-14=35\). 3. The point-to-plane distance is the magnitude of the scalar projection onto the unit normal: \(d=\frac{\lvert 35\rvert}{7}=5\). 4. The shortest distance is \(5\) units.

\(5\) units

- Use the point-to-plane distance formula. - The denominator is the magnitude of the plane's normal vector. - Use an absolute value in the numerator. - Compare the two exact fractions.

1. A normal vector to the plane is \(\mathbf{n}=\langle 2,1,-2\rangle\), with magnitude \(\lVert\mathbf{n}\rVert=3\). 2. For \(P\), \(d(P,E)=\frac{\lvert 2(3)+12-2(1)-12\rvert}{3}=\frac{4}{3}\). 3. For \(Q\), \(d(Q,E)=\frac{\lvert 2(1)+2-2(-3)-12\rvert}{3}=\frac{2}{3}\). 4. Since \(\frac{4}{3}=2\left(\frac{2}{3}\right)\), \(P\) is twice as far from the plane as \(Q\).

a) \(d(P,E)=\frac{4}{3}\) and \(d(Q,E)=\frac{2}{3}\) b) \(d(P,E)=2d(Q,E)\)

- Identify the normal vector and a point on the plane from the point-normal equation. - Form the vector from the plane point to \(A\). - The distance is the absolute scalar projection onto the normal vector. - Divide the dot product by the normal vector's magnitude.

1. The plane's normal vector is \(\mathbf{n}=\langle 4,-12,6\rangle\), and a point on the plane is \(P=(1,1,0)\). 2. The vector from \(P\) to \(A\) is \(\overrightarrow{PA}=\langle 4,-2,4\rangle\). 3. The normal vector has magnitude \(\sqrt{4^2+(-12)^2+6^2}=14\). 4. The distance is the magnitude of the scalar projection of \(\overrightarrow{PA}\) onto the normal: \(d=\frac{\lvert\mathbf{n}\cdot\overrightarrow{PA}\rvert}{\lVert\mathbf{n}\rVert}=\frac{\lvert 4(4)+(-12)(-2)+6(4)\rvert}{14}=\frac{64}{14}=\frac{32}{7}\approx 4.57\).

\(\frac{32}{7}\approx 4.57\) units

- Identify the sphere's center from standard form. - The radius to the point of tangency is perpendicular to the tangent plane. - Use the center and tangent point to find a normal vector. - Substitute the tangent point into the plane equation.

1. The sphere's center is \(M=(3, -1, 2)\). 2. The radius vector to the point of tangency is \(\overrightarrow{MB}=\langle2, 3, 6\rangle\). This vector is normal to the tangent plane. 3. A plane with this normal vector has the form \(2x+3y+6z=d\). 4. Substitute \(B=(5, 2, 8)\): \(d=2(5)+3(2)+6(8)=64\). 5. Therefore, the tangent plane is \(2x+3y+6z=64\).

\(2x+3y+6z=64\)

- Use the direction vector from each line. - Apply the dot-product angle formula. - Use an absolute value to obtain the acute angle. - Convert the cosine value to degrees.

1. The direction vectors are \(\vec{u}=\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}-4\\4\\7\end{pmatrix}\). 2. Their dot product is \(18\), and their magnitudes are \(3\) and \(9\). 3. Thus \(\cos\alpha=\frac{|18|}{3\cdot9}=\frac{2}{3}\). 4. Therefore, \(\alpha=\cos^{-1}\left(\frac{2}{3}\right)\approx48.2^\circ\).

\(\alpha\approx48.2^\circ\)

- Read a normal vector from each plane equation. - The acute angle between planes equals the acute angle between their normals. - Use an absolute value in the dot-product formula. - Convert the cosine value to degrees.

1. Normal vectors are \(\vec{n}_1=\begin{pmatrix}2\\1\\-2\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}3\\-4\\0\end{pmatrix}\). 2. Their dot product is \(2\), and their magnitudes are \(3\) and \(5\). 3. Thus \(\cos\alpha=\frac{|2|}{3\cdot5}=\frac{2}{15}\). 4. Therefore, \(\alpha=\cos^{-1}\left(\frac{2}{15}\right)\approx82.3^\circ\).

\(\alpha\approx82.3^\circ\)

- Multiply corresponding components and add. - Relate the sign of the dot product to the sign of the cosine of the angle.

1. \(\vec{a}\cdot\vec{b}=4\cdot2+(-1)\cdot5+2\cdot(-3)=8-5-6=-3\). 2. Since the dot product is negative, the angle between the nonzero vectors is obtuse.

a) \(\vec{a}\cdot\vec{b}=-3\) b) The angle is obtuse.

- Test every unordered pair. - Orthogonal vectors have dot product \(0\). - Track negative signs carefully.

1. Compute all pairwise dot products: \(\vec{a}\cdot\vec{b}=0\), \(\vec{a}\cdot\vec{c}=0\), \(\vec{a}\cdot\vec{d}=8\), \(\vec{b}\cdot\vec{c}=0\), \(\vec{b}\cdot\vec{d}=4\), and \(\vec{c}\cdot\vec{d}=-10\). 2. The pairs with dot product \(0\) are orthogonal.

\((\vec{a},\vec{b})\), \((\vec{a},\vec{c})\), and \((\vec{b},\vec{c})\)

- Think of vectors along different coordinate axes. - Relate a zero dot product to the angle between nonzero vectors.

1. The statement does not carry over directly to dot products. 2. For example, let \(\vec{u}=\begin{pmatrix}1\\0\\0\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}0\\1\\0\end{pmatrix}\). 3. Both vectors are nonzero, but \(\vec{u}\cdot\vec{v}=0\). 4. Two nonzero vectors have dot product \(0\) exactly when they are orthogonal.

No. For example, \(\begin{pmatrix}1\\0\\0\end{pmatrix}\cdot\begin{pmatrix}0\\1\\0\end{pmatrix}=0\), although both vectors are nonzero. They are perpendicular.

- Identify the type of object produced by a dot product. - Check whether the outer operation has two vector inputs. - Associativity requires both regrouped expressions to be defined in the same set.

1. \(\vec{a}\cdot\vec{b}=1\) and \(\vec{b}\cdot\vec{c}=1\). 2. Each inner dot product is a scalar. The dot product is defined only between two vectors, not between a scalar and a vector. 3. Since neither regrouped expression is defined, an associative law for three vectors cannot be stated for the dot product.

1. \(\vec{a}\cdot\vec{b}=1\), \(\vec{b}\cdot\vec{c}=1\) 2. Both regrouped expressions are undefined because an inner dot product produces a scalar. 3. The dot product is not an operation from two vectors to another vector, so associativity does not apply.

- Track whether each intermediate result is a scalar or a vector. - Check that addition combines objects of the same type. - A dot product requires two vector inputs.

1. The expression is defined. \(\vec{a}\cdot\vec{b}\) is a scalar, and scaling \(\vec{c}\) produces a vector. 2. The expression is undefined because it attempts to add a scalar and a vector. 3. The expression is undefined because division by a vector is not defined. 4. The expression is undefined because \(\vec{b}\cdot\vec{c}\) is a scalar, while a dot product requires two vectors. 5. The expression is defined. The dot product is a scalar, so subtracting the scalar \(k\) produces a scalar.

1. Defined; vector 2. Undefined 3. Undefined 4. Undefined 5. Defined; scalar

- Form the side vectors that meet at each vertex. - A dot product of \(0\) identifies perpendicular vectors. - Stop once you have found a right angle.

1. Form two side vectors meeting at \(A\): \(\overrightarrow{AB}=\begin{pmatrix}4\\0\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\3\\5\end{pmatrix}\). 2. Their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\cdot0+0\cdot3+0\cdot5=0\). 3. Therefore, \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are perpendicular, so the triangle has a right angle at \(A\).

\(\triangle ABC\) has a right angle at \(A\).

- Multiply corresponding components and add. - For part c, first add the two vectors inside the parentheses. - The distributive property provides an alternative check.

1. \(\vec{a}\cdot\vec{b}=4\cdot1+(-3)\cdot2+2\cdot5=8\). 2. \(\vec{c}\cdot\vec{d}=(-2)\cdot3+6\cdot1+1\cdot0=0\). 3. \(\vec{b}+\vec{c}=\begin{pmatrix}-1\\8\\6\end{pmatrix}\). 4. Therefore, \(\vec{a}\cdot(\vec{b}+\vec{c})=4\cdot(-1)+(-3)\cdot8+2\cdot6=-16\).

a) \(8\) b) \(0\) c) \(-16\)

- Track whether each intermediate result is a scalar or a vector. - Check the input types required by each operation. - Recall that \(\vec{a}\cdot\vec{a}=\|\vec{a}\|^2\).

1. Part a is undefined. The inner dot product \(\vec{b}\cdot\vec{c}\) is a scalar, and a dot product requires two vectors. 2. Part b is undefined. The dot product \(\vec{a}\cdot\vec{b}\) is a scalar, which cannot be added to the vector \(\vec{c}\). 3. Part c is defined because both numerator and denominator are scalars and the denominator is nonzero. Since \(\vec{a}\cdot\vec{a}=\|\vec{a}\|^2\), the value is \(\|\vec{a}\|\). 4. The expression \(\sqrt{\vec{a}\cdot\vec{a}}=\|\vec{a}\|\) is a nonnegative real number. It generally cannot equal \(\vec{a}\), which is a vector.

a) Undefined. b) Undefined. c) Defined; its value is \(\|\vec{a}\|\). d) \(\sqrt{\vec{a}\cdot\vec{a}}=\|\vec{a}\|\) is a scalar, not a vector.

- Add \(\vec{b}\) and \(\vec{c}\) component by component. - Apply the component definition of the dot product. - Distribute and regroup the real-number terms.

1. \(\vec{b}+\vec{c}=\begin{pmatrix}b_1+c_1\\b_2+c_2\\b_3+c_3\end{pmatrix}\). 2. Therefore, \(\vec{a}\cdot(\vec{b}+\vec{c})=a_1(b_1+c_1)+a_2(b_2+c_2)+a_3(b_3+c_3)\). 3. Distribute and regroup: \((a_1b_1+a_2b_2+a_3b_3)+(a_1c_1+a_2c_2+a_3c_3)\). 4. These two sums are \(\vec{a}\cdot\vec{b}\) and \(\vec{a}\cdot\vec{c}\), proving the identity.

\(\vec{a}\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}\)

- Write the scaled vector in component form. - Expand the dot product. - Factor the common scalar using real-number properties.

1. \(k\vec{u}=\begin{pmatrix}ku_1\\ku_2\\ku_3\end{pmatrix}\). 2. Then \((k\vec{u})\cdot\vec{v}=(ku_1)v_1+(ku_2)v_2+(ku_3)v_3\). 3. By associativity of multiplication, this is \(k(u_1v_1)+k(u_2v_2)+k(u_3v_3)\). 4. By the distributive property, factor out \(k\): \(k(u_1v_1+u_2v_2+u_3v_3)=k(\vec{u}\cdot\vec{v})\).

\((k\vec{u})\cdot\vec{v}=k(\vec{u}\cdot\vec{v})\). The proof uses the associative property of multiplication and the distributive property.

- A dot product returns a scalar. - Check the sign of each component product. - Recall the exact component formula for a dot product.

1. In part a, the corresponding components were multiplied but not added, producing a vector instead of a scalar. The correct value is \(3\cdot2+2\cdot1+5\cdot0=8\). 2. In part b, \((-1)\cdot2=-2\), not \(2\). The correct value is \(4+6-2=8\). 3. In part c, all components were added instead of multiplying corresponding components and summing. The correct value is \(1\cdot0+2\cdot1+3\cdot2=8\).

a) The dot product must be a scalar; the correct result is \(8\). b) There is a sign error; the correct result is \(8\). c) Corresponding products must be added; the correct result is \(8\).

- Find each direction vector by subtracting endpoint coordinates. - Perpendicular vectors have dot product \(0\). - Keep \(a\) symbolic throughout the calculation.

1. The direction vectors are \(\overrightarrow{OF}=\begin{pmatrix}a\\a\\a\end{pmatrix}\) and \(\overrightarrow{EB}=\begin{pmatrix}0\\a\\-a\end{pmatrix}\). 2. Their dot product is \(\overrightarrow{OF}\cdot\overrightarrow{EB} =a\cdot0+a\cdot a+a\cdot(-a)=0\). 3. Therefore, the two diagonal directions are perpendicular for every \(a>0\).

Yes. Since \(\overrightarrow{OF}\cdot\overrightarrow{EB}=0\), the space diagonal \(OF\) and face diagonal \(EB\) have perpendicular directions.

- Read a normal vector from the equation of \(E\). - Test both direction vectors of \(F\) with the dot product. - Check whether the initial point of \(F\) lies in \(E\). - Distinct parallel planes do not have an intersection line.

1. A normal vector to \(E\) is \(\vec{n}=\langle 1,1,-2\rangle\). 2. Both direction vectors of \(F\) are orthogonal to \(\vec{n}\): \(\langle 2,0,1\rangle\cdot\vec{n}=2-2=0\) and \(\langle 1,1,1\rangle\cdot\vec{n}=1+1-2=0\). 3. Therefore \(F\) is parallel to \(E\). 4. The initial point of \(F\) does not satisfy the equation of \(E\), because \(1+2-2(1)=1\neq2\). 5. The planes are distinct and strictly parallel, so no intersection line exists.

The planes are strictly parallel. There is no intersection line.

- What relationship do the normal vectors of parallel planes have? - Keep the coefficients of \(x\), \(y\), and \(z\) proportional to those of the given plane. - Substitute the point on \(F\) to find the constant.

1. Parallel planes have parallel normal vectors, so \(F\) has the form \(5x-y+3z=d\). 2. Substitute \(P\): \(d=5(2)-6+3(-1)=10-6-3=1\). 3. Therefore \(F: 5x-y+3z=1\).

\(F: 5x-y+3z=1\)

- A beam striking a plane at a right angle points in a normal direction. - A displacement vector within the plane is orthogonal to the normal vector. - Expand the dot product and simplify.

1. Because the beam is perpendicular to the sensor, its direction vector is a normal vector to the plane. 2. For any point \(X\) in the plane, the displacement \(\overrightarrow{PX}=\langle x-5,y+1,z-4\rangle\) is orthogonal to \(\vec{v}\). Therefore \(\langle 2,2,-1\rangle\cdot\langle x-5,y+1,z-4\rangle=0\). 3. Expand: \(2(x-5)+2(y+1)-(z-4)=0\). 4. Simplifying gives \(2x+2y-z=4\).

a) \(\langle 2,2,-1\rangle\cdot\langle x-5,y+1,z-4\rangle=0\) b) \(2x+2y-z=4\)

- Find a direction vector from the two points on the line. - A line perpendicular to a plane points in a normal direction. - Use the given point to determine the constant in the plane equation. - Expand the dot product carefully.

1. A direction vector of \(g\) is \(\overrightarrow{AB}=\langle 2-0,1-4,0-(-2)\rangle=\langle 2,-3,2\rangle\). 2. Because \(E\) is perpendicular to \(g\), this direction vector is a normal vector to \(E\). 3. Use the point-normal condition: \(\langle 2,-3,2\rangle\cdot\langle x-3,y-3,z-3\rangle=0\). 4. Expanding and simplifying gives \(2x-3y+2z=3\).

\(E: 2x-3y+2z=3\)

- Compute the dot product component by component. - In \(\vec{n}\cdot\vec{x}=d\), the normal-vector components become the coefficients of the coordinates. - Let \(\vec{x}=\langle x,y,z\rangle\) and expand.

1. Compute \(d\): \(d=\langle 1,3,-2\rangle\cdot\langle 5,-2,4\rangle=1(5)+3(-2)-2(4)=5-6-8=-9\). 2. Therefore the vector equation is \(\langle 1,3,-2\rangle\cdot\vec{x}=-9\). 3. Writing \(\vec{x}=\langle x,y,z\rangle\) and expanding the dot product gives \(x+3y-2z=-9\).

a) \(d=-9\), so \(\langle 1,3,-2\rangle\cdot\vec{x}=-9\) b) \(x+3y-2z=-9\)

- Use the dot product to calculate \(d\). - The normal-vector components are the coefficients in standard form. - Substitute the coordinates of \(S\) into the equation and compare both sides.

1. Compute \(d=\vec{n}\cdot\vec{q}=\langle 4,-2,5\rangle\cdot\langle 3,0,-1\rangle=4(3)-2(0)+5(-1)=7\). 2. Thus \(\langle 4,-2,5\rangle\cdot\vec{x}=\langle 4,-2,5\rangle\cdot\langle 3,0,-1\rangle=7\). 3. Expanding gives \(4x-2y+5z=7\). 4. For \(S\), the left side is \(4(1)-2(2)+5(0)=0\). Since \(0\neq7\), \(S\) does not lie in \(F\).

a) \(\langle 4,-2,5\rangle\cdot\vec{x}=\langle 4,-2,5\rangle\cdot\langle 3,0,-1\rangle=7\) b) \(4x-2y+5z=7\) c) No, \(S\notin F\).

- A plane perpendicular to a line has a normal vector parallel to the line’s direction vector. - Use the normal-vector components as the variable coefficients. - Substitute a point on each plane to find its constant.

1. The direction vector \(\langle 4,1,-2\rangle\) is a normal vector to both planes. 2. For \(E_1\), substitute \((3, -2, 5)\) into \(4x+y-2z=d_1\): \(d_1=4(3)+(-2)-2(5)=0\). Thus \(E_1: 4x+y-2z=0\). 3. For \(E_2\), substitute \(P(0, 6, 1)\): \(d_2=4(0)+6-2(1)=4\). Thus \(E_2: 4x+y-2z=4\).

a) \(E_1: 4x+y-2z=0\) b) \(E_2: 4x+y-2z=4\)

- How is the direction of a line perpendicular to a plane related to the plane's normal vector? - Which coefficients in the plane equation give a normal vector? - What information is needed to write a parametric equation of a line?

1. A normal vector to plane \(E\) is \(\vec{n}=\begin{pmatrix}4\\-1\\2\end{pmatrix}\). 2. A line perpendicular to a plane has a direction vector parallel to the plane's normal vector. 3. Using \(P\) as the position vector gives \(g:\vec{x}=\begin{pmatrix}3\\-1\\4\end{pmatrix}+r\begin{pmatrix}4\\-1\\2\end{pmatrix}\).

\(g:\vec{x}=\begin{pmatrix}3\\-1\\4\end{pmatrix}+r\begin{pmatrix}4\\-1\\2\end{pmatrix}\), where \(r\in\mathbb{R}\)

- Evaluate each dot product component by component. - For points in the same plane with normal vector \(\vec{n}\), the value of \(\vec{x}\cdot\vec{n}\) is constant. - Substitute the coordinates of \(C\) into the plane equation.

1. \(\overrightarrow{OA}\cdot\vec{n}=\langle 1,2,3\rangle\cdot\langle 1,1,1\rangle=1+2+3=6\). Also, \(\overrightarrow{OB}\cdot\vec{n}=\langle 5,0,1\rangle\cdot\langle 1,1,1\rangle=5+0+1=6\). 2. Since both points give the same constant, the plane equation is \(x+y+z=6\). 3. Substitute \(C\): \(k+3+0=6\), so \(k=3\).

1) Both dot products equal \(6\). 2) \(E: x+y+z=6\) 3) \(k=3\)

- Identify the line’s direction vector and the plane’s two direction vectors. - Use a dot product to test each pair for orthogonality. - Check that the two plane directions are not scalar multiples.

1. The line direction is \(\vec{w}=\langle 2,-2,1\rangle\). The plane directions are \(\vec{u}=\langle 1,1,0\rangle\) and \(\vec{v}=\langle 0,1,2\rangle\). 2. The plane directions are not scalar multiples, so they define two distinct directions in the plane. 3. Compute \(\vec{w}\cdot\vec{u}=2-2+0=0\). 4. Compute \(\vec{w}\cdot\vec{v}=0-2+2=0\). 5. Because the line direction is orthogonal to both nonparallel plane directions, \(g\) is perpendicular to \(E\).

Line \(g\) is perpendicular to plane \(E\).

- Substitute a point’s position vector into the dot product. - A point lies in the plane when the resulting value equals \(8\). - For part b, treat the equation as a linear equation in \(x\). - Solve after substituting the two known coordinates.

1. For \(P\), \(\langle 5,-2,1\rangle\cdot\langle 1,0,3\rangle=5(1)-2(0)+3=8\). Therefore \(P\in E\). 2. For \(Q\), substitute the known coordinates: \(5x-2(4)+1=8\). 3. Simplify: \(5x-7=8\), so \(5x=15\) and \(x=3\).

a) Yes, \(P\in E\). b) \(x=3\)

- Find \(d\) by taking the dot product of the normal vector with the given point’s position vector. - Test each point by evaluating the same dot product. - A point belongs to the plane only when the result equals \(d\). - Keep track of the negative component of the normal vector.

1. Compute the constant using \(A\): \(d=\langle 2,3,-4\rangle\cdot\langle 1,2,1\rangle=2+6-4=4\). Thus \(E: \langle 2,3,-4\rangle\cdot\vec{x}=4\). 2. For \(B\), \(2(4)+3(0)-4(1)=4\), so \(B\in E\). 3. For \(C\), \(2(2)+3(1)-4(1)=3\neq4\), so \(C\notin E\).

a) \(E: \langle 2,3,-4\rangle\cdot\vec{x}=4\) b) \(B\in E\), but \(C\notin E\).

- Write the line in parametric form. - Use a dot product to confirm that the line is not parallel to the plane. - Substitute the line coordinates into the plane equation. - Substitute the solved parameter into the line.

1. Parametrize the line: \(\vec{r}(t)=\langle 5,-2,1\rangle+t\langle 2,1,-1\rangle\). 2. A normal vector to the plane is \(\vec{n}=\langle 3,-1,2\rangle\). Since \(\vec{n}\cdot\vec{d}=6-1-2=3\neq0\), the line intersects the plane once. 3. Substitute the line coordinates into the plane equation: \(3(5+2t)-(-2+t)+2(1-t)=10\). 4. Simplifying gives \(19+3t=10\), so \(t=-3\). 5. The intersection point is \(S=\langle 5,-2,1\rangle-3\langle 2,1,-1\rangle=\langle -1,-5,4\rangle\).

\(S(-1, -5, 4)\)

- Subtract the coordinates of \(L\) from those of \(M\) to get the ray direction. - Parametrize the ray with \(t\ge0\). - Substitute the ray coordinates into the plane equation. - Check the sign of the parameter.

1. The ray direction is \(\overrightarrow{LM}=\langle 2,3,2\rangle\), so \(\vec{r}(t)=\langle 4,-2,1\rangle+t\langle 2,3,2\rangle\), with \(t\ge0\). 2. A normal vector to the wall is \(\vec{n}=\langle 3,-1,2\rangle\). Since \(\vec{n}\cdot\overrightarrow{LM}=6-3+4=7\neq0\), the line intersects the wall once. 3. Substitute into the plane equation: \(3(4+2t)-(-2+3t)+2(1+2t)=30\). 4. This simplifies to \(16+7t=30\), so \(t=2\). 5. Since \(t>0\), the intersection is on the ray. The point is \(S=\langle 4,-2,1\rangle+2\langle 2,3,2\rangle=(8, 4, 5)\).

\(S(8, 4, 5)\)

- Read a normal vector from the plane equation. - Use a dot product to test parallelism. - Test the line’s initial point in the plane equation. - A parallel line has either no common points or infinitely many.

1. A normal vector to \(E\) is \(\vec{n}=\langle 1,2,-2\rangle\). 2. The line direction satisfies \(\langle 2,2,3\rangle\cdot\vec{n}=2+4-6=0\), so \(g\) is parallel to \(E\). 3. Test the initial point: \(4+2(1)-2(1)=4\neq8\). Thus the point is not in the plane. 4. Therefore \(g\) is strictly parallel to \(E\), and they have no common points.

The line is strictly parallel to the plane, so they have no common points.

- Check whether both directions of the parametric plane are orthogonal to the normal of the standard-form plane. - Substitute the initial point of the parametric plane. - Parallel planes are identical only when they share a point.

1. A normal vector to \(E_1\) is \(\vec{n}=\langle 2,1,-2\rangle\). 2. The two directions of \(E_2\) satisfy \(\vec{n}\cdot\langle 1,0,1\rangle=2-2=0\) and \(\vec{n}\cdot\langle 1,-2,0\rangle=2-2=0\). 3. Therefore, \(E_2\) is parallel to every plane in the family \(E_1\). 4. Substitute the initial point \((1, 2, 1)\) of \(E_2\) into the left side of \(E_1\): \(2(1)+2-2(1)=2\). 5. Thus the planes are identical when \(d=2\). 6. For \(d\neq2\), the planes have parallel normals but no common point, so they are strictly parallel.

The planes are identical for \(d=2\). For every \(d\neq2\), they are strictly parallel.

- Read a normal vector from the plane equation. - Find the normal vector's magnitude. - Substitute the point into the point-to-plane distance formula.

1. A normal vector to the plane is \(\mathbf{n}=\langle 4,-4,7\rangle\), with magnitude \(\sqrt{4^2+(-4)^2+7^2}=9\). 2. Apply the point-to-plane distance formula: \(d=\frac{\lvert 4(2)-4(-1)+7(6)-9\rvert}{9}\). 3. Simplify: \(d=\frac{\lvert 45\rvert}{9}=5\).

\(5\) units

- Use the line direction and plane normal. - The sine formula gives the line-plane angle. - Take the absolute value of the dot product.

1. The line direction is \(\vec{u}=\begin{pmatrix}3\\4\\0\end{pmatrix}\), and a plane normal is \(\vec{n}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\). 2. Their dot product is \(-2\), and their magnitudes are \(5\) and \(3\). 3. Thus \(\sin\alpha=\frac{|\vec{u}\cdot\vec{n}|}{\|\vec{u}\|\|\vec{n}\|}=\frac{2}{15}\). 4. Therefore, \(\alpha=\sin^{-1}\left(\frac{2}{15}\right)\approx7.66^\circ\).

\(\alpha\approx7.66^\circ\)

- Use the point-to-plane distance formula in both parts. - A tangent sphere's radius equals the perpendicular distance from its center to the plane.

1. A normal vector to \(G\) is \(\langle 4,-4,7\rangle\), with magnitude \(9\). 2. The distance from \(P\) is \(d(P,G)=\frac{\lvert 4(1)-4(-2)+7(4)-18\rvert}{9}=\frac{22}{9}\approx2.44\). 3. A sphere tangent to a plane has radius equal to the distance from its center to the plane. 4. Therefore, \(r=d(M,G)=\frac{\lvert 4(5)-4(2)+7(2)-18\rvert}{9}=\frac{8}{9}\approx0.89\).

a) \(\frac{22}{9}\approx2.44\) units b) \(r=\frac{8}{9}\approx0.89\) unit

- A point lies in the plane when its coordinates make the equation true. - Substitute the known coordinates and solve the resulting linear equation for \(k\). - Compare the evaluated left side with \(8\); values greater than \(8\) lie toward the chosen normal vector.

1. For \(A\), \(4(1)-2(-1)+2=8\), so \(A\in E\). 2. For \(B\), \(4(2)-2k+4=8\). Thus \(12-2k=8\), so \(k=2\). 3. For \(C\), \(4(3)-2(5)+1=3\). Since \(3<8\), \(C\) lies on the side opposite the direction of \(\vec{n}\). Moving in the direction of \(\vec{n}\) increases the value of \(4x-2y+z\).

a) Yes, \(A\in E\). b) \(k=2\) c) \(C\) lies on the side opposite the direction of \(\vec{n}\).

- The segment joining a point and its reflection is perpendicular to the reflecting plane. - Use the plane’s normal vector as the direction of a perpendicular line. - Find where that line meets the plane. - The intersection point is the midpoint of the original point and its reflection.

1. A normal vector to the plane is \(\vec{n}=\langle 1,1,1\rangle\). The perpendicular line through \(A\) is \(\vec{r}(t)=\langle 1,2,3\rangle+t\langle 1,1,1\rangle\). 2. At the point where this line meets the plane, \((1+t)+(2+t)+(3+t)=12\). 3. Thus \(6+3t=12\), so \(t=2\). The foot of the perpendicular is \(S(3, 4, 5)\). 4. Point \(S\) is the midpoint of \(A\) and its reflection. Therefore \(A'=2S-A=(6, 8, 10)-(1, 2, 3)=(5, 6, 7)\).

\(A'(5, 6, 7)\)

- Find a direction vector for the line. - Read a normal vector from the plane equation. - The line-plane angle is complementary to the angle between the line direction and the plane normal. - Use the sine form of the angle formula.

1. A direction vector for the line is \(\vec{v}=\overrightarrow{AB}=\begin{pmatrix}3\\4\\0\end{pmatrix}\). 2. A normal vector for the plane is \(\vec{n}=\begin{pmatrix}3\\0\\-4\end{pmatrix}\). 3. Their magnitudes are both \(5\), and \(\vec{v}\cdot\vec{n}=9\). 4. For the angle between a line and a plane, \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{9}{25}\). 5. Therefore, \(\alpha=\sin^{-1}\left(\frac{9}{25}\right)\approx21.1^\circ\).

\(\alpha\approx21.1^\circ\)

- What relationship do direction vectors of parallel lines have, and how can a component without \(a\) determine the scale factor? - What dot product value indicates perpendicular vectors? - The line positions do not affect these direction conditions.

1. Parallel direction vectors must be scalar multiples. From \(2=-4c\), the scale factor is \(c=-\frac{1}{2}\). Then \(a=6\cdot\left(-\frac{1}{2}\right)=-3\), and the third coordinate also matches. 2. Perpendicular direction vectors have dot product \(0\). Thus \(\begin{pmatrix}a\\2\\-3\end{pmatrix}\cdot\begin{pmatrix}1\\4\\2\end{pmatrix}=a+8-6=0\), so \(a=-2\).

1. \(a=-3\) 2. \(a=-2\)

- First compare the direction vectors for scalar-multiple relationships. - For parallel directions, test whether one line's point lies on the other line. - Use the dot product to test perpendicular direction vectors. - Perpendicular lines must also intersect.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so \(g\) and \(h\) are parallel or identical. 2. The point \((3, 1, 2)\) on \(h\) lies on \(g\) when \(r=1\). Therefore, \(g\) and \(h\) are identical. 3. For \(g\) and \(k\), the direction-vector dot product is \(\begin{pmatrix}2\\-1\\2\end{pmatrix}\cdot\begin{pmatrix}1\\4\\1\end{pmatrix}=2-4+2=0\). Thus their directions are perpendicular. 4. Since \(g\) and \(k\) share the position point \((1, 2, 0)\), they intersect there at a right angle.

Lines \(g\) and \(h\) are identical. Lines \(g\) and \(k\) are perpendicular and intersect at \((1, 2, 0)\).

- Write a general point on \(g\) using its parameter. - The vector from \(A\) to the intersection point must be perpendicular to the direction vector of \(g\). - Which vector operation tests whether two vectors are orthogonal? - After finding the intersection point, use the vector from \(A\) to that point as the direction of \(h\).

1. A general point on \(g\) is \(F(t)=(1+t, 1+2t, 1+2t)\). 2. The vector from \(A\) to \(F(t)\) is \(\overrightarrow{AF}=\langle t-3, 2t-1, 2t-2\rangle\). 3. At the perpendicular intersection, \(\overrightarrow{AF}\) is orthogonal to the direction vector \(\langle 1, 2, 2\rangle\). Thus, \((t-3)+2(2t-1)+2(2t-2)=0\). 4. Simplifying gives \(9t-9=0\), so \(t=1\). Then \(F=(2, 3, 3)\), and \(\overrightarrow{AF}=\langle -2, 1, 0\rangle\). 5. Therefore, \(h: \mathbf{x}=\langle 4, 2, 3\rangle+s\langle -2, 1, 0\rangle\).

\(h: \mathbf{x}=\langle 4, 2, 3\rangle+s\langle -2, 1, 0\rangle\), or any equivalent parametrization

- Represent \(S\) as a general point on \(g\). - The vector from \(B\) to \(S\) must be orthogonal to the direction vector of \(g\). - Once \(S\) is known, use \(B\) and \(S\) to write the equation of \(h\).

1. A point on \(g\) has coordinates \(S=(2+2k, 2k, 1-k)\). 2. The vector from \(B\) to \(S\) is \(\overrightarrow{BS}=\langle 2k+1, 2k-6, -k-1\rangle\). 3. Since \(h\) meets \(g\) at a right angle, \(\overrightarrow{BS}\cdot\langle 2, 2, -1\rangle=0\). This gives \(2(2k+1)+2(2k-6)-(-k-1)=0\). 4. Simplifying gives \(9k-9=0\), so \(k=1\) and \(S=(4, 2, 0)\). 5. A direction vector for \(h\) is \(\overrightarrow{BS}=\langle 3, -4, -2\rangle\). Therefore, \(h: \mathbf{x}=\langle 1, 6, 2\rangle+s\langle 3, -4, -2\rangle\).

a) \(S(4, 2, 0)\) b) \(h: \mathbf{x}=\langle 1, 6, 2\rangle+s\langle 3, -4, -2\rangle\)

- A unit vector has magnitude \(1\). How can you normalize the direction vector? - The given formula is the vector projection of \(\overrightarrow{OP}\) onto the line. - The point-to-line distance is the length of \(\overrightarrow{PF}\).

1. The direction vector has magnitude \(\sqrt{2^2+2^2+1^2}=3\), so \(\mathbf{u}=\left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right\rangle\). 2. Compute the scalar projection: \(\overrightarrow{OP}\cdot\mathbf{u}=4\left(\frac{2}{3}\right)+1\left(\frac{2}{3}\right)-1\left(\frac{1}{3}\right)=3\). 3. Therefore, \(\overrightarrow{OF}=3\mathbf{u}=\langle 2, 2, 1\rangle\), so \(F=(2, 2, 1)\). 4. The vector from \(P\) to \(F\) is \(\overrightarrow{PF}=\langle -2, 1, 2\rangle\), whose magnitude is \(\sqrt{(-2)^2+1^2+2^2}=3\).

a) \(\mathbf{u}=\left\langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right\rangle\) b) \(F(2, 2, 1)\) c) \(3\) units

- Which relationships between two lines determine a unique plane? - Compare the lines' given points and direction vectors. - Let a normal vector have unknown components and set its dot product with each line direction equal to zero. - Use the normal vector and a point on the plane to write its Cartesian equation.

1. The lines have the same point \((1, 2, 3)\), so they intersect. Their direction vectors \(\begin{pmatrix}1\\-2\\1\end{pmatrix}\) and \(\begin{pmatrix}2\\1\\0\end{pmatrix}\) are not scalar multiples. Therefore, the intersecting lines determine a unique plane. 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be normal to the plane. Perpendicularity to the two line directions gives \(a-2b+c=0\) and \(2a+b=0\). Choose \(\vec{n}=\begin{pmatrix}-1\\2\\5\end{pmatrix}\). 3. Using the point \((1, 2, 3)\), the plane equation is \(-x+2y+5z=18\).

a) The lines intersect at \((1, 2, 3)\), and their direction vectors are linearly independent, so they determine a unique plane. b) One normal vector is \(\begin{pmatrix}-1\\2\\5\end{pmatrix}\), and \(E:-x+2y+5z=18\).

- Form two spanning vectors from one contact point. - Let a normal vector have unknown components and set its dot product with each spanning vector equal to zero. - A point lies in a plane when its coordinates satisfy the plane equation. - Once three contact points fix a plane, the height of a fourth contact point is no longer arbitrary.

1. Two spanning vectors are \(\overrightarrow{AB}=\begin{pmatrix}18\\0\\0.5\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\18\\0.25\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. The dot-product equations are \(18a+0.5c=0\) and \(18b+0.25c=0\). Choosing \(c=-72\) gives \(a=2\) and \(b=1\), so \(\vec{n}=\begin{pmatrix}2\\1\\-72\end{pmatrix}\). Because the plane contains \(A=(0, 0, 0)\), its equation is \(E:2x+y-72z=0\). 3. For \(x=18\) and \(y=18\), \(2\cdot18+18-72z=0\). Therefore, \(72z=54\), so \(z=\frac{3}{4}=0.75\). 4. Three contact points determine a plane. A fourth leg touches the same floor plane only if its contact point has exactly the required height; even a small mismatch can make the table wobble.

a) \(E:2x+y-72z=0\) b) \(z=0.75\,\text{in.}\). A fourth contact point must satisfy the plane equation exactly, so a small difference in leg length or floor height can cause wobbling.

- Find a normal vector orthogonal to both direction vectors of the plane. - Check whether the line’s direction vector is orthogonal to the normal vector. - If the line is parallel to the plane, test its initial point in the plane equation. - Solve the resulting equation for \(a\).

1. Let \(\vec{n}=\langle p,q,w\rangle\) be normal to \(E\). Orthogonality to the two plane directions gives \(2p+w=0\) and \(q-w=0\). 2. Choose \(p=1\). Then \(w=-2\) and \(q=-2\), so \(\vec{n}=\langle 1,-2,-2\rangle\). 3. A standard equation of \(E\) is \(x-2y-2z=d\). Using \((1, 1, 1)\), \(d=1-2-2=-3\). 4. The line direction is parallel to the plane because \(\langle 1,-2,-2\rangle\cdot\langle 2,1,0\rangle=2-2=0\). 5. Therefore the entire line lies in the plane exactly when its initial point lies in the plane. Substitute \((3, 2, a)\): \(3-2(2)-2a=-3\). 6. Thus \(-1-2a=-3\), so \(a=1\).

\(a=1\)

- Read a normal vector from the plane equation. - A line parallel to a plane has a direction vector orthogonal to the plane’s normal vector. - Set the dot product equal to \(0\) and solve for \(a\).

1. A normal vector to \(E\) is \(\vec{n}=\langle 2,-4,2\rangle\). 2. For the line to be parallel to the plane, its direction vector must be orthogonal to \(\vec{n}\): \(\langle 2,a,3\rangle\cdot\langle 2,-4,2\rangle=0\). 3. Compute the dot product: \(4-4a+6=0\). 4. Thus \(10-4a=0\), so \(a=\frac52\). 5. The initial point is not in the plane because \(2(4)-4(1)+2(2)=8\neq5\), so the line is strictly parallel rather than contained in the plane.

\(a=\frac52\)

- Form two connecting vectors from one of the points. - Let the normal vector be perpendicular to both spanning vectors and write two dot-product equations. - Use the normal vector and one point for point-normal form. - Expand the dot product for Cartesian form.

1. Two spanning vectors are \(\overrightarrow{PQ}=\begin{pmatrix}2\\2\\-1\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}-2\\4\\3\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(2a+2b-c=0\) and \(-2a+4b+3c=0\). One solution is \(\vec{n}=\begin{pmatrix}5\\-2\\6\end{pmatrix}\). 3. Using point \(P\), a point-normal equation is \(\left(\vec{x}-\begin{pmatrix}1\\0\\2\end{pmatrix}\right)\cdot\begin{pmatrix}5\\-2\\6\end{pmatrix}=0\). 4. Expanding gives \(E:5x-2y+6z=17\).

Point-normal form: \(\left(\vec{x}-\begin{pmatrix}1\\0\\2\end{pmatrix}\right)\cdot\begin{pmatrix}5\\-2\\6\end{pmatrix}=0\) Cartesian form: \(E:5x-2y+6z=17\)

- Write an unknown normal vector and set its dot product with each plane direction vector equal to \(0\). - Use the initial point after finding a normal vector. - Test both points in the standard equation. - Two distinct points in a plane determine a line contained in that plane.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality to both direction vectors requires \(a+2b+2c=0\) and \(3a+c=0\). 2. From \(c=-3a\), the first equation becomes \(a+2b-6a=0\), so \(2b=5a\). Choose \(a=2\), giving \(b=5\) and \(c=-6\). Thus \(\vec{n}=\langle 2,5,-6\rangle\). 3. Point-normal form is \(\langle 2,5,-6\rangle\cdot\langle x,y-4,z+2\rangle=0\). 4. Expanding gives \(2x+5y-6z=32\). 5. Test \(A\): \(2(4)+5(6)-6(1)=32\), so \(A\in E\). Test \(B\): \(2(7)+5(6)-6(2)=32\), so \(B\in E\). 6. A plane contains the entire line through two distinct points in the plane, so line \(AB\) lies in \(E\).

a) Point-normal form: \(\langle 2,5,-6\rangle\cdot\langle x,y-4,z+2\rangle=0\) Standard form: \(2x+5y-6z=32\) b) Yes, line \(AB\) lies in \(E\).

- Set the dot product of an unknown normal vector with each direction vector equal to \(0\). - Use the plane’s initial point to write point-normal form. - Test both points that define the line. - A single point in the plane is not enough to show that the whole line is contained in it.

1. Let \(\vec{n}=\langle a,b,c\rangle\). The conditions \(\vec{n}\cdot\langle 2,1,0\rangle=0\) and \(\vec{n}\cdot\langle 0,2,-1\rangle=0\) give \(2a+b=0\) and \(2b-c=0\). 2. Choose \(a=-1\). Then \(b=2\) and \(c=4\), so \(\vec{n}=\langle -1,2,4\rangle\). 3. Point-normal form is \(\langle -1,2,4\rangle\cdot\langle x-5,y+1,z-3\rangle=0\). 4. Expanding gives \(-x+2y+4z=5\). 5. For \(P\), \(-1+2+4=5\), so \(P\in E\). For \(Q\), \(-3+4+8=9\neq5\), so \(Q\notin E\). 6. Since one point defining \(g\) is not in the plane, \(g\) is not contained in \(E\).

a) Point-normal form: \(\langle -1,2,4\rangle\cdot\langle x-5,y+1,z-3\rangle=0\) Standard form: \(-x+2y+4z=5\) b) No, \(g\) is not contained in \(E\).

- Let the normal vector have unknown components. - Set its dot product with each direction vector equal to \(0\). - Choose a convenient nonzero value for one component and solve for the others. - Substitute the initial point to find the constant.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality to the two direction vectors gives \(2a+b=0\) and \(a+2c=0\). 2. Choose \(a=2\). Then \(b=-4\) and \(c=-1\), so \(\vec{n}=\langle 2,-4,-1\rangle\). 3. The plane has the form \(2x-4y-z=d\). Substitute \((1, 2, -1)\): \(d=2(1)-4(2)-(-1)=-5\). 4. Therefore the standard equation is \(2x-4y-z=-5\), equivalently \(2x-4y-z+5=0\).

\(E: 2x-4y-z=-5\)

- Find a vector orthogonal to both directions in \(F\). - Identify which coordinate is fixed for every point of \(F\). - Use the x-axis direction and \(\overrightarrow{OP}\) as two directions in \(G\). - Perpendicular planes have orthogonal normal vectors.

1. a) A normal vector to \(F\) must be orthogonal to both \(\langle 0,1,0\rangle\) and \(\langle 0,0,1\rangle\). Choose \(\vec{n}_F=\langle 1,0,0\rangle\). 2. Using the point \((3, 0, 0)\), the standard equation is \(x=3\). 3. b) Plane \(F\) is parallel to the \(yz\)-plane, is \(3\) units from it, and is perpendicular to both the \(xy\)-plane and the \(xz\)-plane. 4. c) Plane \(G\) contains the direction \(\langle 1,0,0\rangle\) of the x-axis and the direction \(\overrightarrow{OP}=\langle 0,1,1\rangle\). 5. A normal vector \(\vec{n}_G=\langle a,b,c\rangle\) must satisfy \(a=0\) and \(b+c=0\). Choose \(\vec{n}_G=\langle 0,-1,1\rangle\). 6. Since the origin lies in \(G\), a point-normal equation is \(\vec{r}\cdot\langle 0,-1,1\rangle=0\), or \(-y+z=0\). 7. Finally, \(\vec{n}_F\cdot\vec{n}_G=0\), so the planes are perpendicular.

a) \(F: x=3\) b) \(F\) is parallel to the \(yz\)-plane, is \(3\) units from it, and is perpendicular to the \(xy\)-plane and the \(xz\)-plane. c) One point-normal equation is \(\vec{r}\cdot\langle 0,-1,1\rangle=0\). The planes are perpendicular because their normal vectors have dot product \(0\).

- Use a direction vector for the \(z\)-axis. - The normal vector must be orthogonal to every vector that lies in the plane. - A plane containing a coordinate axis also contains the origin.

1. Since the plane contains the \(z\)-axis, it contains the origin and the direction vector \(\vec{k}=\langle 0,0,1\rangle\). 2. A normal vector must satisfy \(\vec{n}\cdot\vec{k}=0\), so \(n_z=0\). 3. The vector \(\overrightarrow{OP}=\langle 2,4,1\rangle\) also lies in the plane. Thus \(\langle n_x,n_y,0\rangle\cdot\langle 2,4,1\rangle=0\), which gives \(2n_x+4n_y=0\). 4. One suitable normal vector is \(\langle 2,-1,0\rangle\). Because the plane contains the origin, its equation is \(2x-y=0\). 5. In general, if a plane contains a coordinate axis, the component of its normal vector in that axis’s direction is \(0\). The plane also contains the origin, so the constant term in its standard equation is \(0\).

a) \(E: 2x-y=0\) b) The normal-vector component corresponding to the contained axis is \(0\), and the plane’s standard equation has constant term \(0\).

- Read the normal vector from the coefficients of the plane equation. - Compare that normal vector with a direction vector of the \(z\)-axis. - Check whether the origin lies in every plane.

1. A normal vector to \(E_k\) is \(\vec{n}_k=\langle k,3,0\rangle\). 2. A direction vector of the \(z\)-axis is \(\vec{e}_z=\langle 0,0,1\rangle\). Since \(\vec{n}_k\cdot\vec{e}_z=0\) for every allowed \(k\), the \(z\)-axis direction lies in every plane. 3. Every equation has right side \(0\), so every plane contains the origin. Therefore, every plane contains the entire \(z\)-axis. 4. The common line is \(\vec{r}(t)=\langle 0,0,0\rangle+t\langle 0,0,1\rangle\). Substitution gives \(k(0)+3(0)=0\) for every \(t\) and every allowed \(k\).

a) Every plane contains the \(z\)-axis. b) \(\vec{r}(t)=\langle 0,0,0\rangle+t\langle 0,0,1\rangle\), \(t\in\mathbb{R}\)

- Compare two connecting vectors from the same point. - When do three points determine a unique plane? - Choose a normal vector perpendicular to the line's direction vector. - Substitute one given point to find the constant in the plane equation.

1. The connecting vectors are \(\overrightarrow{PQ}=\begin{pmatrix}3\\3\\3\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}-3\\-3\\-3\end{pmatrix}\). Since \(\overrightarrow{PR}=-\overrightarrow{PQ}\), the three points are collinear. 2. Three collinear points lie on infinitely many planes, so they do not determine a unique plane. 3. A direction vector of their line is \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). Choose a normal vector perpendicular to it, such as \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\). 4. A plane with this normal has equation \(x-y=d\). Substituting \(P\) gives \(d=2\). Thus one possible plane is \(x-y=2\).

a) The points are collinear, so they do not determine a unique plane. b) One possible plane is \(x-y=2\).

- Compare the direction vectors first. - Test whether a point from one line lies on the other line. - How many planes can contain a single line? - For part b, choose a plane normal perpendicular to the line's direction vector.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel or identical. 2. The point \((3, 1, 1)\) on \(h\) lies on \(g\) when \(r=1\). Therefore, the lines are identical. 3. One line lies in infinitely many planes, so the identical lines do not uniquely determine a plane. 4. A normal vector for a plane containing the line must be perpendicular to \(\begin{pmatrix}2\\-1\\1\end{pmatrix}\). Choose \(\vec{n}=\begin{pmatrix}1\\2\\0\end{pmatrix}\), since their dot product is \(0\). 5. Using the point \((1, 2, 0)\), the plane equation is \(x+2y=5\).

a) The lines are identical, so they do not uniquely determine a plane. b) One possible plane is \(E:x+2y=5\).

- A line perpendicular to a plane has a direction vector parallel to the plane normal. - A plane containing a line needs the line direction and a second independent spanning vector. - A line is parallel to a plane when its direction is perpendicular to the plane normal. - Test a point on the line to distinguish containment from strict parallelism.

1. A direction vector for \(g\) is \(\vec{v}=\overrightarrow{AB}=\begin{pmatrix}2\\2\\-1\end{pmatrix}\). Because \(E\) is perpendicular to \(g\), this vector is normal to \(E\). Using \(P\) gives \(E:2x+2y-z=10\). 2. Use point \(A\), the line direction, and any second vector not parallel to it, such as \(\begin{pmatrix}1\\0\\0\end{pmatrix}\). One possible plane is \(F:\vec{x}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+r\begin{pmatrix}2\\2\\-1\end{pmatrix}+s\begin{pmatrix}1\\0\\0\end{pmatrix}\). 3. A normal vector to \(H\) is \(\vec{n}_H=\begin{pmatrix}1\\0\\2\end{pmatrix}\). Since \(\vec{n}_H\cdot\vec{v}=0\), the line is parallel to the plane or lies in it. Point \(A\) does not satisfy the plane equation because \(1+2\cdot3=7\ne10\). Therefore, \(g\) is strictly parallel to \(H\).

1. \(E:2x+2y-z=10\) 2. One possible equation is \(F:\vec{x}=\begin{pmatrix}1\\-2\\3\end{pmatrix}+r\begin{pmatrix}2\\2\\-1\end{pmatrix}+s\begin{pmatrix}1\\0\\0\end{pmatrix}\). 3. Yes. The line direction is perpendicular to the plane normal, and \(A\notin H\), so \(g\) is strictly parallel to \(H\).

- Perpendicular planes have perpendicular normal vectors. - If a plane contains a line, the plane normal is perpendicular to the line direction. - A line is perpendicular to a plane when its direction is parallel to the plane normal. - Use a dot product to rule out parallelism.

1. A plane perpendicular to the \(xy\)-plane may have a normal vector of the form \(\vec{n}=\begin{pmatrix}a\\b\\0\end{pmatrix}\). To contain \(h\), its normal must be perpendicular to the line direction \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), so \(a-b=0\). Choose \(\vec{n}=\begin{pmatrix}1\\1\\0\end{pmatrix}\). Using the point \((0, 4, 2)\) gives \(E:x+y=4\). 2. Choose \(\vec{n}_G=\begin{pmatrix}1\\0\\0\end{pmatrix}\). Its dot product with the line direction is \(1\ne0\), so the line is not parallel to a plane with this normal. The vectors are not scalar multiples, so the line is not perpendicular to that plane.

1. \(E:x+y=4\) 2. One possible normal vector is \(\vec{n}_G=\begin{pmatrix}1\\0\\0\end{pmatrix}\).

- Find each midpoint by averaging the endpoint coordinates. - Use two connecting vectors in the plane and find a normal vector whose dot product with each is zero. - Test the point by substituting its coordinates into the plane equation.

1. The midpoints are \(P=(3, 0, 6)\) and \(Q=(0, 4, 6)\). 2. Using \(R\) as a point, two spanning vectors are \(\overrightarrow{RP}=\begin{pmatrix}3\\0\\2\end{pmatrix}\) and \(\overrightarrow{RQ}=\begin{pmatrix}0\\4\\2\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(3a+2c=0\) and \(4b+2c=0\). One solution is \(\vec{n}=\begin{pmatrix}4\\3\\-6\end{pmatrix}\). 3. Using \(R=(0, 0, 4)\), the plane equation is \(E:4x+3y-6z=-24\). 4. For \(T\), \(4\cdot1.5+3\cdot2-6\cdot6=-24\), so \(T\in E\).

a) \(E:4x+3y-6z=-24\) b) Yes, \(T=(1.5, 2, 6)\) lies in \(E\).

- Find the line’s direction vector. - A line parallel to a plane has a direction vector orthogonal to the plane’s normal vector. - If the dot product is nonzero, solve for the intersection parameter. - Substitute the parameter back into the line equation.

1. A direction vector is \(\vec{d}=\overrightarrow{AB}=\langle 4,2,-4\rangle\), and a normal vector to \(F\) is \(\vec{n}=\langle 1,1,1\rangle\). 2. Compute \(\vec{n}\cdot\vec{d}=4+2-4=2\neq0\). Therefore the line is not parallel to the plane and intersects it once. 3. Parametrize the line: \(\vec{r}(t)=\langle 2,1,5\rangle+t\langle 4,2,-4\rangle\). 4. Substitute into the plane equation: \((2+4t)+(1+2t)+(5-4t)=5\). 5. Thus \(8+2t=5\), so \(t=-\frac32\). 6. Substitution gives \(S=\langle 2,1,5\rangle-\frac32\langle 4,2,-4\rangle=\langle -4,-2,11\rangle\).

The line intersects the plane at \(S(-4, -2, 11)\).

- Use the equal axis intercepts to write the plane equation. - Find the line’s direction vector from the two given points. - Use a dot product to decide whether the line is parallel to the plane. - Substitute the line into the plane equation.

1. The three intercepts give the plane equation \(x+y+z=6\), with normal vector \(\vec{n}=\langle 1,1,1\rangle\). 2. A direction vector of the line is \(\vec{d}=\overrightarrow{PQ}=\langle 2,2,2\rangle\). Since \(\vec{n}\cdot\vec{d}=6\neq0\), the line intersects the plane exactly once. 3. Parametrize the line: \(\vec{r}(t)=\langle 1,1,1\rangle+t\langle 2,2,2\rangle\). 4. Substitute into the plane: \((1+2t)+(1+2t)+(1+2t)=6\). 5. Thus \(3+6t=6\), so \(t=\frac12\). 6. The intersection point is \(S(2, 2, 2)\).

The line intersects the plane at \(S(2, 2, 2)\).

- Compare the line direction with the plane normal using a dot product. - For the value that makes the dot product zero, test the line's initial point in the plane equation. - If the dot product is nonzero, consider how many times the line can meet the plane.

1. A normal vector to \(E\) is \(\vec{n}=\langle 3,-2,1\rangle\), and the direction vector of \(g_k\) is \(\vec{u}=\langle 1,2,k\rangle\). 2. Their dot product is \(\vec{n}\cdot\vec{u}=3-4+k=k-1\). 3. When \(k=1\), the direction vector is orthogonal to the plane normal, so the line is parallel to the plane or lies in it. 4. The initial point \((2, 1, 6)\) satisfies the plane equation because \(3(2)-2(1)+6=10\). Therefore, \(g_1\) lies in \(E\). 5. When \(k\neq1\), \(\vec{n}\cdot\vec{u}\neq0\), so the line intersects the plane at exactly one point.

For \(k=1\), \(g_1\) lies in \(E\). For every \(k\neq1\), \(g_k\) intersects \(E\) at exactly one point.

- The line direction must be orthogonal to the plane normal. - The line's initial point must satisfy the plane equation.

1. The line direction is \(\vec{v}=\langle 1,-2,3\rangle\), and a normal vector to the plane is \(\vec{n}=\langle a,1,2\rangle\). 2. For the line to lie in the plane, its direction must be orthogonal to the plane normal: \(\vec{v}\cdot\vec{n}=a-2+6=a+4=0\). 3. Thus \(a=-4\). 4. The line's initial point \((1, 2, 1)\) must also satisfy the plane equation. Substituting it gives \((-4)(1)+2+2(1)=b\), so \(b=0\).

\(a=-4\) and \(b=0\)

- Compare the line direction with the plane normal using a dot product. - Substitute a general point of the line into the plane equation. - Use the sine formula for the acute angle between a line and a plane.

1. The line direction is \(\vec{v}=\begin{pmatrix}0\\1\\-1\end{pmatrix}\), and a plane normal is \(\vec{n}=\begin{pmatrix}2\\-1\\1\end{pmatrix}\). Since \(\vec{v}\cdot\vec{n}=-2\neq0\), the line intersects the plane. 2. Substituting the line into the plane gives \(2\cdot2-(5+t)+(3-t)=6\), so \(t=-2\). 3. The intersection point is \((2, 3, 5)\). 4. The angle satisfies \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{1}{\sqrt{3}}\). Therefore, \(\alpha\approx35.3^\circ\).

The line intersects the plane at \((2, 3, 5)\) with angle \(\alpha\approx35.3^\circ\).

- Find a normal vector orthogonal to both plane direction vectors. - A line in the plane needs a point in the plane and a direction orthogonal to the normal vector. - A strictly parallel line needs a parallel direction but an initial point outside the plane.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality to the plane directions gives \(2a+b-c=0\) and \(3b+2c=0\). A solution is \(\vec{n}=\langle 5,-4,6\rangle\). 2. The plane equation is \(5x-4y+6z=d\). Using \((4, 1, 0)\), \(d=20-4=16\). 3. Choose \(g: \vec{r}(t)=\langle 4,1,0\rangle+t\langle 2,1,-1\rangle\). Its initial point is in \(E\), and \(\langle 5,-4,6\rangle\cdot\langle 2,1,-1\rangle=10-4-6=0\), so the entire line lies in the plane. 4. Choose \(h: \vec{r}(t)=\langle 0,0,0\rangle+t\langle 0,3,2\rangle\). Its direction satisfies \(\langle 5,-4,6\rangle\cdot\langle 0,3,2\rangle=-12+12=0\), so it is parallel to \(E\). 5. The origin is not in \(E\) because \(0\neq16\). Therefore \(h\) is strictly parallel to the plane.

a) One possible line is \(g: \vec{r}(t)=\langle 4,1,0\rangle+t\langle 2,1,-1\rangle\). b) One possible line is \(h: \vec{r}(t)=t\langle 0,3,2\rangle\).

- A vector parallel to a plane is orthogonal to its normal vector. - Choose one point that satisfies the plane equation. - Choose a second point that does not satisfy the plane equation. - Use the same parallel direction for both lines.

1. A normal vector to \(E\) is \(\vec{n}=\langle 1,2,-2\rangle\). A vector is parallel to the plane when its dot product with \(\vec{n}\) is \(0\). 2. Choose \(\vec{v}=\langle 2,0,1\rangle\), since \(\vec{n}\cdot\vec{v}=2-2=0\). 3. The point \(P(6, 0, 0)\) lies in \(E\). Therefore \(k: \vec{r}(t)=\langle 6,0,0\rangle+t\langle 2,0,1\rangle\) lies in the plane. 4. The origin does not lie in \(E\), since \(0\neq6\). Therefore \(m: \vec{r}(t)=t\langle 2,0,1\rangle\) is parallel to \(E\) and disjoint from it.

a) One possible vector is \(\vec{v}=\langle 2,0,1\rangle\). b) \(k: \vec{r}(t)=\langle 6,0,0\rangle+t\langle 2,0,1\rangle\) \(m: \vec{r}(t)=t\langle 2,0,1\rangle\)

- Use a dot product to make the line direction orthogonal to the plane normal. - After finding \(a\), test the line's initial point in the plane equation.

1. A normal vector to \(E\) is \(\vec{n}=\langle 3,1,-2\rangle\), and the line direction is \(\vec{v}=\langle 1,a,3\rangle\). 2. The line direction is parallel to the plane when \(\vec{n}\cdot\vec{v}=0\): \(3+a-6=0\). 3. Therefore, \(a=3\). 4. Test the initial point \((2, 6, 1)\) in the plane equation: \(3(2)+6-2(1)=10\). 5. The point lies in \(E\), so the entire line \(g_3\) lies in \(E\); it is not strictly parallel to the plane.

\(a=3\), and \(g_3\) lies in \(E\).

- A line is parallel to a plane when its direction is orthogonal to the plane normal. - A line is perpendicular to a plane when its direction is a scalar multiple of the plane normal.

1. A normal vector to \(E\) is \(\vec{n}=\langle 4,-2,1\rangle\), and the line direction is \(\vec{v}=\langle a,4,2\rangle\). 2. a) For the line to be parallel to the plane, require \(\vec{v}\cdot\vec{n}=0\): \(4a-8+2=0\). 3. Thus \(4a=6\), so \(a=\frac32\). 4. b) For the line to be perpendicular to the plane, \(\vec{v}\) must be a scalar multiple of \(\vec{n}\). 5. From the third components, the scalar multiple would have to be \(2\). That would make the second component \(2(-2)=-4\), but the second component of \(\vec{v}\) is \(4\). 6. This contradiction shows that no value of \(a\) makes the line perpendicular to the plane.

a) \(a=\frac32\) b) No value of \(a\) makes \(g_a\) perpendicular to \(E\).

- Use a dot product to test whether the line direction is parallel to the plane. - Test the line's initial point to distinguish containment from strict parallelism. - A line perpendicular to a plane has a direction parallel to the plane normal.

1. A normal vector to \(F\) is \(\vec{n}=\langle 1,2,-2\rangle\), and the line direction is \(\vec{v}=\langle 2,k,-4\rangle\). 2. a) For the line to be parallel to the plane, require \(\vec{v}\cdot\vec{n}=0\): \(2+2k+8=0\). 3. Thus \(2k+10=0\), so \(k=-5\). 4. The initial point \((2, 1, 4)\) does not lie in \(F\), because \(2+2-8=-4\neq10\). Therefore, \(h_{-5}\) is strictly parallel to \(F\). 5. b) For the line to be perpendicular to the plane, \(\vec{v}\) must be a scalar multiple of \(\vec{n}\). 6. Since the first component gives a scalar multiple of \(2\), the second component must be \(k=2(2)=4\). The third component also agrees because \(2(-2)=-4\).

a) \(k=-5\); \(h_{-5}\) is strictly parallel to \(F\). b) \(k=4\)

- Recall the geometric meaning of a plane’s normal vector. - Interpret a zero dot product geometrically. - List the possible numbers of intersection points between a line and a plane. - Distinguish a line parallel to a plane from a line contained in the plane.

1. True. A plane’s normal vector is perpendicular to every direction in the plane. A line whose direction is parallel to that normal is therefore perpendicular to the plane. 2. False. The equation \(\vec{v}\cdot\vec{n}=0\) means the line direction is parallel to the plane. The line then has either no points in common with the plane or infinitely many points in common with it, not exactly one. 3. True. The direction condition makes the line parallel to the plane. If one point of the line is also in the plane, then every point of the line remains in the plane.

1. True. 2. False. 3. True.

- A line is parallel to a plane when its direction vector is orthogonal to the plane’s normal vector. - Use the point-normal condition to decide whether a point lies in a plane. - Recall what “strictly parallel” excludes.

1. a) The line intersects the plane at exactly one point when \(\vec{u}\cdot\vec{n}\neq0\). A zero dot product would make the line direction parallel to the plane, while a nonzero dot product means the line crosses the plane. 2. b) A point \(P\) lies in \(E\) exactly when \((\vec{p}-\vec{a})\cdot\vec{n}=0\). Because a strictly parallel line has no points in common with the plane, every point \(P\) on \(g\) must satisfy \((\vec{p}-\vec{a})\cdot\vec{n}\neq0\).

a) Exactly one intersection occurs when \(\vec{u}\cdot\vec{n}\neq0\). b) For every point \(P\) on the strictly parallel line, \((\vec{p}-\vec{a})\cdot\vec{n}\neq0\).

- A plane's normal vector gives the direction of a perpendicular line. - Write the perpendicular line through \(A\). - Its intersection with the plane is the perpendicular foot. - Use the distance formula between \(A\) and the foot.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle 1,1,0\rangle\). The perpendicular line through \(A\) is \(\mathbf{x}=\langle 7,5,2\rangle+r\langle 1,1,0\rangle\). 2. Substitute into the plane equation: \((7+r)+(5+r)=4\). Thus, \(12+2r=4\), so \(r=-4\). 3. Therefore, \(F=(3,1,2)\). 4. The distance is \(AF=\sqrt{(-4)^2+(-4)^2+0^2}=4\sqrt{2}\approx 5.66\).

a) \(F(3,1,2)\) b) \(4\sqrt{2}\approx 5.66\) units

- Use the plane's normal vector as the direction of the perpendicular line through \(B\). - Find where that line intersects the plane. - The perpendicular foot is the midpoint between a point and its reflection.

1. A normal vector to the plane is \(\mathbf{n}=\langle 2,-2,1\rangle\). The perpendicular line through \(B\) is \(\mathbf{x}=\langle -1,4,3\rangle+r\langle 2,-2,1\rangle\). 2. Substitute into the plane equation: \(2(-1+2r)-2(4-2r)+(3+r)=2\). This simplifies to \(9r-7=2\), so \(r=1\). 3. Therefore, \(L=(1,2,4)\). 4. Since \(L\) is the midpoint of \(\overline{BB'}\), \(B'=2L-B=2(1,2,4)-(-1,4,3)=(3,0,5)\).

a) \(L(1,2,4)\) b) \(B'(3,0,5)\)

- Write a general point on the line. - The vector from \(A\) to the perpendicular foot must be orthogonal to the direction vector. - Use the midpoint relationship between \(A\), the perpendicular foot, and \(A'\).

1. A general point on \(h\) is \(F(t)=(1+t, 1, 1-t)\), so \(\overrightarrow{AF}=\langle t-1, -2, -t-3\rangle\). 2. Require \(\overrightarrow{AF}\cdot\langle 1, 0, -1\rangle=0\): \((t-1)-(-t-3)=2t+2=0\). Therefore, \(t=-1\). 3. The perpendicular foot is \(F=(0, 1, 2)\). 4. Because \(F\) is the midpoint of \(\overline{AA'}\), \(A'=2F-A=2(0, 1, 2)-(2, 3, 4)=(-2, -1, 0)\).

\(A'(-2, -1, 0)\)

- A line perpendicular to a plane has a direction parallel to the plane normal. - For a line to lie in a plane, its direction must be orthogonal to the plane normal and one point of the line must lie in the plane.

1. A normal vector to \(E\) is \(\vec{n}=\langle 2,1,-1\rangle\), and the line direction is \(\vec{v}=\langle k,2,-2\rangle\). 2. a) A line perpendicular to the plane has a direction parallel to \(\vec{n}\). From the second component, the scalar multiple is \(2\). 3. Therefore, \(k=2(2)=4\). The third component also agrees because \(2(-1)=-2\). 4. b) For a line to lie in the plane, its direction must be orthogonal to the plane normal. This would require \(\vec{v}\cdot\vec{n}=2k+2+2=0\), so \(k=-2\). 5. However, the common initial point \((1, 0, 4)\) is not in \(E\), because \(2(1)+0-4-6=-8\neq0\). 6. Therefore, no line in the family lies in \(E\).

a) \(k=4\) b) No line in the family lies entirely in \(E\).

- A line perpendicular to a plane has a direction vector parallel to the plane’s normal vector. - Use a dot product to test whether the line direction is parallel to the plane. - Check whether the initial point satisfies the plane equation. - Notice which coordinates are affected by \(a\).

1. A normal vector to \(F\) is \(\vec{n}=\langle 1,1,0\rangle\), and every line in the family has direction vector \(\vec{d}=\langle 1,-1,2\rangle\). 2. a) A line perpendicular to \(F\) must have a direction vector parallel to \(\vec{n}\). The vectors \(\langle 1,-1,2\rangle\) and \(\langle 1,1,0\rangle\) are not scalar multiples, so no line in the family is perpendicular to \(F\). 3. b) Compute \(\vec{d}\cdot\vec{n}=1-1+0=0\). Thus every line in the family is parallel to \(F\). 4. The initial point \((3, 1, a)\) never lies in \(F\), because \(3+1=4\neq10\), regardless of \(a\). 5. Therefore every line in the family is strictly parallel to \(F\), so none lies in the plane.

a) No line is perpendicular to \(F\), because \(\langle 1,-1,2\rangle\) is not parallel to \(\langle 1,1,0\rangle\). b) No line lies in \(F\). Every line is parallel to \(F\), but its initial point fails the plane equation.

- A line is parallel to a plane when its direction vector is orthogonal to the plane's normal vector. - Test one point on the line to determine whether the line lies in the plane. - If the line is parallel to the plane, use any point on the line in the point-to-plane distance formula.

1. A direction vector for \(g\) is \(\overrightarrow{AB}=\langle 1,-3,0\rangle\). A normal vector to \(E\) is \(\mathbf{n}=\langle 6,2,-3\rangle\). 2. Their dot product is \(\mathbf{n}\cdot\overrightarrow{AB}=6(1)+2(-3)-3(0)=0\), so the line is parallel to the plane or lies in it. 3. Test point \(A\): \(6(1)+2(5)-3(2)=10\ne 12\). Therefore, \(g\) is parallel to and distinct from \(E\). 4. Every point on \(g\) has the same distance from \(E\). Using \(A\), \(d(g,E)=\frac{\lvert 6(1)+2(5)-3(2)-12\rvert}{\sqrt{6^2+2^2+(-3)^2}}=\frac{2}{7}\approx 0.29\).

The line \(g\) is parallel to and distinct from \(E\), and the distance is \(\frac{2}{7}\approx 0.29\) units.

- Check the dot product of the line's direction vector and the plane's normal vector. - Test a point on the line to determine whether the line lies in the plane. - For a parallel line and plane, use any point on the line in the point-to-plane distance formula.

1. The line's direction vector is \(\mathbf{u}=\langle 2,2,1\rangle\), and a normal vector to the plane is \(\mathbf{n}=\langle 1,-2,2\rangle\). 2. Since \(\mathbf{u}\cdot\mathbf{n}=2-4+2=0\), the direction of the line is parallel to the plane. 3. The point \(P=(1,1,4)\) on the line does not lie in the plane because \(1-2(1)+2(4)=7\ne5\). Thus, the line is parallel to and distinct from the plane. 4. The distance equals the distance from \(P\) to the plane: \(d=\frac{\lvert 1-2+8-5\rvert}{\sqrt{1^2+(-2)^2+2^2}}=\frac{2}{3}\approx0.67\).

The line is parallel to and distinct from the plane, and the distance is \(\frac{2}{3}\approx0.67\) units.

- The shortest segment from a point to a line is perpendicular to the line. - Write a general point on \(g\) in terms of \(t\). - Set the dot product of the connecting vector and the direction vector equal to \(0\). - The distance is the magnitude of the perpendicular connecting vector.

1. A general point on \(g\) is \(F(t)=(1+2t, 2-t, 3+2t)\), so \(\overrightarrow{PF}=\langle 2t-6, -t, 2t-3\rangle\). 2. For the shortest segment, \(\overrightarrow{PF}\cdot\langle 2, -1, 2\rangle=0\). Thus, \(2(2t-6)+t+2(2t-3)=9t-18=0\), so \(t=2\). 3. The perpendicular foot is \(F=(5, 0, 7)\). 4. Therefore, the distance is \(PF=\sqrt{(-2)^2+(-2)^2+1^2}=3\).

\(3\) units

- Substitute the coordinates of \(P\) into the point-to-plane distance formula. - The normal vector's magnitude is the denominator. - The absolute-value equation produces two linear cases.

1. A normal vector to the plane is \(\langle 6,2,-3\rangle\), with magnitude \(7\). 2. Apply the distance formula: \(\frac{\lvert 6k+2(1)-3(-3)-5\rvert}{7}=4\). 3. Simplify: \(\lvert 6k+6\rvert=28\). 4. Therefore, \(6k+6=28\) or \(6k+6=-28\). 5. The solutions are \(k=\frac{11}{3}\) and \(k=-\frac{17}{3}\).

\(k=\frac{11}{3}\) or \(k=-\frac{17}{3}\)

- First write the plane equation using the given point and normal vector. - Find the magnitude of the normal vector. - Substitute \(S\) into the point-to-plane distance formula. - Points can lie the same distance from the plane on opposite sides.

1. The point-normal equation is \(12(y-1)-5(z-1)=0\), which simplifies to \(12y-5z-7=0\). 2. The normal vector has magnitude \(\sqrt{12^2+(-5)^2}=13\). 3. Substitute \(S(2,y,4)\) into the distance formula: \(\frac{\lvert 12y-5(4)-7\rvert}{13}=3\). 4. Thus, \(\lvert 12y-27\rvert=39\). 5. The two cases give \(12y-27=39\) or \(12y-27=-39\), so \(y=\frac{11}{2}\) or \(y=-1\).

\(y=\frac{11}{2}\) or \(y=-1\)

- The shortest segment from \(Q\) to the line is perpendicular to the line. - Express a general point on \(g\) in terms of \(\lambda\). - Use a dot product to impose orthogonality. - Find the magnitude of the resulting connecting vector.

1. A general point on \(g\) is \(P=(1+\lambda, 2+\lambda, 3+\lambda)\), so \(\overrightarrow{QP}=\langle \lambda-2, \lambda+2, \lambda+3\rangle\). 2. For the closest point, \(\overrightarrow{QP}\cdot\langle 1, 1, 1\rangle=0\). Thus, \((\lambda-2)+(\lambda+2)+(\lambda+3)=3\lambda+3=0\), so \(\lambda=-1\). 3. Therefore, \(P=(0, 1, 2)\). 4. The minimum distance is \(QP=\sqrt{(-3)^2+1^2+2^2}=\sqrt{14}\approx 3.74\).

a) \(P(0, 1, 2)\) b) \(\sqrt{14}\approx 3.74\) units

- For parallel lines, use any point on one line and find its distance to the other. - Write a general point on \(f\). - The minimum connecting vector must be orthogonal to the common direction vector. - Use the vector's magnitude for the distance.

1. Because the lines are parallel, their distance equals the distance from \(S\) to \(f\). 2. A general point on \(f\) is \(F(t)=(2+t, 3-2t, 2t)\), so \(\overrightarrow{SF}=\langle t+1, 3-2t, 2t-2\rangle\). 3. The shortest connecting vector is orthogonal to the common direction vector: \(\overrightarrow{SF}\cdot\langle 1, -2, 2\rangle=0\). This gives \((t+1)-2(3-2t)+2(2t-2)=9t-9=0\), so \(t=1\). 4. Then \(\overrightarrow{SF}=\langle 2, 1, 0\rangle\), so the distance is \(\sqrt{2^2+1^2}=\sqrt{5}\approx 2.24\).

\(\sqrt{5}\approx 2.24\) units

- An altitude is perpendicular to the opposite side. - Represent a general point on line \(AB\) with a parameter. - Use the dot product to find the perpendicular foot. - Then apply \(A=\frac{1}{2}bh\).

1. Line \(AB\) can be written as \(\mathbf{x}=\langle 2, 2, 1\rangle+r\langle 4, 0, 3\rangle\). A general point is \(L=(2+4r, 2, 1+3r)\). 2. Require \(\overrightarrow{CL}\cdot\langle 4, 0, 3\rangle=0\): \(4(4r-2)+3(3r-4)=25r-20=0\). Thus, \(r=\frac{4}{5}\). 3. Therefore, \(L=\left(\frac{26}{5}, 2, \frac{17}{5}\right)=(5.2, 2, 3.4)\). 4. The altitude length is \(CL=\sqrt{1.2^2+(-3)^2+(-1.6)^2}=\sqrt{13}\approx 3.61\). 5. Since \(AB=\sqrt{4^2+0^2+3^2}=5\), the area is \(\frac{1}{2}(5)(\sqrt{13})=\frac{5\sqrt{13}}{2}\approx 9.01\).

a) \(L\left(\frac{26}{5}, 2, \frac{17}{5}\right)\) b) \(\sqrt{13}\approx 3.61\) units c) \(\frac{5\sqrt{13}}{2}\approx 9.01\) square units

- The three base points are the x-, y-, and z-intercepts of the plane. - Use intercept form to write the plane equation. - Apply the point-to-plane distance formula. - Interpret the perpendicular distance from the apex to the base plane.

1. Because the plane intersects the coordinate axes at \(2\), \(4\), and \(6\), its intercept form is \(\frac{x}{2}+\frac{y}{4}+\frac{z}{6}=1\). 2. Multiplying by \(12\) gives \(E: 6x+3y+2z=12\). 3. A normal vector is \(\langle 6,3,2\rangle\), with magnitude \(7\). 4. The distance from \(S\) is \(\frac{\lvert 6(5)+3(5)+2(5)-12\rvert}{7}=\frac{43}{7}\approx6.14\). 5. This distance is the height of the tetrahedron relative to base \(PQR\).

a) \(E: 6x+3y+2z=12\) b) \(\frac{43}{7}\approx6.14\) units; it is the tetrahedron's height above base \(PQR\).

- Find a vector orthogonal to both direction vectors of \(E\). - Parallel planes have proportional normal vectors. - Only the constant changes when a plane is translated parallel to itself. - Use the distance formula for parallel planes.

1. Let \(\mathbf{u}=\langle 1,3,0\rangle\) and \(\mathbf{v}=\langle 0,3,2\rangle\). A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) satisfies \(a+3b=0\) and \(3b+2c=0\). Choose \(\mathbf{n}=\langle 6,-2,3\rangle\). 2. Its magnitude is \(7\). Using point \((2,1,0)\), the plane equation is \(6x-2y+3z=10\). 3. A parallel plane has equation \(6x-2y+3z=k\). Its distance from \(E\) is \(\frac{\lvert k-10\rvert}{7}\). 4. Set \(\frac{\lvert k-10\rvert}{7}=10\). Then \(\lvert k-10\rvert=70\), so \(k=80\) or \(k=-60\).

\(F_1: 6x-2y+3z=80\) and \(F_2: 6x-2y+3z=-60\)

- The shortest segment from a point to a line is perpendicular to the line. - Write a general point on the robot's path. - Use a dot product to find when the connecting vector is perpendicular to the direction vector. - Find the magnitude of the perpendicular vector.

1. A general point on the rail is \(L(t)=(5+t, 1+2t, 3-2t)\), so \(\overrightarrow{QL}=\langle 3+t, 2+2t, -1-2t\rangle\). 2. For the minimum distance, \(\overrightarrow{QL}\cdot\langle 1, 2, -2\rangle=0\). Thus, \((3+t)+2(2+2t)-2(-1-2t)=9t+9=0\), so \(t=-1\). 3. Then \(\overrightarrow{QL}=\langle 2, 0, 1\rangle\). 4. The minimum distance is \(\sqrt{2^2+0^2+1^2}=\sqrt{5}\approx 2.24\,\text{m}\).

\(\sqrt{5}\,\text{m}\approx 2.24\,\text{m}\)

- Represent a general point on the laser beam. - The vector from the sensor to the closest point is perpendicular to the beam's direction vector. - After finding the point, use the three-dimensional distance formula.

1. A general point on the beam is \(L(k)=(-2+2k, 5-k, 3+2k)\), so \(\overrightarrow{SL}=\langle 2k-7, -k-5, 2k-9\rangle\). 2. At the closest point, \(\overrightarrow{SL}\cdot\langle 2, -1, 2\rangle=0\). Thus, \(2(2k-7)-(-k-5)+2(2k-9)=9k-27=0\), so \(k=3\). 3. Therefore, \(L=(4, 2, 9)\). 4. The distance is \(SL=\sqrt{(4-5)^2+(2-10)^2+(9-12)^2}=\sqrt{74}\approx 8.60\).

a) \(L(4, 2, 9)\) b) \(\sqrt{74}\approx 8.60\) units

- The shortest segment from \(A\) to the line is perpendicular to the line. - Use a parameter to write a general point on \(g\). - Set the dot product of the connecting vector and the direction vector equal to \(0\). - The vector's magnitude gives the distance.

1. A general point on \(g\) is \(F(t)=(1+t, 1, 1+t)\), so \(\overrightarrow{AF}=\langle t-2, -1, t\rangle\). 2. Require \(\overrightarrow{AF}\cdot\langle 1, 0, 1\rangle=0\): \((t-2)+t=2t-2=0\), so \(t=1\). 3. Therefore, \(F=(2, 1, 2)\). 4. The distance is \(AF=\sqrt{(-1)^2+(-1)^2+1^2}=\sqrt{3}\approx 1.73\).

The perpendicular foot is \(F(2, 1, 2)\), and the distance is \(\sqrt{3}\approx 1.73\) units.

- Read a normal vector from the equation of \(E\). - Use two directions in \(G_a\) and solve zero dot-product conditions to find a normal vector. - A zero dot product of normals corresponds to what plane angle? - Set the cosine expression equal to \(\frac{1}{2}\).

1. A normal vector to \(E\) is \(\vec{n}_E=\begin{pmatrix}1\\1\\0\end{pmatrix}\). In \(G_a\), use \(\overrightarrow{OP}=\begin{pmatrix}1\\0\\0\end{pmatrix}\) and \(\overrightarrow{OQ_a}=\begin{pmatrix}0\\1\\a\end{pmatrix}\). If \(\vec{n}_a=\begin{pmatrix}x\\y\\z\end{pmatrix}\) is perpendicular to both, then \(x=0\) and \(y+az=0\). Choose \(\vec{n}_a=\begin{pmatrix}0\\-a\\1\end{pmatrix}\). 2. When \(a=0\), the normal vectors have dot product \(0\), so the planes meet at \(90^\circ\). 3. In general, \(\cos\alpha=\frac{|a|}{\sqrt{2a^2+2}}\). Setting this equal to \(\cos60^\circ=\frac{1}{2}\) and squaring gives \(a^2=1\). 4. Thus \(a=1\) or \(a=-1\).

a) \(90^\circ\) b) \(a=1\) or \(a=-1\)

- Read the sphere's center and radius from standard form. - Compare the center-to-plane distance with the sphere's radius for tangency. - For a circular cross-section, use a right triangle formed by the sphere radius, the cross-section radius, and the center-to-plane distance. - Use the plane's normal vector to calculate the perpendicular distance.

1. The sphere has center \(M=(2, -1, 4)\) and radius \(5\). The planes have normal vector \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 2. The perpendicular distance from \(M\) to \(E_k\) is \(d=\frac{|2(2)-2(-1)+4-k|}{3}=\frac{|10-k|}{3}\). 3. For tangency, \(d=5\). Thus, \(|10-k|=15\), giving \(k=-5\) or \(k=25\). 4. For a cross-sectional circle of radius \(4\), the sphere radius, plane distance, and circle radius satisfy \(d^2+4^2=5^2\). Hence, \(d=3\). 5. Therefore, \(\frac{|10-k|}{3}=3\), so \(|10-k|=9\), giving \(k=1\) or \(k=19\).

a) \(k=-5\) or \(k=25\) b) \(k=1\) or \(k=19\)

- Read the center and radius from the sphere equation. - Use the plane's normal vector to calculate the perpendicular distance from the center. - Compare that distance with the sphere's radius. - Use the right triangle formed by the sphere radius, the plane distance, and the cross-section radius.

1. The sphere has center \(M=(2, -3, 1)\) and radius \(R=10\). 2. The plane has normal vector \(\mathbf{n}=\langle6, -3, 2\rangle\), whose magnitude is \(7\). The distance from \(M\) to the plane is \(d=\frac{|6(2)-3(-3)+2(1)+15|}{7}=\frac{38}{7}\). 3. Since \(\frac{38}{7}<10\), the plane intersects the sphere in a circle. 4. The cross-section radius is \(\rho=\sqrt{R^2-d^2}=\sqrt{100-\left(\frac{38}{7}\right)^2}=\frac{24\sqrt6}{7}\approx8.40\).

The plane intersects the sphere in a circle with radius \(\rho=\frac{24\sqrt6}{7}\approx8.40\).

- Complete the square to identify the sphere's center and radius. - Find a vector perpendicular to both direction vectors of the plane. - Compare the center-to-plane distance with the sphere's radius. - When the plane misses the sphere, subtract the radius from the center-to-plane distance.

1. Complete the square: \((x-1)^2+(y-2)^2+(z+2)^2=16\). Thus, the sphere has center \(M=(1, 2, -2)\) and radius \(R=4\). 2. A normal vector to the plane must be perpendicular to both direction vectors. The vector \(\mathbf{n}=\langle0, -3, 4\rangle\) satisfies \(\mathbf{n}\cdot\langle1, 0, 0\rangle=0\) and \(\mathbf{n}\cdot\langle0, 4, 3\rangle=0\). 3. Using the point \((5, 0, 6)\), the plane equation is \(-3y+4z-24=0\). 4. The distance from \(M\) to the plane is \(d=\frac{|-3(2)+4(-2)-24|}{5}=\frac{38}{5}\). 5. Since \(\frac{38}{5}>4\), the plane does not intersect the sphere. The shortest distance from the sphere's surface to the plane is \(d-R=\frac{38}{5}-4=\frac{18}{5}=3.6\).

The plane does not intersect the sphere. The shortest distance from the sphere's surface to the plane is \(\frac{18}{5}=3.6\).

- Complete the square to find the sphere's center. - The radius from the center to the tangent point is normal to the tangent plane. - A normal vector may be replaced by a nonzero scalar multiple. - Substitute the tangent point to determine the constant in the plane equation.

1. Complete the square: \((x+5)^2+(y-4)^2+(z+1)^2=144\). Thus, the sphere's center is \(M=(-5, 4, -1)\). 2. The radius vector to the point of tangency is \(\overrightarrow{MB}=\langle8, 8, 4\rangle\), which is parallel to \(\langle2, 2, 1\rangle\). 3. This radius vector is normal to the tangent plane, so write the plane as \(2x+2y+z=d\). 4. Substitute \(B=(3, 12, 3)\): \(d=2(3)+2(12)+3=33\). 5. Therefore, one equation of the tangent plane is \(2x+2y+z=33\).

\(2x+2y+z=33\)

- Complete the square in each variable. - The right side of the standard sphere equation is \(r^2\). - Use the plane's normal vector to find the perpendicular distance from the center. - Compare that distance with the sphere's radius.

1. Complete the square: \((x-2)^2+(y+3)^2+(z-1)^2=25\). Therefore, \(M=(2, -3, 1)\) and \(r=5\). 2. The plane has normal vector \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 3. The distance from \(M\) to the plane is \(d=\frac{|2(2)-2(-3)+1-25|}{3}=\frac{14}{3}\approx4.67\). 4. Since \(\frac{14}{3}<5\), the plane intersects the sphere in a circle.

a) \(M=(2, -3, 1)\) and \(r=5\) b) \(d=\frac{14}{3}\approx4.67\). Since \(d<r\), the plane intersects the sphere.

- A plane is tangent when its distance from the sphere's center equals the radius. - The radius to the point of tangency is perpendicular to the plane. - For a plane parallel to the xy-plane, only the z-coordinate determines the distance. - Identify the z-values covered by the sphere.

1. The distance from \(M=(4, 0, -2)\) to the plane \(x=10\) is \(|4-10|=6\), equal to the sphere's radius. Therefore, the plane is tangent. 2. The radius to the tangent point is parallel to the x-axis and extends from \(x=4\) to \(x=10\), so \(B=(10, 0, -2)\). 3. The distance from \(M\) to the plane \(z=k\) is \(|k+2|\). The plane misses the sphere when this distance is greater than \(6\). 4. Solve \(|k+2|>6\): \(k<-8\) or \(k>4\).

a) The center-to-plane distance is \(6\), so \(E\) is tangent at \(B=(10, 0, -2)\). b) \(k<-8\) or \(k>4\)

- Substitute the point into the sphere equation. - The radius vector to the tangent point is normal to the tangent plane. - Parallel planes have parallel normal vectors. - Use the point-to-plane distance and set it equal to \(10\).

1. Substitute \(P\) into the sphere equation: \((2-4)^2+(2+2)^2+(-3-1)^2=4+16+16=36\). Thus, \(P\) lies on the sphere. 2. The center is \(M=(4, -2, 1)\). The vector \(\overrightarrow{MP}=\langle-2, 4, -4\rangle\) is normal to the tangent plane and is parallel to \(\langle1, -2, 2\rangle\). 3. A plane through \(P\) with this normal vector is \(T:x-2y+2z+8=0\). 4. Every plane parallel to \(T\) has the form \(x-2y+2z+k=0\). Its distance from \(M\) is \(\frac{|4-2(-2)+2(1)+k|}{3}=\frac{|10+k|}{3}\). 5. Set this distance equal to \(10\): \(|10+k|=30\). Thus, \(k=20\) or \(k=-40\). 6. The planes are \(F_1:x-2y+2z+20=0\) and \(F_2:x-2y+2z-40=0\).

a) Substitution gives \(36=36\), so \(P\) lies on the sphere. b) \(T:x-2y+2z+8=0\) c) \(F_1:x-2y+2z+20=0\) and \(F_2:x-2y+2z-40=0\)

- Compare the center-to-plane distance with the sphere's radius. - Use the plane's normal vector to calculate the perpendicular distance. - The sphere radius, plane distance, and cross-section radius form a right triangle. - The circle's center is the perpendicular projection of the sphere's center onto the plane.

1. The plane has normal vector \(\mathbf{n}=\langle2, -1, 2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(1)-2+2(3)+12|}{3}=6\). 2. Since \(6<7\), the plane intersects the sphere in a circle. 3. The cross-section radius is \(r_s=\sqrt{7^2-6^2}=\sqrt{13}\approx3.61\). 4. The circle's center is the perpendicular projection of \(M\) onto the plane. Write \(M_s=(1, 2, 3)+t\langle2, -1, 2\rangle\). 5. Substitution into the plane equation gives \(18+9t=0\), so \(t=-2\). Therefore, \(M_s=(-3, 4, -1)\).

a) The plane intersects the sphere in a circle. b) \(r_s=\sqrt{13}\approx3.61\) and \(M_s=(-3, 4, -1)\)

- Read the sphere's center and radius from its vector equation. - Use the plane's normal vector to calculate the center-to-plane distance. - Compare that distance with the sphere's radius. - Use the Pythagorean relationship for the cross-section radius. - Project the sphere's center onto the plane along the normal vector.

1. The sphere has center \(M=(5, -2, 3)\) and radius \(R=15\). 2. The plane's normal vector is \(\mathbf{n}=\langle6, -3, 2\rangle\), whose magnitude is \(7\). The distance from \(M\) to the plane is \(d=\frac{|6(5)-3(-2)+2(3)+7|}{7}=7\). 3. Since \(7<15\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r=\sqrt{15^2-7^2}=\sqrt{176}=4\sqrt{11}\approx13.27\). 5. Its center is the perpendicular projection of \(M\) onto the plane. Write \(M_S=M+t\mathbf{n}\). Substitution into the plane equation gives \(49+49t=0\), so \(t=-1\). 6. Therefore, \(M_S=(5, -2, 3)-\langle6, -3, 2\rangle=(-1, 1, 1)\).

The plane intersects the sphere because \(7<15\). \(M_S=(-1, 1, 1)\) \(r=4\sqrt{11}\approx13.27\)

- At a point of tangency, the radius is perpendicular to the tangent line. - Express a general point on \(g\) using a parameter. - Use a dot product to impose the perpendicularity condition. - The radius is the distance from the center to the point of tangency.

1. A direction vector for \(g\) is \(\mathbf{u}=\langle2, 1, 1\rangle\), so \(B=(4, 1, 1)+t\langle2, 1, 1\rangle\). 2. At the point of tangency, \(\overrightarrow{MB}\) is perpendicular to \(\mathbf{u}\). Since \(\overrightarrow{MB}=\langle2+2t, -4+t, 2+t\rangle\), set \(\overrightarrow{MB}\cdot\mathbf{u}=0\). 3. This gives \(2(2+2t)+(-4+t)+(2+t)=6t+2=0\), so \(t=-\frac13\). 4. Therefore, \(B=\left(\frac{10}{3}, \frac{2}{3}, \frac{2}{3}\right)\). 5. The radius is \(MB=\sqrt{\left(\frac43\right)^2+\left(-\frac{13}{3}\right)^2+\left(\frac53\right)^2}=\frac{\sqrt{210}}{3}\approx4.83\).

The radius is \(r=\frac{\sqrt{210}}{3}\approx4.83\), and the point of tangency is \(B=\left(\frac{10}{3}, \frac{2}{3}, \frac{2}{3}\right)\).

- Express a general point on the line using its parameter. - At the point of tangency, the radius vector is perpendicular to the line's direction vector. - The radius equals the shortest distance from the center to the line.

1. Write the point of tangency as \(S=(3+s, 2s, 1-s)\). Then \(\overrightarrow{MS}=\langle2+s, 2s-1, -s-3\rangle\). 2. At the point of tangency, \(\overrightarrow{MS}\) is perpendicular to the line's direction vector \(\mathbf{v}=\langle1, 2, -1\rangle\). 3. Set the dot product equal to zero: \((2+s)+2(2s-1)-(-s-3)=6s+3=0\). Thus, \(s=-\frac12\). 4. Substitution gives \(S=\left(\frac52, -1, \frac32\right)\). 5. The radius is \(MS=\sqrt{\left(\frac32\right)^2+(-2)^2+\left(-\frac52\right)^2}=\frac{5\sqrt2}{2}\approx3.54\).

The radius is \(r=\frac{5\sqrt2}{2}\approx3.54\), and the point of tangency is \(S=\left(\frac52, -1, \frac32\right)\).

- Parallel planes have parallel normal vectors. - A tangent plane is one radius from the sphere's center. - Move from the center in both unit-normal directions to locate the tangent points. - Substitute each tangent point into a plane equation with the given normal vector.

1. The sphere has center \(M=(2, 0, -1)\) and radius \(6\). The given plane has normal vector \(\mathbf{n}=\langle2, 1, -2\rangle\), whose magnitude is \(3\). 2. The tangent points lie \(6\) units from \(M\) in the two normal directions. Since \(\frac{6}{3}\mathbf{n}=2\mathbf{n}=\langle4, 2, -4\rangle\), the points are \(B_1=M+2\mathbf{n}=(6, 2, -5)\) and \(B_2=M-2\mathbf{n}=(-2, -2, 3)\). 3. Parallel tangent planes have the form \(2x+y-2z=d\). 4. Substituting \(B_1\) gives \(d=24\), and substituting \(B_2\) gives \(d=-12\). 5. Therefore, the tangent planes are \(T_1:2x+y-2z=24\) and \(T_2:2x+y-2z=-12\).

Points of tangency: \(B_1=(6, 2, -5)\) and \(B_2=(-2, -2, 3)\) Tangent planes: \(T_1:2x+y-2z=24\) and \(T_2:2x+y-2z=-12\)

- Complete the square to find the sphere's center and radius. - A plane perpendicular to a vector can use that vector as a normal vector. - Move one radius from the center in both normal directions. - A plane parallel to the y-axis has a normal vector with zero y-component.

1. Complete the square: \((x+2)^2+(y-1)^2+(z+3)^2=25\). Thus, the sphere has center \(M=(-2, 1, -3)\) and radius \(5\). 2. Because each plane is perpendicular to \(\mathbf{v}\), use \(\mathbf{v}\) as its normal vector. Its zero y-component also makes each plane parallel to the y-axis. 3. Since \(\|\mathbf{v}\|=5\), the tangent points are \(B_1=M+\mathbf{v}=(2, 1, -6)\) and \(B_2=M-\mathbf{v}=(-6, 1, 0)\). 4. The tangent planes have the form \(4x-3z=d\). Substitution gives \(d=26\) at \(B_1\) and \(d=-24\) at \(B_2\). 5. Therefore, the planes are \(T_1:4x-3z=26\) and \(T_2:4x-3z=-24\).

Points of tangency: \(B_1=(2, 1, -6)\) and \(B_2=(-6, 1, 0)\) Tangent planes: \(T_1:4x-3z=26\) and \(T_2:4x-3z=-24\)

- Verify that the stated point lies on both spheres. - Compare the distance between the centers with the sum of the radii. - The tangent plane is perpendicular to the radius at the point of tangency. - Use the line of centers as a normal direction for the plane.

1. The sphere centers are \(M_1=(2, 1, 4)\) and \(M_2=(10, 7, 4)\), and both radii are \(5\). 2. For \(B\), \(BM_1^2=4^2+3^2+0^2=25\) and \(BM_2^2=(-4)^2+(-3)^2+0^2=25\), so \(B\) lies on both spheres. 3. The center distance is \(M_1M_2=\sqrt{8^2+6^2}=10\), equal to the sum of the radii. Thus, the spheres are externally tangent at \(B\). 4. The common tangent plane is perpendicular to the line of centers. A normal vector is \(\overrightarrow{M_1M_2}=\langle8, 6, 0\rangle\), which is parallel to \(\langle4, 3, 0\rangle\). 5. The plane through \(B\) is \(4(x-6)+3(y-4)=0\), or \(4x+3y-36=0\).

The spheres are externally tangent at \(B=(6, 4, 4)\). Their common tangent plane is \(4x+3y-36=0\).

- Use the point-to-plane distance formula in part a). - Substitute a point's coordinates into the plane equation to test whether it lies in the plane. - Use the midpoint relationship \(B=\frac{A+C}{2}\). - Apply the distance formula again to \(C\).

1. A normal vector to \(E\) is \(\langle 6,3,2\rangle\), with magnitude \(7\). Therefore, \(d(A,E)=\frac{\lvert -14\rvert}{7}=2\). 2. Substitute \(B\) into the plane equation: \(6(1)+3(2)+2(1)-14=0\). Thus, \(B\) lies in \(E\). 3. Because \(B\) is the midpoint of \(\overline{AC}\), \(C=2B-A\). Therefore, \(C=(2,4,2)\). 4. Its distance from the plane is \(d(C,E)=\frac{\lvert 6(2)+3(4)+2(2)-14\rvert}{7}=\frac{14}{7}=2\). 5. Therefore, \(A\) and \(C\) are both \(2\) units from \(E\), on opposite sides of the plane.

a) \(2\) units b) \(B\) lies in \(E\), and \(C=(2,4,2)\). c) \(d(C,E)=2\) units, so the distances are equal.

- For each point, find the point on \(g\) that makes the connecting vector perpendicular to the direction vector. - Use a dot product to impose orthogonality. - Treat the origin as the point \((0,0,0)\). - The distance is the magnitude of the perpendicular vector.

1. For part a), a general point on \(g\) is \(F(t)=(1+2t, 2+2t, 3+t)\). The orthogonality condition \((F-P)\cdot\langle 2, 2, 1\rangle=0\) gives \(t=2\). 2. Then \(F=(5, 6, 5)\), so the distance from \(P\) to \(g\) is \(\sqrt{1^2+(-2)^2+2^2}=3\). 3. For part b), use the origin \(O=(0,0,0)\). The condition \((F-O)\cdot\langle 2, 2, 1\rangle=0\) gives \(t=-1\). 4. The corresponding point is \((-1, 0, 2)\), so the distance from the origin to \(g\) is \(\sqrt{(-1)^2+0^2+2^2}=\sqrt{5}\approx 2.24\).

a) \(3\) units b) \(\sqrt{5}\approx 2.24\) units

- A plane’s normal vector gives the direction of every line perpendicular to the plane. - A line parallel to a plane has a direction vector orthogonal to the plane’s normal vector. - To distinguish strict parallelism from containment, test a point of the line in the plane equation. - Check that the stated intersection point in part a) actually lies in the plane.

1. A normal vector to \(E\) is \(\vec{n}=\langle 4,-1,8\rangle\). 2. a) A line perpendicular to the plane can use \(\vec{n}\) as its direction vector. Since \(4(4)-6+8(1)=18\), point \(P\) lies in the plane. One equation is \(g: \vec{r}(t)=\langle 4,6,1\rangle+t\langle 4,-1,8\rangle\). 3. b) A line parallel to \(E\) needs a direction vector orthogonal to \(\vec{n}\). Choose \(\vec{v}=\langle 1,4,0\rangle\), since \(\vec{v}\cdot\vec{n}=4-4=0\). Thus one possible line is \(h: \vec{r}(t)=t\langle 1,4,0\rangle\). The origin is not in \(E\), so this line is strictly parallel to the plane. 4. c) For \(k\), \(\langle 2,0,-1\rangle\cdot\langle 4,-1,8\rangle=8-8=0\), so the line is parallel to the plane. 5. Its initial point lies in \(E\), because \(4(1)-2+8(2)=18\). Therefore \(k\) lies entirely in \(E\).

a) One possible equation is \(g: \vec{r}(t)=\langle 4,6,1\rangle+t\langle 4,-1,8\rangle\). b) One possible equation is \(h: \vec{r}(t)=t\langle 1,4,0\rangle\). c) Line \(k\) lies entirely in plane \(E\).

- Use the plane’s normal vector as the direction of a perpendicular line. - Substitute a general point of the line into the plane equation. - A line parallel to a plane has direction orthogonal to the plane’s normal. - A line perpendicular to a plane has direction parallel to the normal.

1. A normal vector to \(F\) is \(\vec{n}=\langle 1,2,-2\rangle\). 2. a) A line perpendicular to the plane can use the normal vector as its direction. Since it passes through the origin, \(g: \vec{r}(t)=t\langle 1,2,-2\rangle\). 3. b) Substitute \((t, 2t, -2t)\) into the plane equation: \(t+2(2t)-2(-2t)=12\). 4. Thus \(9t=12\), so \(t=\frac43\). Therefore \(S=\left(\frac43, \frac83, -\frac83\right)\). 5. c) The direction of \(\ell\) satisfies \(\langle 1,1,1\rangle\cdot\langle 1,2,-2\rangle=1\neq0\), so \(\ell\) is not parallel to the plane. 6. The vectors \(\langle 1,1,1\rangle\) and \(\langle 1,2,-2\rangle\) are not scalar multiples, so \(\ell\) is not perpendicular to the plane.

a) \(g: \vec{r}(t)=t\langle 1,2,-2\rangle\) b) \(S\left(\frac43, \frac83, -\frac83\right)\) c) Line \(\ell\) is neither parallel nor perpendicular to \(F\).

- Use the plane's normal vector to find the perpendicular distance from the center. - Compare that distance with the sphere's radius. - A plane that passes through a sphere without containing its center creates a circular cross-section. - Use the right triangle formed by the sphere radius, the plane distance, and the cross-section radius.

1. The plane's normal vector is \(\mathbf{n}=\langle2, -2, 1\rangle\), whose magnitude is \(3\). 2. The distance from \(M\) to the plane is \(d=\frac{|2(2)-2(-1)+4-1|}{3}=3\). 3. Since \(3<6\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r_S=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt3\approx5.20\).

a) The plane intersects the sphere in a circle because \(3<6\). b) \(r_S=3\sqrt3\approx5.20\)

- Read the center and radius from the sphere equation. - A tangent plane is exactly one radius from the sphere's center. - Use the plane's normal vector to write the distance expression. - An absolute-value equation may produce two values.

1. The sphere has center \(M=(1, -2, 3)\) and radius \(5\). 2. The plane's normal vector is \(\mathbf{n}=\langle4, 0, -3\rangle\), whose magnitude is \(5\). 3. The distance from \(M\) to \(F_c\) is \(d=\frac{|4(1)-3(3)+c|}{5}=\frac{|c-5|}{5}\). 4. Tangency requires \(d=5\), so \(|c-5|=25\). 5. Therefore, \(c=30\) or \(c=-20\).

\(c=30\) or \(c=-20\)

- Read the sphere's center and radius from standard form. - Compare the center-to-plane distance with the radius. - Use the Pythagorean relationship for the cross-section radius. - Project the sphere's center onto the plane along the normal vector.

1. The sphere has center \(M=(1, 2, 3)\) and radius \(R=7\). 2. The plane's normal vector is \(\mathbf{n}=\langle2, -1, 2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(1)-2+2(3)-12|}{3}=2\). 3. Since \(2<7\), the plane intersects the sphere in a circle. 4. The circle's radius is \(r=\sqrt{7^2-2^2}=\sqrt{45}=3\sqrt5\). 5. Its center is the perpendicular projection of \(M\) onto the plane. Write \(M_s=M+t\mathbf{n}\). Substitution gives \(6+9t=12\), so \(t=\frac23\). 6. Therefore, \(M_s=\left(\frac73, \frac43, \frac{13}{3}\right)\).

The plane intersects the sphere in a circle. \(r=3\sqrt5\) \(M_s=\left(\frac73, \frac43, \frac{13}{3}\right)\)

- Use \(\langle 1,1,1\rangle\) as the direction vector of the line. - To make a plane parallel to the line, choose a normal vector orthogonal to that direction. - Make sure the plane does not contain the origin. - In part b), represent an equal-coordinate point as \((a, a, a)\).

1. a) One possible plane is \(H: x-y=1\), with normal vector \(\vec{n}=\langle 1,-1,0\rangle\). 2. Since \(\langle 1,1,1\rangle\cdot\vec{n}=0\), line \(g\) is parallel to \(H\). The origin is not in \(H\), because \(0\neq1\). Therefore \(g\) and \(H\) have no common points. 3. b) A point with equal coordinates has the form \((a, a, a)\). Substitution into \(E_d\) gives \(a+a+a=d\). 4. Thus \(3a=d\), so \(a=\frac d3\). This equation has exactly one solution for every real \(d\). 5. The unique point is \(P\left(\frac d3, \frac d3, \frac d3\right)\).

a) One possible plane is \(H: x-y=1\). b) \(P\left(\frac d3, \frac d3, \frac d3\right)\)

- Points in the mirror plane do not move under reflection. - The line and its image make equal angles with the plane. - The acute angle between two lines cannot exceed \(90^\circ\). - Perpendicular lines meet at \(90^\circ\).

1. Every point in the mirror plane is fixed by the reflection. Since \(S\in E\) and \(S\in g\), its image is the same point and lies on \(g'\). 2. The original and image lines are symmetric about the plane, so the larger geometric angle formed by their directions is \(2\alpha\). 3. For \(\alpha=35^\circ\), \(2\alpha=70^\circ\), so \(\varphi=70^\circ\). 4. For \(\alpha=70^\circ\), \(2\alpha=140^\circ\). The acute angle between the lines is \(180^\circ-140^\circ=40^\circ\). 5. The lines are perpendicular when \(2\alpha=90^\circ\), so \(\alpha=45^\circ\).

a) \(S\) is fixed by reflection because \(S\in E\), so \(S\in g'\). b) \(70^\circ\) c) \(40^\circ\) d) \(45^\circ\)

- Use the plane normal and the line direction vector. - Apply the sine formula for the angle between a line and a plane. - Squaring removes the absolute value and radical. - Check the result in the unsquared equation.

1. A normal vector to \(E\) is \(\vec{n}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\), with magnitude \(3\). The line direction is \(\vec{v}_a=\begin{pmatrix}a\\1\\1\end{pmatrix}\), with magnitude \(\sqrt{a^2+2}\). 2. The line-plane angle formula gives \(\frac{|2a+1|}{3\sqrt{a^2+2}}=\sin45^\circ=\frac{\sqrt{2}}{2}\). 3. Squaring and simplifying gives \(4(2a+1)^2=18(a^2+2)\), which reduces to \((a-4)^2=0\). 4. Thus \(a=4\), which satisfies the original equation.

\(a=4\)

- A line perpendicular to a plane has a direction vector parallel to the plane’s normal vector. - Use the given point and the normal vector in a point-normal equation. - Substitute the line’s parametric coordinates into the plane equation. - Compare the resulting point with the three-dimensional midpoint formula.

1. A direction vector of the line is \(\overrightarrow{AB}=\langle 2,-2,2\rangle\). Since the line is perpendicular to the plane, use the parallel normal vector \(\vec{n}=\langle 1,-1,1\rangle\). 2. The plane through \(C\) satisfies \(\langle 1,-1,1\rangle\cdot\langle x-5,y-3,z-3\rangle=0\). Expanding gives \(x-y+z=5\). 3. Parametrize the line as \(\vec{r}(t)=\langle 1,2,3\rangle+t\langle 2,-2,2\rangle\). 4. Substitute into the plane equation: \((1+2t)-(2-2t)+(3+2t)=5\). Thus \(6t+2=5\), so \(t=\frac{1}{2}\). 5. Therefore \(S=(2, 1, 4)\). 6. The midpoint of \(A\) and \(B\) is \(\left(\frac{1+3}{2},\frac{2+0}{2},\frac{3+5}{2}\right)=(2, 1, 4)\), so \(S\) is the midpoint.

a) \(E: x-y+z=5\) b) \(S(2, 1, 4)\), which is also the midpoint of \(\overline{AB}\)

- A line perpendicular to a plane has a direction vector parallel to the plane’s normal. - Use the given plane point to find the constant. - Substitute the line into the plane equation. - The segment corresponds to parameter values from \(0\) through \(1\).

1. A direction vector of \(h\) is \(\overrightarrow{PQ}=\langle 2,2,1\rangle\). 2. Because \(F\) is perpendicular to \(h\), this direction vector can be used as a normal vector to \(F\). 3. Using point \(R\), the plane equation is \(2x+2y+z=2(7)+2(3)+3=23\). 4. Parametrize the line: \(\vec{r}(t)=\langle 1,1,1\rangle+t\langle 2,2,1\rangle\). 5. Substitution into the plane gives \(2(1+2t)+2(1+2t)+(1+t)=23\). 6. Thus \(5+9t=23\), so \(t=2\). The intersection point is \(S(5, 5, 3)\). 7. Segment \(\overline{PQ}\) corresponds to \(0\le t\le1\). Since \(t=2\), \(S\) is not on the segment.

a) \(F:2x+2y+z=23\) b) \(S(5, 5, 3)\), and \(S\notin\overline{PQ}\).

- A reflection plane is the perpendicular-bisector plane of the segment joining a point and its image. - For a half-turn axis, use the midpoint and choose a direction perpendicular to \(\overrightarrow{AB}\). - Use a dot product to check the perpendicular direction. - Substitute the coordinates of \(C\) into the plane equation.

1. The midpoint of \(\overline{AB}\) is \(M\left(\frac{4+0}{2},\frac{-2+4}{2},\frac{1+5}{2}\right)=(2, 1, 3)\). 2. The vector \(\overrightarrow{AB}=\langle -4,6,4\rangle\) is normal to the reflection plane. Use the parallel normal vector \(\vec{n}=\langle -2,3,2\rangle\). 3. Since \(E\) passes through \(M\), \(-2x+3y+2z=-2(2)+3(1)+2(3)=5\). Thus \(E: -2x+3y+2z=5\). 4. An axis for a half-turn mapping \(A\) to \(B\) must pass through \(M\) and be perpendicular to \(\overrightarrow{AB}\). Choose \(\vec{v}=\langle 3,2,0\rangle\), because \(\langle -2,3,2\rangle\cdot\langle 3,2,0\rangle=0\). 5. One possible axis is \(\vec{r}(t)=\langle 2,1,3\rangle+t\langle 3,2,0\rangle\). 6. Test \(C\): \(-2(1)+3(1)+2(3)=7\neq5\), so \(C\) is not in \(E\).

a) \(E: -2x+3y+2z=5\) b) One possible line is \(\vec{r}(t)=\langle 2,1,3\rangle+t\langle 3,2,0\rangle\), \(t\in\mathbb{R}\). c) No, \(C\notin E\).

- Form the three side vectors. - Use their magnitudes for the side lengths. - Test the two side vectors meeting at each candidate vertex with a dot product.

1. The side vectors are \(\overrightarrow{PQ}=\begin{pmatrix}2\\3\\0\end{pmatrix}\), \(\overrightarrow{PR}=\begin{pmatrix}-3\\2\\3\end{pmatrix}\), and \(\overrightarrow{QR}=\begin{pmatrix}-5\\-1\\3\end{pmatrix}\). 2. Their lengths are \(PQ=\sqrt{13}\), \(PR=\sqrt{22}\), and \(QR=\sqrt{35}\). 3. At \(P\), \(\overrightarrow{PQ}\cdot\overrightarrow{PR} =2\cdot(-3)+3\cdot2+0\cdot3=0\). 4. Therefore, triangle \(PQR\) has a right angle at \(P\).

a) \(PQ=\sqrt{13}\), \(PR=\sqrt{22}\), \(QR=\sqrt{35}\) b) The triangle is right at \(P\).

- Use the two side vectors that meet at each vertex. - A zero dot product indicates a right angle. - Compare side lengths to test for an isosceles triangle.

1. \(\overrightarrow{AB}=\begin{pmatrix}2\\1\\-2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-2\\2\\-1\end{pmatrix}\), and \(\overrightarrow{BC}=\begin{pmatrix}-4\\1\\1\end{pmatrix}\). 2. \(\overrightarrow{AB}\cdot\overrightarrow{AC}=-4+2+2=0\), so the angle at \(A\) is right. 3. The side lengths are \(AB=3\), \(AC=3\), and \(BC=3\sqrt{2}\). 4. Since \(AB=AC\), the triangle is isosceles.

The triangle is a right isosceles triangle with the right angle at \(A\) and \(AB=AC=3\).

- Use a dot product on the two sides meeting at \(A\). - Compare the magnitudes of those same side vectors.

1. \(\overrightarrow{AB}=\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\). 2. Their dot product is \(1\cdot2+2\cdot(-2)+2\cdot1=0\), so the angle at \(A\) is right. 3. Their magnitudes are both \(3\), so \(AB=AC\). 4. Therefore, the triangle is both right and isosceles.

The triangle is right at \(A\) and isosceles with \(AB=AC=3\).

- Identify the two vectors that meet at \(D\). - Set their dot product equal to zero. - Solve the resulting linear equation.

1. The vectors forming the angle at \(D\) are \(\overrightarrow{DE}=\begin{pmatrix}2\\1\\k-2\end{pmatrix}\) and \(\overrightarrow{DF}=\begin{pmatrix}1\\2\\-2\end{pmatrix}\). 2. A right angle requires \(\overrightarrow{DE}\cdot\overrightarrow{DF}=0\). 3. The dot product is \(2\cdot1+1\cdot2+(k-2)(-2)=8-2k\). 4. Solving \(8-2k=0\) gives \(k=4\).

\(k=4\)

- Write the dot product in terms of \(k\). - Set the resulting expression equal to \(16\). - Solve the linear equation.

1. Set up the dot product equation: \(5k+3k+(-2)\cdot4=16\). 2. Simplify: \(8k-8=16\). 3. Solve: \(8k=24\), so \(k=3\).

\(k=3\)

- Set the dot product equal to zero. - Use distributivity in part b). - Reuse the orthogonality result.

1. Orthogonality requires \(\vec{u}\cdot\vec{v}=2\cdot3+1\cdot(-4)+2z=0\). 2. Thus, \(2+2z=0\), so \(z=-1\). 3. By distributivity, \(\vec{u}\cdot(\vec{u}+\vec{v}) =\vec{u}\cdot\vec{u}+\vec{u}\cdot\vec{v}\). 4. The second term is \(0\), and \(\vec{u}\cdot\vec{u}=2^2+1^2+(-1)^2=6\).

a) \(z=-1\) b) \(6\)

- Use the magnitude formula for each vector. - For part c, first simplify the two vectors inside the parentheses. - Multiply corresponding components and add.

1. \(\|\vec{a}\|=\sqrt{1^2+2^2+(-2)^2}=3\) and \(\|\vec{b}\|=\sqrt{4^2+0^2+3^2}=5\). 2. \(\vec{a}\cdot\vec{b}=1\cdot4+2\cdot0+(-2)\cdot3=-2\). 3. \(\vec{a}-\vec{b}=\begin{pmatrix}-3\\2\\-5\end{pmatrix}\) and \(2\vec{a}+\vec{b}=\begin{pmatrix}6\\4\\-1\end{pmatrix}\). 4. Their dot product is \((-3)\cdot6+2\cdot4+(-5)\cdot(-1)=-18+8+5=-5\).

a) \(\|\vec{a}\|=3\), \(\|\vec{b}\|=5\) b) \(\vec{a}\cdot\vec{b}=-2\) c) \(-5\)

- Orthogonality means a zero dot product. - Simplify to a quadratic equation. - Factor the quadratic.

1. Set the dot product equal to zero: \(k^2+5\cdot(-6)+k=0\). 2. Thus, \(k^2+k-30=0\). 3. Factor: \((k+6)(k-5)=0\). 4. Therefore, \(k=-6\) or \(k=5\).

\(k=-6\) or \(k=5\)

- Compute the dot product componentwise. - In the angle formula, identify which factor can be zero for nonzero vectors.

1. \(2\cdot3+4y+(-1)\cdot2=0\). 2. Thus, \(4+4y=0\), so \(y=-1\). 3. If both vectors are nonzero, then their magnitudes are positive. The product can still be \(0\) when \(\cos\varphi=0\). 4. This occurs when \(\varphi=90^\circ\), so the vectors are orthogonal.

a) \(y=-1\) b) The dot product is zero when the angle is \(90^\circ\), because \(\cos90^\circ=0\).

- Set the dot product equal to zero. - Solve the resulting quadratic equation.

1. Orthogonality requires \(\vec{a}\cdot\vec{b}=0\). 2. The dot product is \(t^2-5t+4\). 3. Factor: \(t^2-5t+4=(t-1)(t-4)\). 4. Therefore, \(t=1\) or \(t=4\).

\(t=1\) or \(t=4\)

- Write one dot-product equation for each orthogonality condition. - Solve the resulting system of two equations.

1. Orthogonality gives \(\vec{n}\cdot\vec{u}=2x-y=0\) and \(\vec{n}\cdot\vec{v}=x+2y-5=0\). 2. From the first equation, \(y=2x\). 3. Substitute into the second: \(x+4x-5=0\), so \(x=1\). 4. Therefore, \(y=2\).

\(x=1\), \(y=2\)

- Apply the distributive and commutative properties of the dot product. - Interpret a zero dot product geometrically. - Identify \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) as diagonal vectors.

1. Distribute the dot product: \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\vec{a}\cdot\vec{a}-\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{a}-\vec{b}\cdot\vec{b}\). 2. Since \(\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}\), the middle terms cancel. 3. Using \(\vec{x}\cdot\vec{x}=\|\vec{x}\|^2\) gives \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|^2-\|\vec{b}\|^2\). 4. The dot product is \(0\) exactly when \(\|\vec{a}\|=\|\vec{b}\|\). 5. Thus, the parallelogram has equal adjacent side lengths and is a rhombus. Equivalently, its diagonal vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are perpendicular.

a) \((\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|^2-\|\vec{b}\|^2\) b) The parallelogram must be a rhombus.

- Divide \(\vec{a}\) by its magnitude to create a unit vector. - Use the vector projection formula. - Subtract the parallel component from \(\vec{b}\) to obtain the perpendicular component.

1. \(\|\vec{a}\|=\sqrt{3^2+4^2}=5\), so \(\hat{a}=\frac{\vec{a}}{\|\vec{a}\|}=\begin{pmatrix}\frac35\\\frac45\\0\end{pmatrix}\). 2. The projection is \(\operatorname{proj}_{\vec{a}}\vec{b}=\frac{\vec{a}\cdot\vec{b}}{\vec{a}\cdot\vec{a}}\vec{a}\). 3. Since \(\vec{a}\cdot\vec{b}=11\) and \(\vec{a}\cdot\vec{a}=25\), \(\vec{b}_{\parallel}=\frac{11}{25}\vec{a}=\begin{pmatrix}\frac{33}{25}\\\frac{44}{25}\\0\end{pmatrix}\). 4. The perpendicular component is \(\vec{b}_{\perp}=\vec{b}-\vec{b}_{\parallel}=\begin{pmatrix}-\frac{8}{25}\\\frac{6}{25}\\2\end{pmatrix}\). 5. A check gives \(\vec{a}\cdot\vec{b}_{\perp}=0\).

1. \(\hat{a}=\begin{pmatrix}\frac35\\\frac45\\0\end{pmatrix}\) 2. \(\operatorname{proj}_{\vec{a}}\vec{b}=\begin{pmatrix}\frac{33}{25}\\\frac{44}{25}\\0\end{pmatrix}\) 3. \(\vec{b}_{\parallel}=\begin{pmatrix}\frac{33}{25}\\\frac{44}{25}\\0\end{pmatrix}\), \(\vec{b}_{\perp}=\begin{pmatrix}-\frac{8}{25}\\\frac{6}{25}\\2\end{pmatrix}\)

- Use the vector projection formula for a line through the origin. - The projected vector is the position vector of the perpendicular foot. - The distance is the magnitude of the difference between the original vector and its projection.

1. Since \(\mathbf{v}\cdot\mathbf{v}=1^2+2^2+2^2=9\) and \(\mathbf{p}\cdot\mathbf{v}=3+6+6=15\), the projection is \(\operatorname{proj}_{\mathbf{v}}\mathbf{p}=\frac{15}{9}\mathbf{v}=\frac{5}{3}\langle 1, 2, 2\rangle=\left\langle \frac{5}{3}, \frac{10}{3}, \frac{10}{3}\right\rangle\). 2. Therefore, \(F=\left(\frac{5}{3}, \frac{10}{3}, \frac{10}{3}\right)\). 3. The perpendicular vector is \(\overrightarrow{PF}=\left\langle -\frac{4}{3}, \frac{1}{3}, \frac{1}{3}\right\rangle\), so the distance is \(\sqrt{\frac{16}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{2}\approx 1.41\).

1. \(F\left(\frac{5}{3}, \frac{10}{3}, \frac{10}{3}\right)\) 2. \(\sqrt{2}\approx 1.41\) units

- Use the right angle subtended by a diameter. - Form the two vectors that meet at \(C\). - Set their dot product equal to zero.

1. By Thales's theorem, \(\angle ACB=90^\circ\). 2. \(\overrightarrow{CA}=\begin{pmatrix}-x\\-4\\-2\end{pmatrix}\) and \(\overrightarrow{CB}=\begin{pmatrix}4-x\\-2\\2\end{pmatrix}\). 3. Set their dot product equal to zero: \((-x)(4-x)+(-4)\cdot(-2)+(-2)\cdot2=0\). 4. This simplifies to \(x^2-4x+4=0\), or \((x-2)^2=0\). 5. Therefore, \(x=2\).

\(x=2\)

- Write the squared magnitude as a sum of squared components. - Expand each binomial. - For perpendicular vectors, substitute \(\vec{a}\cdot\vec{b}=0\).

1. Expand the left side: \(\|\vec{a}+\vec{b}\|^2=(a_1+b_1)^2+(a_2+b_2)^2+(a_3+b_3)^2\). 2. Expanding and grouping gives \((a_1^2+a_2^2+a_3^2)+2(a_1b_1+a_2b_2+a_3b_3)+(b_1^2+b_2^2+b_3^2)\). 3. These terms are \(\|\vec{a}\|^2\), \(2(\vec{a}\cdot\vec{b})\), and \(\|\vec{b}\|^2\), respectively. 4. If \(\vec{a}\perp\vec{b}\), then \(\vec{a}\cdot\vec{b}=0\), so \(\|\vec{a}+\vec{b}\|^2=\|\vec{a}\|^2+\|\vec{b}\|^2\). 5. This is the Pythagorean theorem for the right triangle whose legs are represented by \(\vec{a}\) and \(\vec{b}\).

1. \(\|\vec{a}+\vec{b}\|^2=\|\vec{a}\|^2+2(\vec{a}\cdot\vec{b})+\|\vec{b}\|^2\) 2. For perpendicular vectors, the identity becomes the Pythagorean theorem.

- Verify one pair of opposite-side vectors is equal. - Compare adjacent side lengths. - Use a dot product to test for a right angle.

1. \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\1\end{pmatrix}\) and \(\overrightarrow{DC}=\begin{pmatrix}2\\2\\1\end{pmatrix}\), so the quadrilateral is a parallelogram. 2. \(\|\overrightarrow{AB}\|=3\), and \(\overrightarrow{AD}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\) also has magnitude \(3\). 3. A parallelogram with equal adjacent sides is a rhombus. 4. However, \(\overrightarrow{AB}\cdot\overrightarrow{AD}=4\ne0\), so adjacent sides are not perpendicular. Therefore, the rhombus is not a square.

It is a rhombus because it is a parallelogram with side length \(3\), but it is not a square because adjacent sides have dot product \(4\), not \(0\).

- Translate each perpendicularity condition into a zero dot product. - Solve the resulting system to describe all common perpendicular vectors. - Scale the direction vector to the required magnitude, remembering both possible directions.

1. Let \(\vec{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Orthogonality gives \(x+2y+2z=0\) and \(2x+y-2z=0\). 2. Solving the system gives \(x=2z\) and \(y=-2z\), so every common perpendicular vector has the form \(\vec{x}=z\begin{pmatrix}2\\-2\\1\end{pmatrix}\). 3. The direction vector has magnitude \(\sqrt{2^2+(-2)^2+1^2}=3\). 4. To obtain magnitude \(18\), \(|z|=\frac{18}{3}=6\). 5. Therefore, the two vectors are \(\begin{pmatrix}12\\-12\\6\end{pmatrix}\) and \(\begin{pmatrix}-12\\12\\-6\end{pmatrix}\).

\(\begin{pmatrix}12\\-12\\6\end{pmatrix}\) and \(\begin{pmatrix}-12\\12\\-6\end{pmatrix}\)

- Write each perpendicularity condition as a zero dot product. - Let the unknown vector have components \(x,y,z\). - Use the third dot-product condition to determine the remaining scale.

1. Let \(\vec{w}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). 2. The three conditions give \(3x+z=0\), \(x+2y-z=0\), and \(x+y+z=16\). 3. From \(3x+z=0\), \(z=-3x\). Substituting into the second equation gives \(2x+y=0\), so \(y=-2x\). 4. Substitute into the final equation: \(x-2x-3x=16\), so \(-4x=16\) and \(x=-4\). 5. Therefore, \(y=8\), \(z=12\), and \(\vec{w}=\begin{pmatrix}-4\\8\\12\end{pmatrix}\).

\(\vec{w}=\begin{pmatrix}-4\\8\\12\end{pmatrix}\)

- Express each diagonal as a sum or difference of side vectors. - Use \(\|\vec{v}\|^2=\vec{v}\cdot\vec{v}\). - Track what happens to the mixed dot-product terms. - In a rhombus, both side-vector magnitudes equal \(s\).

1. The diagonal vectors are \(\overrightarrow{AC}=\vec{a}+\vec{b}\) and \(\overrightarrow{BD}=\vec{b}-\vec{a}\). 2. Expand their squared magnitudes: \(\|\vec{a}+\vec{b}\|^2=\|\vec{a}\|^2+2\vec{a}\cdot\vec{b}+\|\vec{b}\|^2\), and \(\|\vec{b}-\vec{a}\|^2=\|\vec{b}\|^2-2\vec{a}\cdot\vec{b}+\|\vec{a}\|^2\). 3. Adding cancels the mixed terms: \(\|\overrightarrow{AC}\|^2+\|\overrightarrow{BD}\|^2 =2\|\vec{a}\|^2+2\|\vec{b}\|^2\). 4. For a rhombus, \(\|\vec{a}\|=\|\vec{b}\|=s\), so \(\|\overrightarrow{AC}\|^2+\|\overrightarrow{BD}\|^2=4s^2\).

a) The expanded mixed terms cancel, leaving \(2(\|\vec{a}\|^2+\|\vec{b}\|^2)\). b) \(\|\overrightarrow{AC}\|^2+\|\overrightarrow{BD}\|^2=4s^2\)

- Use the midpoint relationship between \(\overrightarrow{MB}\) and \(\overrightarrow{MC}\). - Replace squared magnitudes with self-dot-products. - Expand both expressions and add them. - Look for cancellation of the mixed terms.

1. Since \(M\) is the midpoint of \(BC\), \(\overrightarrow{MC}=-\overrightarrow{MB}\). 2. Let \(\vec{m}=\overrightarrow{AM}\) and \(\vec{x}=\overrightarrow{MB}\). Then \(\overrightarrow{AB}=\vec{m}+\vec{x}\) and \(\overrightarrow{AC}=\vec{m}-\vec{x}\). 3. Expand: \(\|\vec{m}+\vec{x}\|^2=\|\vec{m}\|^2+2\vec{m}\cdot\vec{x}+\|\vec{x}\|^2\), and \(\|\vec{m}-\vec{x}\|^2=\|\vec{m}\|^2-2\vec{m}\cdot\vec{x}+\|\vec{x}\|^2\). 4. Adding gives \(\|\overrightarrow{AB}\|^2+\|\overrightarrow{AC}\|^2 =2\|\vec{m}\|^2+2\|\vec{x}\|^2 =2(\|\overrightarrow{AM}\|^2+\|\overrightarrow{BM}\|^2)\).

Expanding the two squared vector sums causes the mixed dot-product terms to cancel, yielding \(2(\|\overrightarrow{AM}\|^2+\|\overrightarrow{BM}\|^2)\).

- Use the required \(y\)-component in the unknown vector. - Set each dot product equal to zero. - Solve the resulting two equations for the other components.

1. Write \(\vec{n}=\begin{pmatrix}x\\11\\z\end{pmatrix}\). 2. Orthogonality to \(\vec{u}\) gives \(4x-22+z=0\). 3. Orthogonality to \(\vec{v}\) gives \(x+3z=0\). 4. From the second equation, \(x=-3z\). Substitution gives \(-12z-22+z=0\), so \(z=-2\). 5. Then \(x=6\), so \(\vec{n}=\begin{pmatrix}6\\11\\-2\end{pmatrix}\).

\(\vec{n}=\begin{pmatrix}6\\11\\-2\end{pmatrix}\)

- Rewrite \(R_t\) in vector-parametric form. - At each specified vertex, form the two side vectors that meet there. - Perpendicular vectors have dot product \(0\). - Set up a separate equation in \(t\) for parts b) and c).

1. Write the position vector of \(R_t\) as \(\overrightarrow{OR_t}=\begin{pmatrix}0\\0\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\0\end{pmatrix}\). Thus, the locus is a line through \((0, 0, 2)\) with direction vector \(\begin{pmatrix}1\\2\\0\end{pmatrix}\). 2. For a right angle at \(P\), require \(\overrightarrow{PQ}\cdot\overrightarrow{PR_t}=0\). Here, \(\overrightarrow{PQ}=\begin{pmatrix}-2\\2\\2\end{pmatrix}\) and \(\overrightarrow{PR_t}=\begin{pmatrix}t-3\\2t-2\\1\end{pmatrix}\). 3. Their dot product is \(-2(t-3)+2(2t-2)+2=2t+4\). Setting it equal to zero gives \(t=-2\). 4. For a right angle at \(Q\), require \(\overrightarrow{QP}\cdot\overrightarrow{QR_t}=0\). Here, \(\overrightarrow{QP}=\begin{pmatrix}2\\-2\\-2\end{pmatrix}\) and \(\overrightarrow{QR_t}=\begin{pmatrix}t-1\\2t-4\\-1\end{pmatrix}\). 5. Their dot product is \(2(t-1)-2(2t-4)+2=8-2t\). Setting it equal to zero gives \(t=4\).

a) The points lie on the line \(\overrightarrow{OR_t}=\begin{pmatrix}0\\0\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\0\end{pmatrix}\). b) \(t=-2\) c) Yes. The triangle has a right angle at \(Q\) when \(t=4\).

- Compute each dot product first. - Use the resulting scalar to scale the remaining vector. - Compare the two vector results.

1. \(\vec{e}_1\cdot\vec{e}_1=1\), so \((\vec{e}_1\cdot\vec{e}_1)\vec{e}_2=1\vec{e}_2=\begin{pmatrix}0\\1\\0\end{pmatrix}\). 2. \(\vec{e}_1\cdot\vec{e}_2=0\), so \((\vec{e}_1\cdot\vec{e}_2)\vec{e}_1=0\vec{e}_1=\vec{0}\). 3. The expressions are different because the dot product and scalar multiplication are different operations. Their inputs and outputs cannot be regrouped as though they were one associative operation.

\((\vec{e}_1\cdot\vec{e}_1)\vec{e}_2=\begin{pmatrix}0\\1\\0\end{pmatrix}\), while \((\vec{e}_1\cdot\vec{e}_2)\vec{e}_1=\vec{0}\).

- First determine whether the two given vectors are parallel. - Write a dot-product equation for a vector perpendicular to \(\vec{u}\). - What geometric set contains all vectors perpendicular to one fixed nonzero vector?

1. Since \(\vec{v}=-\frac{3}{2}\vec{u}\), the two given vectors are parallel and point in opposite directions. 2. A vector \(\vec{w}=\begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix}\) is perpendicular to \(\vec{u}\) when \(2w_1+4w_2-6w_3=0\). 3. Choose \(w_2=1\) and \(w_3=0\). Then \(2w_1+4=0\), so \(w_1=-2\). One possible vector is \(\vec{w}=\begin{pmatrix}-2\\1\\0\end{pmatrix}\). 4. Because \(\vec{v}\) is a scalar multiple of \(\vec{u}\), every vector perpendicular to \(\vec{u}\) is also perpendicular to \(\vec{v}\). 5. The vectors perpendicular to the line in the direction of \(\vec{u}\) form a plane through the origin. That plane contains infinitely many vector directions.

a) One possible vector is \(\vec{w}=\begin{pmatrix}-2\\1\\0\end{pmatrix}\). b) Because \(\vec{v}=-\frac{3}{2}\vec{u}\), the given vectors are parallel. Their common perpendicular vectors form a plane through the origin, so infinitely many directions are possible.

- Find each vector by subtracting the initial point from the terminal point. - Equal opposite-side vectors identify a parallelogram. - Use a dot product to test whether adjacent sides are perpendicular.

1. Subtract endpoint coordinates: \(\overrightarrow{AB}=\begin{pmatrix}3\\1\\2\end{pmatrix}\) and \(\overrightarrow{DC}=\begin{pmatrix}3\\1\\2\end{pmatrix}\). 2. Similarly, \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\0\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}1\\3\\0\end{pmatrix}\). 3. Both pairs of opposite sides have equal vectors, so \(ABCD\) is a parallelogram. 4. Test adjacent sides for perpendicularity: \(\overrightarrow{AB}\cdot\overrightarrow{AD}=3\cdot1+1\cdot3+2\cdot0=6\). 5. Since the dot product is not zero, adjacent sides are not perpendicular. Therefore, \(ABCD\) is not a rectangle.

a) \(\overrightarrow{AB}=\begin{pmatrix}3\\1\\2\end{pmatrix}\), \(\overrightarrow{DC}=\begin{pmatrix}3\\1\\2\end{pmatrix}\), \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\0\end{pmatrix}\), and \(\overrightarrow{BC}=\begin{pmatrix}1\\3\\0\end{pmatrix}\). b) \(ABCD\) is a parallelogram, but it is not a rectangle because \(\overrightarrow{AB}\cdot\overrightarrow{AD}=6\ne0\).

- Find two side vectors from the same vertex. - Check whether those side vectors are perpendicular. - For the normal vector, impose a zero dot product with each side vector, then scale to the required magnitude.

1. \(\overrightarrow{AB}=\begin{pmatrix}4\\0\\3\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\2\\0\end{pmatrix}\). 2. Their dot product is \(0\), so the side vectors are perpendicular. Their magnitudes are \(5\) and \(2\). 3. Therefore, the triangle's area is \(A=\frac12\cdot5\cdot2=5\). 4. Let \(\vec{n}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Perpendicularity gives \(4x+3z=0\) and \(2y=0\). 5. Thus, \(\vec{n}=t\begin{pmatrix}-3\\0\\4\end{pmatrix}\). The direction vector has magnitude \(5\), so \(|t|=3\) for magnitude \(15\). 6. A positive \(z\)-component requires \(t=3\), giving \(\vec{n}=\begin{pmatrix}-9\\0\\12\end{pmatrix}\).

a) \(5\) square units b) \(\vec{n}=\begin{pmatrix}-9\\0\\12\end{pmatrix}\)

- Write an unknown vector with three components. - Set its dot product with each given vector equal to zero. - Express the solution set using one free parameter.

1. Let \(\vec{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). 2. The perpendicularity conditions are \(4x-y+2z=0\) and \(-2x+3y+z=0\). 3. Solving gives \(x=-\frac{7}{10}z\) and \(y=-\frac45z\). 4. Let \(z=10r\). Then \(\vec{x}=r\begin{pmatrix}-7\\-8\\10\end{pmatrix}\), where \(r\in\mathbb{R}\).

\(\vec{x}=r\begin{pmatrix}-7\\-8\\10\end{pmatrix}\), where \(r\in\mathbb{R}\)

- Translate perpendicularity into a dot-product equation. - Translate the magnitude condition into a second equation. - Solve one equation for one variable, then substitute into the other.

1. Perpendicularity requires \(\vec{u}\cdot\vec{v}=0\): \(x-4+2y=0\), so \(x=4-2y\). 2. The magnitude condition gives \(\sqrt{x^2+2^2+y^2}=3\). After squaring, \(x^2+y^2=5\). 3. Substitute \(x=4-2y\): \((4-2y)^2+y^2=5\). 4. Simplifying gives \(5y^2-16y+11=0\), which factors as \((y-1)(5y-11)=0\). 5. If \(y=1\), then \(x=2\). If \(y=\frac{11}{5}\), then \(x=-\frac{2}{5}\).

The two solutions are \((x, y)=(2, 1)\) and \((x, y)=\left(-\frac{2}{5}, \frac{11}{5}\right)\).

- Use the dot product to express perpendicularity. - Use the magnitude formula for a second equation. - Substitute one variable in terms of the other. - Remember that a squared equation can produce two signs.

1. Perpendicularity gives \(\vec{n}\cdot\vec{w}=3a-4b=0\), so \(a=\frac{4}{3}b\). 2. The magnitude condition gives \(\sqrt{a^2+b^2+12^2}=13\). After squaring, \(a^2+b^2=25\). 3. Substitute \(a=\frac{4}{3}b\): \(\frac{16}{9}b^2+b^2=25\). 4. Thus, \(\frac{25}{9}b^2=25\), so \(b^2=9\) and \(b=\pm3\). 5. The corresponding values of \(a\) are \(a=\pm4\), with matching signs.

The possible ordered pairs are \((a, b)=(4, 3)\) and \((a, b)=(-4, -3)\).

- Determine the type of each parenthesized result first. - Vector addition produces a vector. - A dot product of two vectors produces a scalar, while scalar multiplication produces a vector.

1. In part a, the quantity in parentheses must be a vector so that it can be dotted with \(\vec{w}\). Therefore, \(\square=+\). 2. In part b, adding two vectors produces a vector, while their dot product produces a scalar. Therefore, \(\square=+\). 3. In part c, the quantity in parentheses is a scalar. Scalar multiplication by \(\vec{w}\) produces a vector, so \(\square=\cdot\). 4. In part d, both parenthesized expressions are vectors. Their dot product produces a scalar, so \(\square=\cdot\).

a) \(+\) b) \(+\) c) \(\cdot\) d) \(\cdot\)

- What dot-product value indicates perpendicular vectors? - Compute the dot product in terms of \(k\). - Solve the resulting equation.

1. Perpendicular vectors have dot product \(0\). 2. Compute \(\vec{u}\cdot\vec{v}=2(k+1)+3k-8=5k-6\). 3. Set the expression equal to zero: \(5k-6=0\). 4. Therefore, \(k=\frac{6}{5}\).

\(k=\frac{6}{5}\)

- Write the dot product as an expression in \(k\). - Set it equal to \(11\). - Solve the resulting linear equation.

1. Form the dot product equation: \(3k+4k+(-2)\cdot5=11\). 2. Simplify: \(7k-10=11\). 3. Solve: \(7k=21\), so \(k=3\).

\(k=3\)

- Express perpendicularity with a dot product. - Set the resulting expression equal to zero. - Check whether the equation has more than one solution.

1. Perpendicularity requires \(\vec{u}\cdot\vec{v}=0\). 2. Compute the dot product: \(t^2+2\cdot(-3)+1\cdot5=t^2-1\). 3. Solve \(t^2-1=0\). 4. Factoring gives \((t-1)(t+1)=0\), so \(t=1\) or \(t=-1\).

\(t=-1\) or \(t=1\).

- Assign convenient values to two components and solve for the third. - Recognize the standard form of a plane equation. - Compare the number of solutions with the uniqueness required of an inverse operation.

1. Writing \(\vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\) gives \(2x_1+x_2-2x_3=12\). 2. Two solutions are \(\vec{x}_1=\begin{pmatrix}6\\0\\0\end{pmatrix}\) and \(\vec{x}_2=\begin{pmatrix}0\\12\\0\end{pmatrix}\). 3. The solution set is the plane \(2x_1+x_2-2x_3=12\), whose normal vector is \(\begin{pmatrix}2\\1\\-2\end{pmatrix}\). 4. An inverse operation would need to produce a unique \(\vec{x}\). Because infinitely many vectors on this plane have the same dot product \(12\), no unique vector quotient can be defined.

a) One possible pair is \(\vec{x}_1=\begin{pmatrix}6\\0\\0\end{pmatrix}\) and \(\vec{x}_2=\begin{pmatrix}0\\12\\0\end{pmatrix}\). b) The solutions form a plane in \(\mathbb{R}^3\). c) The dot product equation has infinitely many vector solutions, so an inverse based on vector division would not be unique.

- Use a dot product for part a). - Use the magnitude formula for part b). - Scale \(\vec{x}\) by the ratio of the desired magnitude to its current magnitude. - Use the linearity of the dot product for part d).

1. Perpendicularity requires \(\vec{v}\cdot\vec{x}=0\): \(2\cdot1-2\cdot2+x_3=0\). Thus, \(x_3=2\), and \(\vec{x}=\begin{pmatrix}1\\2\\2\end{pmatrix}\). 2. Its magnitude is \(\|\vec{x}\|=\sqrt{1^2+2^2+2^2}=3\). 3. To obtain magnitude \(12\) without changing direction, multiply by \(4\): \(\vec{y}=4\vec{x}=\begin{pmatrix}4\\8\\8\end{pmatrix}\). 4. By linearity of the dot product, \(\vec{v}\cdot\vec{y}=\vec{v}\cdot(4\vec{x})=4(\vec{v}\cdot\vec{x})=0\). Therefore, \(\vec{y}\perp\vec{v}\).

a) \(x_3=2\) b) \(\|\vec{x}\|=3\) c) \(\vec{y}=\begin{pmatrix}4\\8\\8\end{pmatrix}\) d) Since \(\vec{y}=4\vec{x}\), \(\vec{v}\cdot\vec{y}=4(\vec{v}\cdot\vec{x})=0\).

- Use the dot product to test perpendicularity. - In three dimensions, what geometric set has a fixed vector as its normal? - Compare the dimension of that set with the perpendicular set in two dimensions.

1. Perpendicularity requires \(\vec{a}\cdot\vec{b}=0\). 2. Compute: \(2k+6k-20=0\). 3. Thus, \(8k=20\), so \(k=\frac{5}{2}\). 4. In \(\mathbb{R}^3\), all vectors perpendicular to a fixed nonzero vector \(\vec{n}\) lie in the plane through the origin whose normal vector is \(\vec{n}\). A plane contains infinitely many directions. 5. In \(\mathbb{R}^2\), the perpendicular vectors lie on one line through the origin. The two orientations along that line are opposites of each other.

a) \(k=\frac{5}{2}\) b) In \(\mathbb{R}^3\), the perpendicular vectors form a plane through the origin, so infinitely many directions occur. In \(\mathbb{R}^2\), they form one line through the origin, with two opposite orientations.

- Treat each coordinate axis as a line through the origin. - For each pair of lines, find a direction perpendicular to both. - Project the vector between the given points onto that perpendicular direction.

1. For the x-axis, a common perpendicular direction to \(\langle 0,1,1\rangle\) and \(\langle 1,0,0\rangle\) is \(\langle 0,1,-1\rangle\). Therefore, \(d_x=\frac{\lvert\langle 3,0,4\rangle\cdot\langle 0,1,-1\rangle\rvert}{\sqrt{2}}=2\sqrt{2}\approx2.83\). 2. For the y-axis, a common perpendicular direction to \(\langle 0,1,1\rangle\) and \(\langle 0,1,0\rangle\) is \(\langle 1,0,0\rangle\). Thus, \(d_y=\lvert\langle 3,0,4\rangle\cdot\langle 1,0,0\rangle\rvert=3\). 3. For the z-axis, a common perpendicular direction to \(\langle 0,1,1\rangle\) and \(\langle 0,0,1\rangle\) is also \(\langle 1,0,0\rangle\). Thus, \(d_z=3\).

x-axis: \(2\sqrt{2}\approx2.83\) units y-axis: \(3\) units z-axis: \(3\) units

- Form the two side vectors that meet at \(P\). - A dot product of \(0\) indicates a right angle. - Add all three side lengths for the perimeter. - Use the legs of the right triangle to find its area.

1. The side vectors meeting at \(P\) are \(\overrightarrow{PQ}=\begin{pmatrix}2\\2\\1\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}-1\\2\\-2\end{pmatrix}\). 2. Their dot product is \(\overrightarrow{PQ}\cdot\overrightarrow{PR}=2\cdot(-1)+2\cdot2+1\cdot(-2)=0\). Therefore, the angle at \(P\) is \(90^\circ\). 3. The side lengths are \(\|\overrightarrow{PQ}\|=3\), \(\|\overrightarrow{PR}\|=3\), and \(\|\overrightarrow{QR}\|=\sqrt{18}=3\sqrt{2}\). 4. The perimeter is \(3+3+3\sqrt{2}=6+3\sqrt{2}\). 5. Since the perpendicular sides have lengths \(3\) and \(3\), the area is \(\frac{1}{2}\cdot3\cdot3=\frac{9}{2}\).

a) \(\overrightarrow{PQ}\cdot\overrightarrow{PR}=0\), so the triangle has a right angle at \(P\). b) The perimeter is \(6+3\sqrt{2}\), approximately \(10.24\). c) The area is \(\frac{9}{2}\) square units.

- Form the two vectors that meet at \(E\). - Translate the right-angle condition into a dot-product equation. - For part b), compare the three squared side lengths.

1. The vectors meeting at \(E\) are \(\overrightarrow{ED}=\begin{pmatrix}-2\\-1\\k-1\end{pmatrix}\) and \(\overrightarrow{EF}=\begin{pmatrix}-2\\3\\1\end{pmatrix}\). 2. For a right angle at \(E\), set their dot product equal to zero: \(\overrightarrow{ED}\cdot\overrightarrow{EF}=4-3+(k-1)=k\). 3. Thus, the right-angle condition is \(k=0\). 4. When \(k=4\), \(DE^2=2^2+1^2+3^2=14\), \(EF^2=(-2)^2+3^2+1^2=14\), and \(DF^2=0^2+4^2+(-2)^2=20\). 5. Since \(DE=EF=\sqrt{14}\), the triangle is isosceles when \(k=4\).

a) \(k=0\) b) Yes. When \(k=4\), \(DE=EF=\sqrt{14}\), so the triangle is isosceles.

- Compare the vectors for opposite sides. - Use a dot product to test adjacent sides for perpendicularity. - A square also requires equal adjacent side lengths.

1. Compute adjacent and opposite side vectors: \(\overrightarrow{AB}=\begin{pmatrix}4\\0\\0\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}0\\2\\3\end{pmatrix}\), \(\overrightarrow{DC}=\begin{pmatrix}4\\0\\0\end{pmatrix}\), and \(\overrightarrow{AD}=\begin{pmatrix}0\\2\\3\end{pmatrix}\). 2. Both pairs of opposite sides have equal vectors, so \(ABCD\) is a parallelogram. 3. Also, \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\). Therefore, adjacent sides are perpendicular, and the parallelogram is a rectangle. 4. The adjacent side lengths are \(\|\overrightarrow{AB}\|=4\) and \(\|\overrightarrow{BC}\|=\sqrt{13}\). 5. Since these lengths are unequal, the rectangle is not a square.

The quadrilateral is a rectangle because its opposite-side vectors are equal and adjacent sides have dot product \(0\). It is not a square because its adjacent side lengths are \(4\) and \(\sqrt{13}\).

- Use the equal opposite-side vectors of a parallelogram to locate \(D\). - Equal adjacent side lengths make a parallelogram a rhombus. - A dot product tests whether adjacent sides form a right angle.

1. In a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\). 2. Since \(\overrightarrow{BC}=\begin{pmatrix}0\\3\\4\end{pmatrix}\), \(\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{BC}=\begin{pmatrix}2\\4\\3\end{pmatrix}\). Thus, \(D(2, 4, 3)\). 3. The adjacent side vectors are \(\overrightarrow{AB}=\begin{pmatrix}3\\4\\0\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}0\\3\\4\end{pmatrix}\). 4. Both have magnitude \(5\), so the parallelogram is a rhombus. 5. Their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{BC}=12\ne0\). Therefore, it is not a rectangle.

a) \(D(2, 4, 3)\) b) The parallelogram is a rhombus because adjacent sides both have length \(5\). It is not a rectangle because \(\overrightarrow{AB}\cdot\overrightarrow{BC}=12\ne0\).

- Use \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}\). - Average the two vectors from \(A\) to locate the midpoint of \(BC\). - Compare the two side magnitudes after checking the dot product.

1. The vector from \(B\) to \(C\) is \(\vec{w}=\vec{v}-\vec{u} =\begin{pmatrix}-7\\4\\-3\end{pmatrix}\). 2. The midpoint of \(BC\) has position relative to \(A\) equal to \(\vec{s}_a=\frac{1}{2}(\vec{u}+\vec{v}) =\begin{pmatrix}\frac{1}{2}\\2\\\frac{3}{2}\end{pmatrix}\). 3. Compute \(\vec{s}_a\cdot\vec{w} =\frac{1}{2}(-7)+2\cdot4+\frac{3}{2}(-3)=0\). Thus, the median is perpendicular to \(BC\). 4. Also, \(\|\vec{u}\|=5\) and \(\|\vec{v}\|=5\). 5. Therefore, \(AB=AC\), so the triangle is isosceles. The median to its base is also an altitude.

a) \(\vec{w}=\begin{pmatrix}-7\\4\\-3\end{pmatrix}\) and \(\vec{s}_a=\begin{pmatrix}\frac{1}{2}\\2\\\frac{3}{2}\end{pmatrix}\) b) \(\vec{s}_a\cdot\vec{w}=0\), so the median is perpendicular to \(BC\). c) Since \(AB=AC=5\), the triangle is isosceles, and its median to the base is also an altitude.

- Write one dot-product equation for each given vector. - Two equations in three unknowns leave a free parameter. - Interpret a scalar multiple of one fixed vector geometrically.

1. The two perpendicularity conditions are \(3x_1+x_2+2x_3=0\) and \(x_1-x_2=0\). 2. From the second equation, \(x_1=x_2\). 3. Substitution into the first equation gives \(4x_1+2x_3=0\), so \(x_3=-2x_1\). 4. Let \(t=x_1\). Then \(\vec{x}=t\begin{pmatrix}1\\1\\-2\end{pmatrix},\qquad t\in\mathbb{R}\). 5. These vectors form a line through the origin. Every vector on this line has zero dot product with both given vectors, so the solution is not unique.

a) \(\vec{x}=t\begin{pmatrix}1\\1\\-2\end{pmatrix}\), where \(t\in\mathbb{R}\) b) The solution set is a line through the origin. It is not unique because every scalar multiple of the direction vector satisfies both dot-product equations.

- Use a zero dot product for each perpendicularity condition. - Substitute the known middle component before solving. - Solve the resulting two linear equations.

1. Let \(\vec{c}=\begin{pmatrix}x\\15\\z\end{pmatrix}\). 2. From \(\vec{a}\cdot\vec{c}=0\), \(3x-15+2z=0\), so \(3x+2z=15\). 3. From \(\vec{b}\cdot\vec{c}=0\), \(x+30-z=0\), so \(x-z=-30\). 4. Substitute \(x=z-30\) into the first equation: \(3(z-30)+2z=15\), so \(5z=105\) and \(z=21\). 5. Then \(x=-9\), giving \(\vec{c}=\begin{pmatrix}-9\\15\\21\end{pmatrix}\).

\(\vec{c}=\begin{pmatrix}-9\\15\\21\end{pmatrix}\)

- A normal vector is perpendicular to each spanning vector. - Write two dot-product equations. - Solve the resulting linear system.

1. A vector perpendicular to the plane must be perpendicular to both spanning vectors. 2. Therefore, \(\vec{w}\cdot\vec{u}=w_1+w_2+6=0\) and \(\vec{w}\cdot\vec{v}=2w_1-w_2+18=0\). 3. Add the equations: \(3w_1+24=0\), so \(w_1=-8\). 4. Substitute into the first equation: \(-8+w_2+6=0\), so \(w_2=2\). 5. Thus, \(\vec{w}=\begin{pmatrix}-8\\2\\6\end{pmatrix}\).

\(w_1=-8\) and \(w_2=2\)

- Write a general point on the line. - The vector from \(A\) to the perpendicular foot is orthogonal to the direction vector. - Use a dot product to find the parameter. - The distance is the magnitude of the resulting perpendicular vector.

1. A general point on \(h\) is \(L(t)=(1+t,1+2t,1+2t)\), so \(\overrightarrow{AL}=\langle t,2t-5,2t-4\rangle\). 2. Require \(\overrightarrow{AL}\cdot\langle 1,2,2\rangle=0\): \(t+2(2t-5)+2(2t-4)=9t-18=0\), so \(t=2\). 3. Therefore, \(L=(3,5,5)\). 4. The distance is \(AL=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5}\approx 2.24\).

The perpendicular foot is \(L(3,5,5)\), and the distance is \(\sqrt{5}\approx 2.24\) units.

- The shared position point gives the intersection immediately. - Use the dot-product angle formula for part a. - An isosceles triangle with base \(PQ\) has \(SP=SQ\). - For the reflection, first find the perpendicular projection of \(P\) onto \(h\).

1. The lines share the position point \(S=(2, -1, 5)\), so this is their intersection. 2. Their direction vectors have dot product \(8\) and both have magnitude \(3\). Thus \(\cos\alpha=\frac{8}{9}\), so \(\alpha\approx27.3^\circ\). 3. For an isosceles triangle with base \(PQ\), \(SP=SQ\). Since \(\overrightarrow{SP}=\begin{pmatrix}9\\18\\18\end{pmatrix}\), \(SP=27\). A point on \(h\) has \(SQ=3|s|\), so \(s=9\) or \(s=-9\). 4. These values give \(Q_1=(20, 17, 14)\) and \(Q_2=(-16, -19, -4)\). 5. The perpendicular projection of \(P\) onto \(h\) occurs at \(s=\frac{\overrightarrow{SP}\cdot\vec{v}_h}{\|\vec{v}_h\|^2}=8\), giving foot \(L=(18, 15, 13)\). Therefore, \(P'=2L-P=(25, 13, 3)\).

a) \(S=(2, -1, 5)\), and \(\alpha\approx27.3^\circ\) b) \(Q_1=(20, 17, 14)\) and \(Q_2=(-16, -19, -4)\) c) \(P'=(25, 13, 3)\)

- Form two direction vectors in the plane. - Find a vector orthogonal to both directions using dot-product equations. - Write the plane equation and substitute \(S(k, k, 2)\).

1. Two directions in the plane are \(\overrightarrow{PQ}=\langle -1,2,2\rangle\) and \(\overrightarrow{PR}=\langle -2,-1,4\rangle\). 2. Let \(\vec{n}=\langle a,b,c\rangle\) be a normal vector. The dot-product conditions are \(-a+2b+2c=0\) and \(-2a-b+4c=0\). 3. Solving gives one normal vector \(\vec{n}=\langle 2,0,1\rangle\). 4. Using point \(P\), the plane equation is \(2x+z=4\). 5. Substitute \(S(k, k, 2)\): \(2k+2=4\), so \(k=1\).

\(k=1\)

- Find a vector orthogonal to both directions in \(E\). - Test whether the directions in \(F\) are also orthogonal to that vector. - Use a point from \(F\) to distinguish identical planes from distinct parallel planes.

1. Let \(\vec{n}=\langle a,b,c\rangle\) be normal to \(E\). The equations \(a+c=0\) and \(2a+b=0\) give one normal vector \(\vec{n}=\langle -1,2,1\rangle\). 2. Both direction vectors of \(F\) are orthogonal to this normal: \(\langle 3,1,1\rangle\cdot\vec{n}=-3+2+1=0\) and \(\langle 1,1,-1\rangle\cdot\vec{n}=-1+2-1=0\). 3. Therefore \(E\) and \(F\) are parallel. 4. A standard equation of \(E\) is \(-x+2y+z=d\). Using \((3, 1, 2)\), \(d=-3+2+2=1\). 5. The initial point of \(F\) does not lie in \(E\), because \(-0+2(-1)+1=-1\neq1\). 6. Thus the planes are distinct and strictly parallel.

Planes \(E\) and \(F\) are strictly parallel.

- Write an unknown normal vector and set its dot product with each direction vector equal to \(0\). - Parallel planes have parallel normal vectors. - Use the point on \(H\) to determine the constant.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Since \(\vec{n}\) is orthogonal to both direction vectors, \(2a+c=0\) and \(a-b+3c=0\). 2. Choose \(a=1\). Then \(c=-2\), and the second equation gives \(1-b-6=0\), so \(b=-5\). Thus \(\vec{n}=\langle 1,-5,-2\rangle\). 3. A plane parallel to \(E\) has a parallel normal vector, so \(H\) has the form \(x-5y-2z=d\). 4. Substitute \(Q(3, 0, 2)\): \(d=3-5(0)-2(2)=-1\). 5. Therefore \(H: x-5y-2z=-1\).

\(H: x-5y-2z=-1\)

- Read a normal vector from the plane equation. - Use the dot product to decide whether each line is parallel to the plane. - If a line is parallel, test its initial point in the plane equation. - If a line is not parallel, substitute its coordinates into the plane equation.

1. A normal vector to \(E\) is \(\vec{n}=\langle 1,2,-1\rangle\). 2. a) The direction of \(g\) satisfies \(\langle 2,-1,0\rangle\cdot\vec{n}=2-2=0\), so \(g\) is parallel to the plane. 3. Its initial point lies in \(E\), because \(1+2(1)-(-1)=4\). Therefore \(g\) lies entirely in \(E\). 4. b) The direction of \(h\) satisfies \(\langle 1,1,1\rangle\cdot\vec{n}=2\neq0\), so \(h\) intersects the plane once. 5. Substitute \((3+s, 1+s, 2+s)\) into the plane equation: \((3+s)+2(1+s)-(2+s)=4\). 6. Simplifying gives \(3+2s=4\), so \(s=\frac12\). The intersection point is \(S\left(\frac72, \frac32, \frac52\right)\).

a) Line \(g\) lies entirely in plane \(E\). b) \(S\left(\frac72, \frac32, \frac52\right)\)

- Find a vector orthogonal to both direction vectors of the plane. - Compare the line direction with that normal vector using a dot product. - Convert the plane to a point-normal equation. - Test the line’s initial point to distinguish containment from strict parallelism.

1. Let \(\vec{n}=\langle a,b,c\rangle\) be normal to \(E\). Orthogonality to the plane directions gives \(2a+b=0\) and \(a+c=0\). One solution is \(\vec{n}=\langle 1,-2,-1\rangle\). 2. The line direction satisfies \(\langle 3,1,1\rangle\cdot\vec{n}=3-2-1=0\), so \(g\) is parallel to \(E\). 3. The plane has equation \(x-2y-z=d\). Using \((1, 2, 1)\), \(d=1-4-1=-4\). 4. The initial point of \(g\) does not lie in the plane, because \(2-2(2)-3=-5\neq-4\). 5. Therefore \(g\) is strictly parallel to \(E\).

Line \(g\) is strictly parallel to plane \(E\).

- Find a vector orthogonal to both direction vectors of the parametric plane. - Parallel planes have parallel normal vectors. - Test one point to distinguish identical planes from strictly parallel planes.

1. Let \(\vec{n}_2=\langle p,q,r\rangle\) be a normal vector to \(E_2\). It must satisfy \(\vec{n}_2\cdot\langle 2,1,0\rangle=0\) and \(\vec{n}_2\cdot\langle 3,0,2\rangle=0\). 2. Solving \(2p+q=0\) and \(3p+2r=0\) gives one normal vector \(\vec{n}_2=\langle 2,-4,-3\rangle\). 3. A normal vector to \(E_1\) is \(\vec{n}_1=\langle 2,-4,a\rangle\). For the planes to be parallel, the normals must be scalar multiples. 4. The first two components already agree, so the scalar is \(1\). Therefore, \(a=-3\). 5. Test the initial point \((1, 0, 1)\) of \(E_2\) in \(E_1\): \(2(1)-4(0)-3(1)=-1\neq8\). 6. Thus the planes are strictly parallel.

\(a=-3\); the planes are strictly parallel.

- Read the normal vector from the coefficients and compare it with a direction vector of the \(y\)-axis. - Substitute each axis-intercept point into the general equation. - Choose a convenient common value for \(d\). - A plane that contains the \(y\)-axis must contain the origin.

1. A normal vector is \(\vec{n}=\langle a,0,c\rangle\). Since \(\vec{n}\cdot\langle 0,1,0\rangle=0\), every plane is parallel to the \(y\)-axis. Because \(d\neq0\), none of these planes contains the origin, so the \(y\)-axis is not contained in them. 2. The intercepts give \(4a=d\) and \(6c=d\). Choose \(d=12\); then \(a=3\) and \(c=2\). One equation is \(3x+2z=12\). 3. Keeping \(a=3\) and \(c=2\), choose \(d=0\). Then every point \((0, y, 0)\) satisfies \(3(0)+2(0)=0\), so the plane contains the \(y\)-axis.

1) Every plane is parallel to the \(y\)-axis. 2) One choice is \(a=3\), \(c=2\), and \(d=12\), giving \(3x+2z=12\). 3) \(d=0\)

- Read the normal vector from the coefficients. - A direction vector in the plane is orthogonal to the normal vector. - Choose simple component values that make the dot product zero. - Find one point satisfying the plane equation.

1. A normal vector is \(\vec{n}=\langle 2,5,-3\rangle\). Every direction vector in the plane must satisfy \(\vec{n}\cdot\vec{w}=0\). 2. Choose \(\vec{u}=\langle 5,-2,0\rangle\), since \(2(5)+5(-2)-3(0)=0\). 3. Choose \(\vec{v}=\langle 3,0,2\rangle\), since \(2(3)+5(0)-3(2)=0\). The vectors are linearly independent because neither is a scalar multiple of the other. 4. Setting \(y=z=0\) gives \(x=5\), so \(A(5, 0, 0)\) is on the plane. 5. One vector equation is \(\vec{r}(s,t)=\langle 5,0,0\rangle+s\langle 5,-2,0\rangle+t\langle 3,0,2\rangle\).

a) \(\vec{n}=\langle 2,5,-3\rangle\); every direction vector satisfies \(\vec{n}\cdot\vec{w}=0\). b) One choice is \(\vec{u}=\langle 5,-2,0\rangle\) and \(\vec{v}=\langle 3,0,2\rangle\). c) \(\vec{r}(s,t)=\langle 5,0,0\rangle+s\langle 5,-2,0\rangle+t\langle 3,0,2\rangle\)

- A vector parallel to a plane is orthogonal to the plane’s normal vector. - Write an unknown vector and impose two dot-product equations. - A two-parameter plane equation needs one point and two linearly independent direction vectors.

1. \(\vec{n}\cdot\vec{u}=\langle 1,0,2\rangle\cdot\langle 0,1,0\rangle=0\), so \(\vec{u}\) is parallel to the plane. 2. Let \(\vec{v}=\langle a,b,c\rangle\). Orthogonality to \(\vec{u}\) gives \(b=0\), and orthogonality to \(\vec{n}\) gives \(a+2c=0\). Choose \(c=1\), so \(a=-2\). Thus \(\vec{v}=\langle -2,0,1\rangle\). 3. The plane is \(\vec{r}(s,t)=\langle 3,-2,4\rangle+s\langle 0,1,0\rangle+t\langle -2,0,1\rangle\).

a) \(\vec{n}\cdot\vec{u}=0\) b) \(\vec{v}=\langle -2,0,1\rangle\), or any nonzero scalar multiple c) \(\vec{r}(s,t)=\langle 3,-2,4\rangle+s\langle 0,1,0\rangle+t\langle -2,0,1\rangle\)

- A vector joining the endpoints is normal to their perpendicular-bisector plane. - Use the three-dimensional midpoint formula. - Combine the midpoint and normal vector in a point-normal equation.

1. Compute \(\overrightarrow{AB}=\langle 5-1,0-2,-1-3\rangle=\langle 4,-2,-4\rangle\). Use the parallel normal vector \(\vec{n}=\langle 2,-1,-2\rangle\). 2. The midpoint is \(M\left(\frac{1+5}{2},\frac{2+0}{2},\frac{3+(-1)}{2}\right)=(3, 1, 1)\). 3. Point-normal form is \(\langle 2,-1,-2\rangle\cdot\langle x-3,y-1,z-1\rangle=0\). 4. Expanding gives \(2x-y-2z=3\).

Point-normal form: \(\langle 2,-1,-2\rangle\cdot\langle x-3,y-1,z-1\rangle=0\) Standard form: \(2x-y-2z=3\)

- The coefficients of \(x\), \(y\), and \(z\) form a normal vector. - Scalar multiplication changes a vector’s length and possibly its orientation, but not the line it spans. - Set two coordinates equal to \(0\) to find a convenient point. - Consider multiplying the entire equation by \(2\).

1. A normal vector is \(\vec{n}=\langle 2,-1,2\rangle\), so \(\langle 2,-1,2\rangle\cdot\vec{x}=12\). 2. Every nonzero scalar multiple of \(\vec{n}\) is also normal to the plane. The set is \(\{k\langle 2,-1,2\rangle:k\in\mathbb{R}\setminus\{0\}\}\). 3. Setting \(y=z=0\) gives \(2x=12\), so \(P(6, 0, 0)\) lies on the plane. One point-normal equation is \(\langle 2,-1,2\rangle\cdot\langle x-6,y,z\rangle=0\). 4. The representation is not unique. The normal vector and the constant may be multiplied by the same nonzero scalar, and any point on the plane may be used in point-normal form.

a) \(\langle 2,-1,2\rangle\cdot\vec{x}=12\) b) \(\{k\langle 2,-1,2\rangle:k\in\mathbb{R}\setminus\{0\}\}\) c) For example, \(P(6, 0, 0)\) and \(\langle 2,-1,2\rangle\cdot\langle x-6,y,z\rangle=0\) d) No. Nonzero scalar multiples of the equation and different points on the plane give equivalent representations.

- Which two vectors can you form from the three points in the plane? - How can you find a vector perpendicular to both plane direction vectors? - What position vector represents the origin?

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\4\\4\end{pmatrix}\). 2. Let a normal vector be \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\). Orthogonality requires \(\vec{n}\cdot\overrightarrow{AB}=2a+2b=0\) and \(\vec{n}\cdot\overrightarrow{AC}=4b+4c=0\). Thus \(a=-b\) and \(c=-b\). Choosing \(b=-1\) gives \(\vec{n}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\). 3. A line perpendicular to \(E\) can use this normal vector as its direction vector. Since the line passes through the origin, \(h:\vec{x}=t\begin{pmatrix}1\\-1\\1\end{pmatrix}\).

\(h:\vec{x}=t\begin{pmatrix}1\\-1\\1\end{pmatrix}\), where \(t\in\mathbb{R}\)

- Use the given point in the dot-product equation. - Test a point by comparing its dot product with \(d\). - Write \(\overrightarrow{PQ}\) as a difference of position vectors. - Apply the distributive property of the dot product.

1. Compute \(d=\langle 2,-1,4\rangle\cdot\langle 3,2,-2\rangle=2(3)+(-1)(2)+4(-2)=-4\). 2. For \(B\), \(\langle 4,-3,1\rangle\cdot\langle 3,2,-2\rangle=12-6-2=4\). Since \(4\neq-4\), \(B\notin E\). 3. Because \(P,Q\in E\), \(\overrightarrow{OP}\cdot\vec{n}=d\) and \(\overrightarrow{OQ}\cdot\vec{n}=d\). Since \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\), distributivity gives \(\overrightarrow{PQ}\cdot\vec{n}=\overrightarrow{OQ}\cdot\vec{n}-\overrightarrow{OP}\cdot\vec{n}=d-d=0\). Therefore every vector joining two points in the plane is orthogonal to the normal vector.

1) \(d=-4\) 2) No, \(B\notin E\). 3) \(\overrightarrow{PQ}\cdot\vec{n}=d-d=0\), so \(\overrightarrow{PQ}\perp\vec{n}\).

- Apply the distributive and scalar properties of the dot product. - Substitute the two given zero dot products. - What dimension is spanned by two collinear vectors?

1. For any \(\vec{x}=r\vec{u}+s\vec{v}\), distributivity of the dot product gives \(\vec{n}\cdot\vec{x}=\vec{n}\cdot(r\vec{u}+s\vec{v})=r(\vec{n}\cdot\vec{u})+s(\vec{n}\cdot\vec{v})\). 2. Since \(\vec{n}\cdot\vec{u}=0\) and \(\vec{n}\cdot\vec{v}=0\), it follows that \(\vec{n}\cdot\vec{x}=0\) for every \(r,s\in\mathbb{R}\). Thus \(\vec{n}\) is perpendicular to every direction in the plane. 3. If \(\vec{u}\) and \(\vec{v}\) were collinear, they would span only one line, not a two-dimensional plane direction space. Being perpendicular to that one direction would not specify perpendicularity to a particular plane.

1. \(\vec{n}\cdot(r\vec{u}+s\vec{v})=r(\vec{n}\cdot\vec{u})+s(\vec{n}\cdot\vec{v})=0\), so \(\vec{n}\) is perpendicular to every vector in the direction space. 2. Noncollinear vectors span a plane; collinear vectors span only a line.

- Expand the dot product component by component. - Test each point by substituting its coordinates. - For part c, set \(z=0\) and choose nonzero values for both \(x\) and \(y\). - Many answers are possible for part c.

1. Expanding the dot product gives \(x+3y-2z=6\). 2. For \(R\), \(1+3(1)-2(-1)=6\), so \(R\in E\). For \(S\), \(2+3(2)-2(2)=4\neq6\), so \(S\notin E\). 3. Set \(z=0\). Then \(x+3y=6\). Choose \(y=1\), which gives \(x=3\). Thus \(T(3, 1, 0)\) is in the plane and is not on a coordinate axis.

a) \(x+3y-2z=6\) b) \(R\in E\), but \(S\notin E\). c) One possible point is \(T(3, 1, 0)\).

- Treat two of the three coordinates as free parameters. - Separate the constant vector from the coefficient vectors of the parameters. - The coefficients in the standard equation form a normal vector. - Use dot products to verify orthogonality.

1. Solve for \(x\): \(-x+3y+2z=6\) gives \(x=3y+2z-6\). 2. Let \(y=s\) and \(z=t\). Then \(x=-6+3s+2t\), so \(\vec{r}(s,t)=\langle -6,0,0\rangle+s\langle 3,1,0\rangle+t\langle 2,0,1\rangle\). 3. A normal vector is \(\vec{n}=\langle -1,3,2\rangle\). 4. Check the first direction vector: \(\langle -1,3,2\rangle\cdot\langle 3,1,0\rangle=-3+3+0=0\). 5. Check the second direction vector: \(\langle -1,3,2\rangle\cdot\langle 2,0,1\rangle=-2+0+2=0\). Both direction vectors are orthogonal to the normal vector.

a) \(x=3y+2z-6\) b) \(\vec{r}(s,t)=\langle -6,0,0\rangle+s\langle 3,1,0\rangle+t\langle 2,0,1\rangle\) c) \(\vec{n}=\langle -1,3,2\rangle\), and its dot product with each direction vector is \(0\).

- Write an unknown normal vector and require it to be orthogonal to both direction vectors. - Use the plane’s initial point to determine the constant. - Test each point by substitution into the standard equation. - A point lies in the plane only when both sides are equal.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality to the direction vectors gives \(a+2b=0\) and \(b+2c=0\). 2. Choose \(c=1\). Then \(b=-2\) and \(a=4\), so \(\vec{n}=\langle 4,-2,1\rangle\). 3. The plane has the form \(4x-2y+z=d\). Substitute \((2, 1, -1)\): \(d=4(2)-2(1)-1=5\). Thus \(E: 4x-2y+z=5\). 4. For \(P\), \(4(3)-2(4)+1=5\), so \(P\in E\). 5. For \(Q\), \(4(1)-2(1)+1=3\neq5\), so \(Q\notin E\).

a) \(E: 4x-2y+z=5\) b) \(P\in E\), but \(Q\notin E\).

- Which coordinate is fixed for every point on the plane? - Read a normal vector from the equation \(y=5\). - Use a dot product to characterize vectors parallel to the plane. - Choose two linearly independent coordinate-direction vectors.

1. All points in \(F\) have a constant \(y\)-coordinate of \(5\). Therefore \(F\) is parallel to the \(xz\)-plane and is \(5\) units from it. 2. A normal vector to \(F\) is \(\vec{n}=\langle 0,1,0\rangle\). For any \(\vec{v}=\langle v_x,0,v_z\rangle\), \(\vec{n}\cdot\vec{v}=0\), so \(\vec{v}\) is parallel to the plane. The vector must be nonzero to serve as a direction vector. 3. Use the point \((0, 5, 0)\) and the independent direction vectors \(\langle 1,0,0\rangle\) and \(\langle 0,0,1\rangle\): \(\vec{r}(s,t)=\langle 0,5,0\rangle+s\langle 1,0,0\rangle+t\langle 0,0,1\rangle\).

a) \(F\) is parallel to the \(xz\)-plane and is \(5\) units from it. b) \(\langle 0,1,0\rangle\cdot\langle v_x,0,v_z\rangle=0\), so every nonzero vector of that form is parallel to \(F\). c) \(\vec{r}(s,t)=\langle 0,5,0\rangle+s\langle 1,0,0\rangle+t\langle 0,0,1\rangle\)

- Form two connecting vectors from one point. - Find a normal vector by requiring zero dot products with both connecting vectors. - The normal components become the coefficients in Cartesian form. - Substitute a known point to find the constant.

1. Two spanning vectors are \(\overrightarrow{PQ}=\begin{pmatrix}3\\-2\\2\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}0\\-3\\4\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(3a-2b+2c=0\) and \(-3b+4c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\12\\9\end{pmatrix}\). 3. Using point \(P\), \(2\cdot2+12\cdot3+9\cdot0=40\). Therefore, \(E:2x+12y+9z=40\).

\(E:2x+12y+9z=40\)

- Let the normal vector have unknown components and make it orthogonal to both direction vectors. - Use the initial point to find the constant. - For each axis intercept, set the other two coordinates equal to \(0\). - Keep the intercepts as exact fractions.

1. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality to the direction vectors gives \(a-2b+2c=0\) and \(2a+b=0\). 2. From \(b=-2a\), the first equation becomes \(a+4a+2c=0\), so \(2c=-5a\). Choose \(a=-2\), giving \(b=4\) and \(c=5\). Thus \(\vec{n}=\langle -2,4,5\rangle\). 3. The plane has the form \(-2x+4y+5z=d\). Substitute \((2, 2, 1)\): \(d=-4+8+5=9\). Therefore \(E: -2x+4y+5z=9\). 4. For the \(x\)-intercept, set \(y=z=0\): \(-2x=9\), so \(\left(-\frac{9}{2}, 0, 0\right)\). 5. For the \(y\)-intercept, set \(x=z=0\): \(4y=9\), so \(\left(0, \frac{9}{4}, 0\right)\). 6. For the \(z\)-intercept, set \(x=y=0\): \(5z=9\), so \(\left(0, 0, \frac{9}{5}\right)\).

a) \(E: -2x+4y+5z=9\) b) \(\left(-\frac{9}{2}, 0, 0\right)\), \(\left(0, \frac{9}{4}, 0\right)\), and \(\left(0, 0, \frac{9}{5}\right)\)

- Read the plane’s normal vector from its equation. - Use a dot product to test each line direction. - For a parallel line, test its initial point in the plane equation. - For a nonparallel line, solve for the intersection parameter.

1. A normal vector to \(E\) is \(\vec{n}=\langle 2,-4,3\rangle\). 2. For \(g\), \(\vec{n}\cdot\langle 2,1,0\rangle=4-4=0\), so the line is parallel to the plane. 3. Its initial point lies in \(E\), because \(2(1)-4(2)+3(6)=12\). Therefore \(g\) lies entirely in \(E\). 4. For \(h\), \(\vec{n}\cdot\langle 1,1,1\rangle=2-4+3=1\neq0\), so the line intersects the plane once. 5. Substitute the line into the plane equation: \(2(5+s)-4(1+s)+3(2+s)=12\). 6. This simplifies to \(12+s=12\), so \(s=0\). The intersection point is \(S(5, 1, 2)\).

Line \(g\) lies entirely in plane \(E\). Line \(h\) intersects \(E\) at \(S(5, 1, 2)\).

- Compare the common line direction with the plane normal using a dot product. - Test the initial point to find when a line lies in the plane. - Parameterize the line through the origin and substitute it into the plane equation.

1. a) A normal vector to \(E\) is \(\vec{n}=\langle 1,1,2\rangle\), and the common line direction is \(\vec{u}=\langle 2,-2,0\rangle\). 2. Since \(\vec{n}\cdot\vec{u}=2-2+0=0\), every line direction is parallel to the plane. Therefore, each line either lies in \(E\) or is strictly parallel to it. 3. b) The initial point \((a, 2, 1)\) lies in \(E\) when \(a+2+2(1)=6\). Thus \(a=2\). 4. c) Write \(h\) as \(\vec{r}(s)=s\langle 1,1,1\rangle\). Substitution into \(E\) gives \(s+s+2s=6\). 5. Hence \(s=\frac32\), and the intersection point is \(S\left(\frac32, \frac32, \frac32\right)\).

a) Every \(g_a\) is parallel to \(E\) or lies in \(E\). b) \(a=2\) c) \(S\left(\frac32, \frac32, \frac32\right)\)

- Three points determine a unique plane when they are not collinear. - Find a normal vector by requiring zero dot products with two connecting vectors. - At an axis intercept, the other two coordinates are zero.

1. The vectors \(\overrightarrow{PQ}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and \(\overrightarrow{PR}=\begin{pmatrix}-1\\2\\-2\end{pmatrix}\) are not scalar multiples, so the points are not collinear. 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both connecting vectors. Then \(2a-b+2c=0\) and \(-a+2b-2c=0\). One solution is \(\vec{n}=\begin{pmatrix}-2\\2\\3\end{pmatrix}\). Using \(P\), the plane equation is \(E:-2x+2y+3z=11\). 3. On the \(x\)-axis, \(y=z=0\), giving \(x=-\frac{11}{2}\). On the \(y\)-axis, \(x=z=0\), giving \(y=\frac{11}{2}\). On the \(z\)-axis, \(x=y=0\), giving \(z=\frac{11}{3}\).

a) The two connecting vectors are not scalar multiples. b) \(E:-2x+2y+3z=11\) c) \(\left(-\frac{11}{2}, 0, 0\right)\), \(\left(0, \frac{11}{2}, 0\right)\), and \(\left(0, 0, \frac{11}{3}\right)\)

- Use a dot product to determine whether the line is parallel to the plane. - Write each coordinate of a general point on the line in terms of \(t\). - Substitute those coordinates into the plane equation. - Substitute the solved parameter back into the line.

1. A normal vector to \(E\) is \(\vec{n}=\langle 2,2,-1\rangle\). 2. The line direction satisfies \(\langle 1,-1,3\rangle\cdot\vec{n}=2-2-3=-3\neq0\), so the line intersects the plane exactly once. 3. Substitute \((t, 4-t, -2+3t)\) into the plane equation: \(2t+2(4-t)-(-2+3t)=5\). 4. Simplify: \(10-3t=5\), so \(t=\frac53\). 5. The intersection point is \(S\left(\frac53, 4-\frac53, -2+3\left(\frac53\right)\right)=S\left(\frac53, \frac73, 3\right)\).

The line intersects the plane at \(S\left(\frac53, \frac73, 3\right)\).

- Use a dot product between the line direction and the plane normal. - Test the line's initial point in the plane equation. - Substitute the coordinates of \(g_0\) into the plane equation.

1. A normal vector to \(E\) is \(\vec{n}=\langle 2,1,-1\rangle\), and the line direction is \(\vec{v}=\langle a,2,4\rangle\). 2. a) For the line to be parallel to the plane, require \(\vec{v}\cdot\vec{n}=0\): \(2a+2-4=0\). 3. Thus \(a=1\). 4. b) The initial point \((1, 1, 1)\) is not in \(E\), because \(2(1)+1-1=2\neq5\). Therefore, \(g_1\) is strictly parallel to \(E\). 5. c) A point on \(g_0\) is \((1, 1+2t, 1+4t)\). Substitute into \(E\): \(2+(1+2t)-(1+4t)=5\). 6. This gives \(2-2t=5\), so \(t=-\frac32\). 7. The intersection point is \(S(1, -2, -5)\).

a) \(a=1\) b) \(g_1\) is strictly parallel to \(E\). c) \(S(1, -2, -5)\)

- Find a normal vector orthogonal to both plane direction vectors. - Use the dot product of the line direction and the normal vector. - Write the plane in point-normal form. - Substitute the line into the resulting plane equation.

1. Let \(\vec{n}=\langle a,b,c\rangle\) be normal to \(F\). Orthogonality to the two plane directions gives \(2a+b=0\) and \(b-c=0\). One solution is \(\vec{n}=\langle -1,2,2\rangle\). 2. The line direction satisfies \(\langle 1,2,-1\rangle\cdot\vec{n}=-1+4-2=1\neq0\). Therefore the line intersects the plane exactly once. 3. A standard equation of the plane is \(-x+2y+2z=d\). Using \((1, -1, 2)\), \(d=-1-2+4=1\). 4. Substitute the line into the plane equation: \(-(4+\lambda)+2(1+2\lambda)+2(3-\lambda)=1\). 5. Simplifying gives \(4+\lambda=1\), so \(\lambda=-3\). 6. The intersection point is \(S=\langle 4,1,3\rangle-3\langle 1,2,-1\rangle=\langle 1,-5,6\rangle\).

The line intersects the plane at \(S(1, -5, 6)\).

- Form two direction vectors in the roof plane. - Find a vector orthogonal to both using dot products. - Parametrize the beam from the two points. - Substitute the beam into the plane equation.

1. Two directions in the roof plane are \(\overrightarrow{AB}=\langle -8,8,0\rangle\) and \(\overrightarrow{AC}=\langle -8,0,8\rangle\). 2. Let \(\vec{n}=\langle a,b,c\rangle\). The equations \(-8a+8b=0\) and \(-8a+8c=0\) give \(a=b=c\). Choose \(\vec{n}=\langle 1,1,1\rangle\). 3. Using point \(C\), the plane equation is \(x+y+z=12\). 4. The beam direction is \(\overrightarrow{PQ}=\langle 1,1,3\rangle\), so \(\vec{r}(t)=\langle 1,1,0\rangle+t\langle 1,1,3\rangle\). 5. Substitute into the plane equation: \((1+t)+(1+t)+3t=12\). 6. Thus \(2+5t=12\), so \(t=2\). The intersection point is \(D(3, 3, 6)\).

\(D(3, 3, 6)\)

- Form two direction vectors from the three plane points. - Find a vector orthogonal to both plane directions. - Use a dot product to test the line direction. - Compare the line’s initial point with the points defining the plane.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle 2,-1,1\rangle\) and \(\overrightarrow{AC}=\langle 1,-4,2\rangle\). 2. Let \(\vec{n}=\langle a,b,c\rangle\) be normal to the plane. The equations \(2a-b+c=0\) and \(a-4b+2c=0\) have the solution \(\vec{n}=\langle 2,-3,-7\rangle\). 3. The line direction satisfies \(\langle 3,-5,3\rangle\cdot\vec{n}=6+15-21=0\), so \(g\) is parallel to \(E\). 4. The initial point of \(g\) is \(B\), one of the points defining the plane. Thus the line has one point in the plane. 5. A line parallel to a plane and passing through a point of the plane lies entirely in it. Therefore \(g\) and \(E\) have infinitely many common points.

Line \(g\) lies entirely in plane \(E\), so they have infinitely many common points.

- Use the point-to-plane distance formula. - A point on the z-axis has x- and y-coordinates equal to \(0\). - The absolute-value equation has two solutions. - Use the condition that \(Q\) is not the origin.

1. A normal vector is \(\langle 6,3,-2\rangle\), with magnitude \(7\). Therefore, \(d(O,F)=\frac{\lvert -28\rvert}{7}=4\). 2. A point on the z-axis has the form \(Q(0,0,t)\). Set its distance from the plane equal to \(4\): \(\frac{\lvert -2t-28\rvert}{7}=4\). 3. Thus, \(\lvert -2t-28\rvert=28\), so \(-2t-28=28\) or \(-2t-28=-28\). 4. The solutions are \(t=-28\) and \(t=0\). Because \(Q\) is not the origin, \(t=-28\). 5. Therefore, \(Q=(0,0,-28)\).

1) \(4\) units 2) \(Q=(0,0,-28)\)

- Divide the plane equation by the magnitude of its normal vector. - Use the normalized equation to compute distance. - A point on the x-axis has y- and z-coordinates equal to \(0\). - Solve both cases of the absolute-value equation.

1. A normal vector is \(\langle 2,-2,1\rangle\), with magnitude \(3\). Therefore, a normalized equation is \(\frac{2x-2y+z-9}{3}=0\). 2. The distance from \(P\) is \(\frac{\lvert 2(1)-2(1)+15-9\rvert}{3}=2\). 3. A point on the x-axis has the form \((t,0,0)\). Set \(\frac{\lvert 2t-9\rvert}{3}=5\). 4. Then \(\lvert 2t-9\rvert=15\), so \(t=12\) or \(t=-3\).

a) \(\frac{2x-2y+z-9}{3}=0\) b) \(2\) units c) \((12,0,0)\) and \((-3,0,0)\)

- Find the magnitude of the plane's normal vector. - Use the normalized constant to find the distance from the origin. - Parallel planes have the same normal vector. - Their distance depends on the difference between their constants.

1. A normal vector is \(\langle 4,8,-8\rangle\), with magnitude \(12\). Thus, a normalized equation is \(\frac{4x+8y-8z-15}{12}=0\). 2. The distance from the origin is \(\frac{15}{12}=\frac{5}{4}=1.25\). 3. The distance between the parallel planes is \(\frac{\lvert k-15\rvert}{12}\). 4. Set \(\frac{\lvert k-15\rvert}{12}=3\). Then \(\lvert k-15\rvert=36\), so \(k=51\) or \(k=-21\).

a) \(\frac{4x+8y-8z-15}{12}=0\) b) \(\frac{5}{4}=1.25\) units c) \(k=51\) or \(k=-21\)

- Find a vector orthogonal to both direction vectors of the plane. - Form the vector from a point in the plane to \(P\). - Project that vector onto the plane's normal direction.

1. Let \(\mathbf{u}=\langle 1,2,0\rangle\) and \(\mathbf{v}=\langle 0,2,1\rangle\). A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) must satisfy \(a+2b=0\) and \(2b+c=0\). Choose \(\mathbf{n}=\langle 2,-1,2\rangle\). 2. The normal vector has magnitude \(3\). 3. From the plane point \(A(3,0,0)\) to \(P\), \(\overrightarrow{AP}=\langle 3,1,5\rangle\). 4. The distance is the magnitude of the scalar projection onto the normal direction: \(d=\frac{\lvert\overrightarrow{AP}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{\lvert 3(2)+1(-1)+5(2)\rvert}{3}=5\).

\(5\) units

- Each outer plane is half the total distance from the midplane. - Parallel planes have the same normal vector. - Use the difference between the constants to express the distance.

1. A normal vector to all three planes is \(\langle 8,-4,8\rangle\), with magnitude \(12\). 2. Because \(E\) is halfway between \(H_1\) and \(H_2\), each plane is \(5\) units from \(E\). 3. Write a parallel plane as \(8x-4y+8z=k\). Then \(\frac{\lvert k-27\rvert}{12}=5\). 4. Thus, \(\lvert k-27\rvert=60\), so \(k=87\) or \(k=-33\).

\(H_1: 8x-4y+8z=87\) and \(H_2: 8x-4y+8z=-33\)

- Choose two faces that share an edge. - Find a normal vector for each face. - The acute angle between faces equals the acute angle between their normals. - Use the dot-product angle formula.

1. Choose adjacent faces through \(P_1,P_3,P_5\) and \(P_2,P_3,P_5\). 2. Their plane equations are \(x+y+z=1\) and \(-x+y+z=1\), with normal vectors \(\vec{n}_1=\begin{pmatrix}1\\1\\1\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}-1\\1\\1\end{pmatrix}\). 3. Thus \(\cos\alpha=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{1}{3}\). 4. Therefore, \(\alpha=\cos^{-1}\left(\frac{1}{3}\right)\approx70.5^\circ\).

\(\alpha\approx70.5^\circ\)

- Read a normal vector from the coordinate equation. - For the parametric plane, solve two zero dot-product conditions to find a vector perpendicular to both direction vectors. - Apply the acute-angle formula to the normal vectors. - Use an absolute value in the numerator.

1. A normal vector to \(E_1\) is \(\vec{n}_1=\begin{pmatrix}4\\0\\3\end{pmatrix}\). 2. For \(E_2\), let \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\). Perpendicularity to \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\) and \(\begin{pmatrix}0\\1\\1\end{pmatrix}\) gives \(a-b=0\) and \(b+c=0\). Choose \(\vec{n}_2=\begin{pmatrix}-1\\-1\\1\end{pmatrix}\). 3. Their magnitudes are \(5\) and \(\sqrt{3}\), and their dot product is \(-1\). 4. Thus \(\cos\alpha=\frac{1}{5\sqrt{3}}\), so \(\alpha\approx83.4^\circ\).

\(\alpha\approx83.4^\circ\)

- For each plane, find a vector perpendicular to both direction vectors by solving two zero dot-product conditions. - What does a zero dot product of the normals imply? - The angle between planes equals the angle between their normals, taken between \(0^\circ\) and \(90^\circ\).

1. For \(E_1\), a normal \(\vec{n}_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\) must satisfy \(a+b=0\) and \(b+c=0\). Choose \(\vec{n}_1=\begin{pmatrix}1\\-1\\1\end{pmatrix}\). 2. For \(E_2\), a normal \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\) must satisfy \(a+c=0\) and \(a+b+c=0\). Choose \(\vec{n}_2=\begin{pmatrix}-1\\0\\1\end{pmatrix}\). 3. Their dot product is \(0\), so the normal vectors are perpendicular. 4. Therefore, the planes meet at \(90^\circ\).

\(90^\circ\)

- Find the midpoint of \(GH\). - Use \(\overrightarrow{AM}\) as the line direction. - Find a normal to the plane through \(B\), \(D\), and \(E\). - Apply the sine formula for a line-plane angle.

1. Since \(G=(6, 6, 6)\) and \(H=(0, 6, 6)\), the midpoint is \(M=(3, 6, 6)\). A direction vector for \(g\) is \(\vec{v}=\begin{pmatrix}1\\2\\2\end{pmatrix}\). 2. A normal vector to the plane through \(B\), \(D\), and \(E\) is \(\vec{n}=\begin{pmatrix}1\\1\\1\end{pmatrix}\). 3. The line-plane angle satisfies \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{5}{3\sqrt{3}}\). 4. Therefore, \(\alpha\approx74.2^\circ\).

\(\alpha\approx74.2^\circ\)

- Use two edge vectors from each face. - Solve two zero dot-product conditions to find a normal to each face. - The acute angle between the faces equals the acute angle between their normals. - Apply the dot-product angle formula.

1. For face \(BCS\), use \(\overrightarrow{BC}=\begin{pmatrix}-10\\0\\0\end{pmatrix}\) and \(\overrightarrow{BS}=\begin{pmatrix}-5\\-5\\12\end{pmatrix}\). A normal \(\vec{n}_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\) must satisfy \(a=0\) and \(-5b+12c=0\), so choose \(\vec{n}_1=\begin{pmatrix}0\\12\\5\end{pmatrix}\). 2. For face \(CDS\), use \(\overrightarrow{CD}=\begin{pmatrix}0\\-10\\0\end{pmatrix}\) and \(\overrightarrow{CS}=\begin{pmatrix}5\\-5\\12\end{pmatrix}\). A normal \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\) must satisfy \(b=0\) and \(5a+12c=0\), so choose \(\vec{n}_2=\begin{pmatrix}-12\\0\\5\end{pmatrix}\). 3. Their dot product is \(25\), and each has magnitude \(13\). 4. Therefore, \(\cos\gamma=\frac{25}{169}\), so \(\gamma\approx81.5^\circ\).

\(\gamma\approx81.5^\circ\)

- Set the other two coordinates equal to zero to find an axis intercept. - Use the plane normal and each axis direction vector. - Apply the sine formula for a line-plane angle. - What does a missing variable in a plane equation imply geometrically?

1. On the x-axis, \(y=z=0\), so \(3x=12\) and the intercept is \((4, 0, 0)\). On the y-axis, \(x=z=0\), so \(-4y=12\) and the intercept is \((0, -3, 0)\). 2. A normal vector is \(\vec{n}=\begin{pmatrix}3\\-4\\0\end{pmatrix}\), with magnitude \(5\). For the x-axis direction \(\vec{e}_x\), \(\sin\alpha_x=\frac{3}{5}\), so \(\alpha_x\approx36.9^\circ\). For the y-axis direction \(\vec{e}_y\), \(\sin\alpha_y=\frac{4}{5}\), so \(\alpha_y\approx53.1^\circ\). 3. The equation contains no \(z\)-term, so the plane is parallel to the z-axis. Their angle is \(0^\circ\).

1. \((4, 0, 0)\) and \((0, -3, 0)\) 2. With the x-axis: \(36.9^\circ\); with the y-axis: \(53.1^\circ\) 3. The plane is parallel to the z-axis, so the angle is \(0^\circ\).

- Use the intercept form of a plane. - Read the plane normal from the coordinate equation. - Use coordinate-axis unit vectors as normals to coordinate planes. - Apply the acute-angle formula for two planes.

1. The intercept form is \(\frac{x}{6}+\frac{y}{6}+\frac{z}{3}=1\), or \(x+y+2z=6\). Thus a normal vector is \(\vec{n}=\begin{pmatrix}1\\1\\2\end{pmatrix}\). 2. The \(xy\)-plane has normal \(\begin{pmatrix}0\\0\\1\end{pmatrix}\). Therefore, \(\cos\alpha=\frac{2}{\sqrt{6}}\), so \(\alpha\approx35.3^\circ\). 3. The \(xz\)-plane has normal \(\begin{pmatrix}0\\1\\0\end{pmatrix}\). Therefore, \(\cos\beta=\frac{1}{\sqrt{6}}\), so \(\beta\approx65.9^\circ\).

With the \(xy\)-plane: \(35.3^\circ\) With the \(xz\)-plane: \(65.9^\circ\)

- Find a normal vector to each plane. - Use the dot-product formula for the angle between planes. - A normal vector to the \(xy\)-plane is \(\begin{pmatrix}0\\0\\1\end{pmatrix}\). - Use the exact value of \(\cos45^\circ\). - Apply the condition \(k>0\) after solving.

1. A normal vector to \(E_k\) is \(\vec{n}_k=\begin{pmatrix}3\\4\\k\end{pmatrix}\). A normal to the \(xy\)-plane is \(\vec{e}_z=\begin{pmatrix}0\\0\\1\end{pmatrix}\). 2. The plane-angle formula gives \(\frac{k}{\sqrt{25+k^2}}=\cos45^\circ=\frac{\sqrt{2}}{2}\). 3. Squaring gives \(\frac{k^2}{25+k^2}=\frac{1}{2}\), so \(k^2=25\). 4. Since \(k>0\), \(k=5\).

\(k=5\)

- Find a normal vector for each plane. - Use cosine for the angle between planes and sine for the angle between a line and a plane. - Compare the two numerical angles.

1. A normal vector to \(E_1\) is \(\vec{n}_1=\begin{pmatrix}1\\1\\0\end{pmatrix}\). A normal vector to \(E_2\) is \(\vec{n}_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\). 2. Thus \(\cos\gamma=\frac{1}{2}\), so \(\gamma=60^\circ\). 3. For line \(h\), \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}_1|}{\|\vec{v}\|\|\vec{n}_1\|}=\frac{1}{\sqrt{2}}\), so \(\alpha=45^\circ\). 4. Therefore, \(\alpha<\gamma\).

a) \(\gamma=60^\circ\) b) \(\alpha=45^\circ\) c) \(\alpha<\gamma\)

- Find a vector perpendicular to both spanning vectors by solving two zero dot-product conditions. - For a line-plane angle, decide whether the sine or cosine relationship uses the plane normal. - Think about what the line-plane angle represents geometrically. - Compare the three angle measures directly.

1. Let \(\vec{u}=\begin{pmatrix}1\\1\\0\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}0\\1\\1\end{pmatrix}\). A normal \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) must satisfy \(a+b=0\) and \(b+c=0\), so choose \(\vec{n}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\). Using \(A=(2, 1, 0)\), the plane equation is \(x-y+z=1\). 2. \(\sin\beta=\frac{|\vec{w}\cdot\vec{n}|}{\|\vec{w}\|\|\vec{n}\|}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}\). Therefore, \(\beta\approx54.7^\circ\). 3. \(\cos\gamma_1=\frac{|\vec{w}\cdot\vec{u}|}{\|\vec{w}\|\|\vec{u}\|}=\frac{1}{2}\), so \(\gamma_1=60^\circ\). Similarly, \(\cos\gamma_2=\frac{|\vec{w}\cdot\vec{v}|}{\|\vec{w}\|\|\vec{v}\|}=\frac{1}{2}\), so \(\gamma_2=60^\circ\). 4. Here, \(\beta<\gamma_1\) and \(\beta<\gamma_2\). The angle between a line and a plane is the smallest angle between the line and any line in that plane.

1. \(E:x-y+z=1\) 2. \(\beta\approx54.7^\circ\) 3. \(\gamma_1=60^\circ\) and \(\gamma_2=60^\circ\) 4. The line-plane angle is the minimum angle between the line and a line contained in the plane.

- Find a vector orthogonal to both directions in \(E_1\). - Test both directions of \(E_2\) with that normal vector. - Write a standard equation for \(E_1\). - Test the initial point of \(E_2\) in that equation.

1. Let \(\vec{n}=\langle p,q,w\rangle\) be normal to \(E_1\). Orthogonality to its directions gives \(p+w=0\) and \(q-w=0\). One solution is \(\vec{n}=\langle -1,1,1\rangle\). 2. The directions of \(E_2\) are also orthogonal to \(\vec{n}\): \(\langle 1,1,0\rangle\cdot\vec{n}=-1+1=0\) and \(\langle 1,-1,2\rangle\cdot\vec{n}=-1-1+2=0\). 3. Therefore the planes are parallel. 4. A standard equation of \(E_1\) is \(-x+y+z=d\). Using \((1, 2, 0)\), \(d=1\). 5. The initial point of \(E_2\) also satisfies this equation because \(-2+3+0=1\). 6. Parallel planes that share a point are identical.

The two equations describe the same plane.

- Write a parametric equation for the ray using its starting point and direction vector. - A point on the plane must satisfy its Cartesian equation. - Use the ray direction and a normal vector to the plane to find the angle. - For a line-plane angle, determine whether the sine or cosine relationship is appropriate.

1. The ray is \(\vec{x}=\begin{pmatrix}6\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\-1\\2\end{pmatrix}\), where \(r\ge0\). 2. Substitute into the plane equation: \((6+2r)+(2-r)=10\). Thus \(8+r=10\), so \(r=2\). 3. The intersection point is \(S=\begin{pmatrix}6\\2\\1\end{pmatrix}+2\begin{pmatrix}2\\-1\\2\end{pmatrix}=\begin{pmatrix}10\\0\\5\end{pmatrix}\). 4. A normal vector to the plane is \(\vec{n}=\begin{pmatrix}1\\1\\0\end{pmatrix}\). Therefore, \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{1}{3\sqrt{2}}\), so \(\alpha\approx13.6^\circ\).

The ray meets the plane at \(S=(10, 0, 5)\), and \(\alpha\approx13.6^\circ\).

- First write a parametric equation of the line through the two points. - Substitute the line coordinates into the plane equation to find their common point. - Read a normal vector directly from the plane equation. - Relate the line-plane angle to the angle between the line direction and the plane normal.

1. A direction vector for the line is \(\vec{v}=\overrightarrow{AB}=\begin{pmatrix}2\\2\\1\end{pmatrix}\). Thus \(g:\vec{x}=\begin{pmatrix}0\\1\\2\end{pmatrix}+t\begin{pmatrix}2\\2\\1\end{pmatrix}\). 2. Substitute into the plane equation: \(2(2t)-2(1+2t)+(2+t)=2\). This simplifies to \(t=2\). 3. Therefore, \(S=\begin{pmatrix}0\\1\\2\end{pmatrix}+2\begin{pmatrix}2\\2\\1\end{pmatrix}=\begin{pmatrix}4\\5\\4\end{pmatrix}\). 4. A normal vector is \(\vec{n}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\). Then \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{1}{9}\), so \(\alpha\approx6.4^\circ\).

The intersection point is \(S=(4, 5, 4)\), and \(\alpha\approx6.4^\circ\).

- Read the two normal vectors from the plane equations. - A direction in the intersection line is orthogonal to both normals. - Solve the two dot-product equations for a nonzero direction vector. - Find one common point by fixing one coordinate.

1. The normal vectors are \(\vec{n}_1=\langle 1,1,1\rangle\) and \(\vec{n}_2=\langle 2,-1,1\rangle\). 2. Let the line direction be \(\vec{d}=\langle a,b,c\rangle\). It must satisfy \(a+b+c=0\) and \(2a-b+c=0\). 3. Subtracting the equations gives \(a-2b=0\), so \(a=2b\). The first equation then gives \(c=-3b\). Choose \(b=1\), giving \(\vec{d}=\langle 2,1,-3\rangle\). 4. To find a point on both planes, set \(z=0\). Then \(x+y=6\) and \(2x-y=3\), so \(x=3\) and \(y=3\). 5. Therefore \(g: \vec{r}(t)=\langle 3,3,0\rangle+t\langle 2,1,-3\rangle\).

\(g: \vec{r}(t)=\langle 3,3,0\rangle+t\langle 2,1,-3\rangle\)

- The intersection direction lies in both planes. - Therefore, it is orthogonal to both plane normals. - Use the given direction vector in a dot-product equation involving \(k\).

1. A direction vector of the intersection line must be orthogonal to the normal of each plane. 2. The normals are \(\vec{n}_E=\langle 1,2,1\rangle\) and \(\vec{n}_{F_k}=\langle k,-1,2\rangle\). 3. The given vector already satisfies \(\vec{u}\cdot\vec{n}_E=5-2-3=0\). 4. Require it to be orthogonal to the second normal as well: \(\vec{u}\cdot\vec{n}_{F_k}=5k+1-6=0\). 5. Thus \(5k-5=0\), so \(k=1\).

\(k=1\)

- Find a vector orthogonal to both direction vectors of the plane. - Use the given point to determine the constant in the coordinate equation. - Apply the point-to-plane distance formula to each point.

1. Let \(\mathbf{u}=\langle 2,-2,1\rangle\) and \(\mathbf{v}=\langle 1,2,2\rangle\). A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) must satisfy \(2a-2b+c=0\) and \(a+2b+2c=0\). Choose \(\mathbf{n}=\langle 2,1,-2\rangle\). 2. Using point \((4,1,0)\), the coordinate equation is \(2x+y-2z=9\). 3. The normal vector has magnitude \(3\), so the distance from the origin is \(\frac{\lvert -9\rvert}{3}=3\). 4. The distance from \(S\) is \(\frac{\lvert 2(10)+10-2(10)-9\rvert}{3}=\frac{1}{3}\).

a) \(E: 2x+y-2z=9\) b) \(3\) units c) \(\frac{1}{3}\) unit

- Three points determine a unique plane when they are not collinear. - Use two connecting vectors and solve zero dot-product equations to find a normal vector. - Substitute the parametric coordinates of the line into the plane equation. - For a line-plane angle, relate the line direction to the plane normal.

1. The vectors \(\overrightarrow{AB}=\begin{pmatrix}-3\\3\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}-3\\0\\3\end{pmatrix}\) are not scalar multiples, so the points are not collinear and determine a unique plane. 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be normal to the plane. Perpendicularity to \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) gives \(-3a+3b=0\) and \(-3a+3c=0\). Choose \(\vec{n}=\begin{pmatrix}1\\1\\1\end{pmatrix}\). Using point \(A\), the plane equation is \(E:x+y+z=3\). 3. Substitute the line coordinates into the plane equation: \((1+t)+1+t=3\). Thus \(2t+2=3\), so \(t=\frac{1}{2}\). Therefore, \(S=\left(\frac{3}{2},1,\frac{1}{2}\right)\). 4. With line direction \(\vec{v}=\begin{pmatrix}1\\0\\1\end{pmatrix}\) and plane normal \(\vec{n}=\begin{pmatrix}1\\1\\1\end{pmatrix}\), \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{2}{\sqrt{6}}\). Therefore, \(\alpha\approx54.7^\circ\).

a) The points are not collinear, and \(E:x+y+z=3\). b) The line intersects the plane at \(S=\left(\frac{3}{2},1,\frac{1}{2}\right)\). c) \(\alpha\approx54.7^\circ\)

- Parallel planes have parallel normal vectors. - To find the intersection line, choose one coordinate as a parameter and solve for the others. - Use the plane normals to find the angle between the planes. - Apply the point-to-plane distance formula to the Cartesian equation.

1. The normal vectors \(\vec{n}_1=\begin{pmatrix}2\\2\\-1\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}1\\-2\\2\end{pmatrix}\) are not scalar multiples, so the planes intersect in a line. 2. Solving the two plane equations gives one point \((3, 6, 6)\). A direction vector perpendicular to both normals is \(\begin{pmatrix}2\\-5\\-6\end{pmatrix}\). Thus \(s:\vec{x}=\begin{pmatrix}3\\6\\6\end{pmatrix}+\lambda\begin{pmatrix}2\\-5\\-6\end{pmatrix}\). 3. \(\cos\phi=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{4}{9}\), so \(\phi\approx63.6^\circ\). 4. The point-to-plane distance is \(d=\frac{|2\cdot0+2\cdot0-0-12|}{\sqrt{2^2+2^2+(-1)^2}}=\frac{12}{3}=4\).

a) The planes intersect in the line \(s:\vec{x}=\begin{pmatrix}3\\6\\6\end{pmatrix}+\lambda\begin{pmatrix}2\\-5\\-6\end{pmatrix}\). b) \(\phi\approx63.6^\circ\) c) \(4\) units

- Read the normal vector from the standard equation. - Test it against both direction vectors of the parametric plane. - Check whether the initial point satisfies the standard equation. - Parallel planes sharing a point are the same plane.

1. The normal vector of \(E_2\) is \(\vec{n}=\langle -3,-1,2\rangle\). 2. This vector is orthogonal to both directions in \(E_1\): \(\vec{n}\cdot\langle 1,-1,1\rangle=-3+1+2=0\) and \(\vec{n}\cdot\langle 2,0,3\rangle=-6+6=0\). 3. Therefore \(E_1\) is parallel to \(E_2\). 4. The initial point of \(E_1\) lies in \(E_2\), because \(-3(4)-1+2(-2)=-17\). 5. Parallel planes that share a point are identical.

The planes are identical.

- Form two direction vectors from the three points. - Read a normal vector from the equation of \(F\). - Test the normal vector against both directions using dot products. - Check whether one of the given points lies in \(F\).

1. Two directions in \(E\) are \(\overrightarrow{AB}=\langle 2,1,-2\rangle\) and \(\overrightarrow{AC}=\langle -1,2,0\rangle\). 2. A normal vector to \(F\) is \(\vec{n}_F=\langle 2,1,2.5\rangle\). Test it against the directions in \(E\): \(\vec{n}_F\cdot\overrightarrow{AB}=4+1-5=0\) and \(\vec{n}_F\cdot\overrightarrow{AC}=-2+2=0\). 3. Thus \(E\) is parallel to \(F\). 4. Point \(A\) lies in \(F\), because \(2(1)+0+2.5(2)=7\). 5. Therefore the planes are identical.

Yes. Planes \(E\) and \(F\) are identical.

- Parallel planes have parallel normal vectors. - A direction vector in the plane must have dot product \(0\) with the normal vector. - Choose two direction vectors that are not scalar multiples. - Use the given point as the initial point.

1. A normal vector to \(F\), and therefore to the parallel plane \(E\), is \(\vec{n}=\langle 2,-4,3\rangle\). 2. Choose two linearly independent vectors orthogonal to \(\vec{n}\). For \(\vec{u}=\langle 2,1,0\rangle\), \(\vec{n}\cdot\vec{u}=4-4=0\). 3. For \(\vec{v}=\langle 0,3,4\rangle\), \(\vec{n}\cdot\vec{v}=-12+12=0\). The two vectors are not scalar multiples, so they span directions in the plane. 4. Using \(A\) as the initial point, \(E\) is \(\vec{r}(s,t)=\langle 5,-1,2\rangle+s\langle 2,1,0\rangle+t\langle 0,3,4\rangle\).

\(E: \vec{r}(s,t)=\langle 5,-1,2\rangle+s\langle 2,1,0\rangle+t\langle 0,3,4\rangle\)

- Use \(z=0\) to locate the line’s intersection with the \(xy\)-plane. - A line perpendicular to a plane has a direction vector parallel to the plane’s normal vector. - Substitute the intersection point into the standard equation. - Keep the parameter value exact as a fraction.

1. A point in the \(xy\)-plane has \(z=0\). On \(g\), the \(z\)-coordinate is \(-1+5t\), so \(-1+5t=0\) and \(t=\frac{1}{5}\). 2. Substitute \(t=\frac{1}{5}\): \(S=\left(4+\frac{2}{5}, -\frac{1}{5}, 0\right)=\left(\frac{22}{5}, -\frac{1}{5}, 0\right)\). 3. Because \(g\) is perpendicular to \(E\), the direction vector \(\langle 2,-1,5\rangle\) is a normal vector to \(E\). 4. Substitute \(S\) into \(2x-y+5z=d\): \(d=2\left(\frac{22}{5}\right)-\left(-\frac{1}{5}\right)=9\). 5. Therefore \(E: 2x-y+5z=9\).

\(E: 2x-y+5z=9\)

- Substitute the line's parametric coordinates into the plane equation. - For part b), substitute a general point on the line into the point-to-plane distance formula. - The absolute-value equation gives two parameter values.

1. A general point on \(g\) is \((1+t,2,-1+2t)\). 2. Substitute into the plane equation: \(6(1+t)+3(2)-2(-1+2t)=14\). This simplifies to \(14+2t=14\), so \(t=0\). 3. Therefore, the intersection point is \(S=(1,2,-1)\). 4. A normal vector to \(E\) is \(\langle 6,3,-2\rangle\), with magnitude \(7\). The distance from the general line point to \(E\) is \(\frac{\lvert 2t\rvert}{7}\). 5. Set this equal to \(4\): \(\frac{\lvert 2t\rvert}{7}=4\), so \(t=14\) or \(t=-14\). 6. These values give \((15,2,27)\) and \((-13,2,-29)\).

a) \(S=(1,2,-1)\) b) \((15,2,27)\) and \((-13,2,-29)\)

- The shortest segment from a point to a plane follows the plane's normal vector. - Intersect the perpendicular line through \(B\) with the plane. - Use a general point on that line in the distance formula for part c).

1. A normal vector to \(F\) is \(\mathbf{n}=\langle 1,2,2\rangle\), with magnitude \(3\). Thus, \(d(B,F)=\frac{\lvert 2+2(2)+2(6)-9\rvert}{3}=3\). 2. The perpendicular line through \(B\) is \((x,y,z)=(2,2,6)+r\langle 1,2,2\rangle\). 3. Substitute into the plane equation: \((2+r)+2(2+2r)+2(6+2r)=9\). This gives \(r=-1\), so the closest point is \(Q=(1,0,4)\). 4. For a general point on \(h\), the signed plane expression is \(9r+9\), so its distance from \(F\) is \(\frac{\lvert 9r+9\rvert}{3}=\lvert 3r+3\rvert\). 5. Set \(\lvert 3r+3\rvert=6\). The solutions are \(r=1\) and \(r=-3\), producing \((3,4,8)\) and \((-1,-4,0)\).

a) \(3\) units b) \(Q=(1,0,4)\) c) \((3,4,8)\) and \((-1,-4,0)\)

- Compare the line's direction vector with the plane's normal vector. - Use any point on the parallel line in the point-to-plane distance formula. - Test whether one parameter value produces \(M\). - A tangent sphere's radius equals the distance from its center to the plane.

1. The line's direction vector is \(\mathbf{v}=\langle 3,-2,-2\rangle\), and a normal vector to the plane is \(\mathbf{n}=\langle 2,2,1\rangle\). Their dot product is \(3(2)+(-2)(2)+(-2)(1)=0\). 2. The point \((1,4,5)\) does not lie in the plane because \(2(1)+2(4)+5=15\ne6\). Therefore, \(g\) is parallel to and distinct from \(E\). 3. The distance is \(\frac{\lvert 2(1)+2(4)+5-6\rvert}{3}=3\). 4. Substituting \(t=1\) into \(g\) gives \((1,4,5)+\langle 3,-2,-2\rangle=(4,2,3)=M\), so \(M\in g\). 5. Because the sphere is tangent to the plane, its radius is the distance from \(M\) to \(E\), which is also \(3\). Therefore, its equation is \((x-4)^2+(y-2)^2+(z-3)^2=9\).

a) \(g\parallel E\) b) \(3\) units c) \(M\in g\) when \(t=1\), and the sphere is \((x-4)^2+(y-2)^2+(z-3)^2=9\).

- Compare the plane normals with a dot product. - Solve the two plane equations with one free parameter. - The third normal must be orthogonal to both existing normals. - Substitute the intersection line into the third plane.

1. a) The normals are \(\vec{n}_1=\langle 1,1,1\rangle\) and \(\vec{n}_2=\langle 1,-1,0\rangle\). Their dot product is \(1-1=0\), so the planes are perpendicular. 2. From \(E_2\), \(x=y\). Let \(x=y=t\). Then \(z=3-2t\), so \(s:\vec{r}(t)=\langle 0,0,3\rangle+t\langle 1,1,-2\rangle\). 3. b) Let \(\vec{n}_3=\langle a,b,c\rangle\). Perpendicularity to both planes requires \(\vec{n}_3\cdot\vec{n}_1=0\) and \(\vec{n}_3\cdot\vec{n}_2=0\). 4. Thus \(a+b+c=0\) and \(a-b=0\). Hence \(a=b\) and \(c=-2a\). Choose \(\vec{n}_3=\langle 1,1,-2\rangle\). 5. Using \(P\), the equation is \(x+y-2z=0\). 6. c) Substitute \((t, t, 3-2t)\): \(t+t-2(3-2t)=0\), so \(6t-6=0\) and \(t=1\). 7. The intersection point is \(S(1, 1, 1)\).

a) The planes are perpendicular; \(s:\vec{r}(t)=\langle 0,0,3\rangle+t\langle 1,1,-2\rangle\). b) \(E_3:x+y-2z=0\) c) \(S(1, 1, 1)\)

- Use a dot product between plane normals. - Separate the family equation into a constant term and a coefficient of \(t\). - The new plane normal must be orthogonal to both given normals. - Compare the resulting equation with the family.

1. a) The normals are \(\vec{n}_F=\langle 1,1,1\rangle\) and \(\vec{n}_t=\langle 1,t-1,-t\rangle\). 2. Their dot product is \(1+(t-1)-t=0\), so every \(F_t\) is perpendicular to \(F\). The origin satisfies every equation \(F_t\). 3. Rewrite the family equation as \(x-y+t(y-z)=0\). For this equation to hold for every \(t\), require \(x-y=0\) and \(y-z=0\). 4. Thus the common intersection is \(s:\vec{r}(\lambda)=\lambda\langle 1,1,1\rangle\). 5. b) Let \(\vec{n}^*=\langle a,b,c\rangle\). It must be orthogonal to both \(\langle 1,1,1\rangle\) and \(\langle 1,1,-2\rangle\). 6. The equations \(a+b+c=0\) and \(a+b-2c=0\) give \(c=0\) and \(a=-b\). Choose \(\vec{n}^*=\langle 1,-1,0\rangle\). 7. Since the plane passes through the origin, \(E^*:x-y=0\). 8. c) Setting \(t=0\) in the family gives \(F_0:x-y=0\), so \(E^*=F_0\).

a) Every \(F_t\perp F\), and the common line is \(s:\vec{r}(\lambda)=\lambda\langle 1,1,1\rangle\). b) \(E^*:x-y=0\) c) \(E^*=F_0\), so \(t=0\).

- What coordinates must a point on the pyramid's symmetry axis have? - The distance to the base plane equals the point's z-coordinate. - Find a normal vector and equation for one lateral face. - Set the distance to the base equal to the distance to that lateral face.

1. A point on the symmetry axis has the form \(P=(5,5,z)\). Its distance from the base plane \(z=0\) is \(z\). 2. Consider lateral face \(BCS\). A normal vector to this face is \(\langle 0,12,5\rangle\), so the face has equation \(12y+5z=120\). 3. The distance from \(P=(5,5,z)\) to this face is \(\frac{\lvert 12(5)+5z-120\rvert}{\sqrt{12^2+5^2}}=\frac{60-5z}{13}\), since \(P\) lies below the face. 4. Set the two distances equal: \(z=\frac{60-5z}{13}\). Then \(13z=60-5z\), so \(18z=60\) and \(z=\frac{10}{3}\). 5. By symmetry, this distance is the same for all four lateral faces. Therefore, \(P=\left(5,5,\frac{10}{3}\right)\).

\(P\left(5,5,\frac{10}{3}\right)\)

- First find the orthogonal projection \(F\) of \(Q\) onto the line. - What midpoint relationship connects \(Q\), \(F\), and \(Q'\)? - Use the projection formula for a line through the origin.

1. A unit vector in the direction of \(h\) is \(\mathbf{u}=\left\langle \frac{3}{5}, \frac{4}{5}, 0\right\rangle\). 2. Project \(\overrightarrow{OQ}\) onto the line: \(\overrightarrow{OQ}\cdot\mathbf{u}=7\left(\frac{3}{5}\right)+1\left(\frac{4}{5}\right)+5(0)=5\). 3. Thus, \(\overrightarrow{OF}=5\mathbf{u}=\langle 3, 4, 0\rangle\), so \(F=(3, 4, 0)\). 4. Because \(F\) is the midpoint of \(\overline{QQ'}\), \(\overrightarrow{OQ'}=2\overrightarrow{OF}-\overrightarrow{OQ}=2\langle 3, 4, 0\rangle-\langle 7, 1, 5\rangle=\langle -1, 7, -5\rangle\).

\(Q'(-1, 7, -5)\)

- Find the midpoint coordinates first. - Use the stated fraction of \(\overrightarrow{SC}\) to locate \(M\). - A plane normal is orthogonal to two directions in the plane. - Parametrize \(\overline{DS}\) and check that the intersection parameter lies in \([0,1]\).

1. The midpoints are \(K(1, 1, 4)\) and \(L(3, 1, 4)\). 2. Since \(M=S+\frac14(C-S)\), \(M=\langle 2,2,8\rangle+\frac14\langle 2,2,-8\rangle=\left(\frac52, \frac52, 6\right)\). 3. Directions in \(E\) are \(\overrightarrow{KL}=\langle 2,0,0\rangle\) and \(\overrightarrow{KM}=\left\langle\frac32,\frac32,2\right\rangle\). 4. Let \(\vec{n}=\langle a,b,c\rangle\). The equations \(\vec{n}\cdot\overrightarrow{KL}=0\) and \(\vec{n}\cdot\overrightarrow{KM}=0\) give \(a=0\) and \(\frac32b+2c=0\). Choose \(\vec{n}=\langle 0,-4,3\rangle\). 5. Using point \(K\), the plane equation is \(-4y+3z=8\). 6. Parametrize segment \(\overline{DS}\): \(\vec{r}(t)=\langle 0,4,0\rangle+t\langle 2,-2,8\rangle\), where \(0\le t\le1\). 7. Substitute into the plane equation: \(-4(4-2t)+3(8t)=8\). Thus \(-16+32t=8\), so \(t=\frac34\). 8. Therefore \(N=\langle 0,4,0\rangle+\frac34\langle 2,-2,8\rangle=\left(\frac32, \frac52, 6\right)\).

a) \(E:-4y+3z=8\) b) \(N\left(\frac32, \frac52, 6\right)\)

- In the first method, the vector from \(P\) to a general point on \(g\) must be orthogonal to the line's direction vector. - In the projection method, project the vector from a known point on \(g\) to \(P\) onto the direction vector. - The distance is the magnitude of the vector from \(P\) to \(F\). - Orthogonal vectors have dot product \(0\).

1. For the dot-product method, let \(F(t)=(1+2t, 2-t, -1+2t)\). Then \(\overrightarrow{PF}=\langle 2t-2, 4-t, 2t-5\rangle\). 2. Require \(\overrightarrow{PF}\cdot\langle 2, -1, 2\rangle=0\): \(2(2t-2)-(4-t)+2(2t-5)=0\). This simplifies to \(9t-18=0\), so \(t=2\) and \(F=(5, 0, 3)\). 3. For the projection method, use the point \(A=(1, 2, -1)\) on \(g\). Then \(\overrightarrow{AP}=\langle 2, -4, 5\rangle\) and \(\operatorname{proj}_{\mathbf{v}}\overrightarrow{AP}=\frac{\overrightarrow{AP}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\mathbf{v}=\frac{18}{9}\langle 2, -1, 2\rangle=\langle 4, -2, 4\rangle\). 4. Therefore, \(F=A+\operatorname{proj}_{\mathbf{v}}\overrightarrow{AP}=(5, 0, 3)\), confirming the first result. 5. Finally, \(\overrightarrow{PF}=\langle 2, 2, -1\rangle\), so \(d(P,g)=\sqrt{2^2+2^2+(-1)^2}=3\).

The perpendicular foot is \(F(5, 0, 3)\), and \(d(P,g)=3\).

- The line of symmetry is the perpendicular bisector of the base. - Therefore, \(C\) is the reflection of \(B\) across \(s\). - First find the point on \(s\) closest to \(B\).

1. Since \(s\) is the line of symmetry, \(C\) is the reflection of \(B\) across \(s\). 2. A general point on \(s\) is \(F(\lambda)=(1+\lambda, 2-2\lambda, 3+2\lambda)\), so \(\overrightarrow{BF}=\langle \lambda, 6-2\lambda, 2\lambda-3\rangle\). 3. At the perpendicular foot, \(\overrightarrow{BF}\cdot\langle 1, -2, 2\rangle=0\). This gives \(\lambda-2(6-2\lambda)+2(2\lambda-3)=9\lambda-18=0\), so \(\lambda=2\). 4. Thus, \(F=(3, -2, 7)\). Since \(F\) is the midpoint of \(\overline{BC}\), \(C=2F-B=2(3, -2, 7)-(1, -4, 6)=(5, 0, 8)\).

\(C(5, 0, 8)\)

- Separate the terms that contain \(t\) from those that do not. - A point common to every plane must make both resulting expressions zero. - The position vector of the closest point is orthogonal to the line’s direction vector. - Use a dot product to solve for the line parameter.

1. a) Rewrite the family equation as \((x+2z-4)+t(y-z-2)=0\). 2. For a point to lie in every plane, both \(x+2z-4=0\) and \(y-z-2=0\) must hold. 3. Let \(z=\lambda\). Then \(x=4-2\lambda\) and \(y=2+\lambda\), so \(g: \vec{r}(\lambda)=\langle 4,2,0\rangle+\lambda\langle -2,1,1\rangle\). 4. b) At the point closest to the origin, the position vector \(\vec{OP}=\langle 4-2\lambda,2+\lambda,\lambda\rangle\) is orthogonal to the direction \(\vec{d}=\langle -2,1,1\rangle\). 5. Set the dot product equal to zero: \(\langle 4-2\lambda,2+\lambda,\lambda\rangle\cdot\langle -2,1,1\rangle=0\). 6. This gives \(-8+4\lambda+2+\lambda+\lambda=0\), so \(6\lambda-6=0\) and \(\lambda=1\). 7. Therefore \(P=(2, 3, 1)\).

a) \(g: \vec{r}(\lambda)=\langle 4,2,0\rangle+\lambda\langle -2,1,1\rangle\) b) \(P(2, 3, 1)\)

- The vector from \(A\) to its perpendicular foot must be orthogonal to the line's direction vector. - The point-to-line distance is the length of that perpendicular segment. - For the reflection, the perpendicular foot is the midpoint of the original point and its image. - A tangent sphere's radius equals the distance from its center to the tangent line.

1. A general point on \(g\) is \(L(t)=(t, 1+2t, 1)\), so \(\overrightarrow{AL}=\langle t-3, 2t-1, -5\rangle\). 2. Orthogonality requires \(\overrightarrow{AL}\cdot\langle 1, 2, 0\rangle=0\): \((t-3)+2(2t-1)=5t-5=0\). Thus, \(t=1\) and \(L=(1, 3, 1)\). 3. The distance is \(AL=\sqrt{(-2)^2+1^2+(-5)^2}=\sqrt{30}\approx 5.48\). 4. Since \(L\) is the midpoint of \(\overline{AA'}\), \(A'=2L-A=2(1, 3, 1)-(3, 2, 6)=(-1, 4, -4)\). 5. The tangent sphere has radius \(\sqrt{30}\), so its equation is \((x-3)^2+(y-2)^2+(z-6)^2=30\).

a) \(L(1, 3, 1)\) b) \(\sqrt{30}\approx 5.48\) units c) \(A'(-1, 4, -4)\) d) \((x-3)^2+(y-2)^2+(z-6)^2=30\)

- Form two direction vectors in the plane. - Find a vector whose dot product with both direction vectors is \(0\). - Use one of the given points to write the plane equation. - Then apply the point-to-plane distance formula.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle 3,0,4\rangle\) and \(\overrightarrow{AC}=\langle 0,5,0\rangle\). 2. A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) must be orthogonal to both. Orthogonality to \(\overrightarrow{AC}\) gives \(b=0\), and orthogonality to \(\overrightarrow{AB}\) gives \(3a+4c=0\). Choose \(\mathbf{n}=\langle -4,0,3\rangle\). 3. Using point \(A\), the plane equation is \(-4x+3z=-1\), or \(-4x+3z+1=0\). The normal magnitude is \(5\). 4. For \(P\), \(d(P,E)=\frac{\lvert -4(10)+3(3)+1\rvert}{5}=6\). 5. For \(Q\), \(d(Q,E)=\frac{\lvert -4(-5)+3(-4)+1\rvert}{5}=\frac{9}{5}=1.8\).

\(d(P,E)=6\) units and \(d(Q,E)=\frac{9}{5}=1.8\) units

- First test whether the direction vectors are scalar multiples. - Then test whether the lines intersect by equating coordinates. - Find a vector orthogonal to both line directions. - Project the vector between the two lines onto that common perpendicular direction.

1. The direction vectors \(\mathbf{u}=\langle 1,2,2\rangle\) and \(\mathbf{v}=\langle 2,1,-2\rangle\) are not scalar multiples, so the lines are not parallel. 2. Equating coordinates gives \(2+t=1+2s\), \(2t=4+s\), and \(-1+2t=-2s\). The first two equations give \(t=3\) and \(s=2\), but these values do not satisfy the third equation. Thus, the lines are skew. 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both direction vectors. Solving \(a+2b+2c=0\) and \(2a+b-2c=0\) gives a convenient choice \(\mathbf{n}=\langle -2,2,-1\rangle\), with magnitude \(3\). 4. The vector between the lines' given points is \(\mathbf{w}=\langle -1,4,1\rangle\). The distance is the magnitude of its scalar projection onto the common perpendicular direction: \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{\lvert 2+8-1\rvert}{3}=3\).

The lines are skew, and their distance is \(3\) units.

- The point-to-line distance is the length of the perpendicular segment from \(R\) to \(g\). - Find the perpendicular foot using a dot product. - For the parallelogram, use base times height. - The base length is \(PQ\), and the height is the distance found in part a).

1. A direction vector for \(g\) is \(\overrightarrow{PQ}=\langle 2, 4, 0\rangle\). A general point on \(g\) is \(L(t)=(1+2t, 1+4t, 2)\). 2. The vector from \(R\) to \(L(t)\) is \(\overrightarrow{RL}=\langle 2t-5, 4t, -3\rangle\). Require \(\overrightarrow{RL}\cdot\overrightarrow{PQ}=0\): \(2(2t-5)+4(4t)=20t-10=0\), so \(t=\frac{1}{2}\). 3. The perpendicular foot is \(L=(2, 3, 2)\), and the height is \(RL=\sqrt{(-4)^2+2^2+(-3)^2}=\sqrt{29}\). 4. The base length is \(PQ=\sqrt{2^2+4^2}=2\sqrt{5}\). 5. Therefore, the parallelogram's area is \((2\sqrt{5})(\sqrt{29})=2\sqrt{145}\approx 24.08\).

a) \(\sqrt{29}\approx 5.39\) units b) \(2\sqrt{145}\approx 24.08\) square units

- Check whether the two base vectors are perpendicular. - Use \(\frac{1}{2}bh\) for the base triangle. - Find a vector orthogonal to both base directions. - The pyramid's height is the component of \(\overrightarrow{AS}\) in the normal direction.

1. \(\overrightarrow{AB}=\langle 2,0,-2\rangle\) and \(\overrightarrow{AC}=\langle 0,3,0\rangle\). Their dot product is \(0\), so they are perpendicular. 2. Their lengths are \(AB=2\sqrt{2}\) and \(AC=3\). Therefore, the base area is \(G=\frac{1}{2}(2\sqrt{2})(3)=3\sqrt{2}\approx4.24\). 3. A normal vector to the base is \(\mathbf{n}=\langle 1,0,1\rangle\), and \(\overrightarrow{AS}=\langle 3,0,3\rangle\). The height is the magnitude of the scalar projection: \(h=\frac{\lvert\overrightarrow{AS}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{6}{\sqrt{2}}=3\sqrt{2}\). 4. The base plane has equation \(x+z=2\). The point-to-plane distance from \(S\) is \(h=\frac{\lvert 4+4-2\rvert}{\sqrt{1^2+1^2}}=3\sqrt{2}\approx4.24\), confirming the first method.

a) \(3\sqrt{2}\approx4.24\) square units b) \(3\sqrt{2}\approx4.24\) units by both methods

- Use two edge vectors in each face and solve two zero dot-product conditions to obtain a face normal. - The base normal points vertically. - The acute angle between planes equals the acute angle between their normal vectors. - Use the dot-product angle formula.

1. A normal vector to the base is \(\vec{n}_G=\begin{pmatrix}0\\0\\1\end{pmatrix}\). For face \(BCS\), use \(\overrightarrow{BC}=\begin{pmatrix}-24\\0\\0\end{pmatrix}\) and \(\overrightarrow{BS}=\begin{pmatrix}-12\\-12\\16\end{pmatrix}\). If \(\vec{n}_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\) is perpendicular to both, the dot-product equations give \(a=0\) and \(-12b+16c=0\). Choose \(\vec{n}_1=\begin{pmatrix}0\\4\\3\end{pmatrix}\). 2. The angle \(\alpha\) between the side face and the base satisfies \(\cos\alpha=\frac{|\vec{n}_1\cdot\vec{n}_G|}{\|\vec{n}_1\|\|\vec{n}_G\|}=\frac{3}{5}\). Thus \(\alpha\approx53.1^\circ\). 3. For face \(ABS\), use \(\overrightarrow{AB}=\begin{pmatrix}0\\24\\0\end{pmatrix}\) and \(\overrightarrow{AS}=\begin{pmatrix}-12\\12\\16\end{pmatrix}\). Requiring \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\) to be perpendicular to both gives \(b=0\) and \(-12a+16c=0\). Choose \(\vec{n}_2=\begin{pmatrix}4\\0\\3\end{pmatrix}\). 4. The acute angle \(\beta\) between the side faces satisfies \(\cos\beta=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{9}{25}\). Thus \(\beta\approx68.9^\circ\).

a) \(53.1^\circ\) b) \(68.9^\circ\)

- Use two edge vectors in each face and solve two zero dot-product conditions to obtain a face normal. - The base normal is vertical. - The acute angle between planes equals the acute angle between their normal vectors. - Use the dot-product angle formula.

1. A normal vector to the base is \(\vec{n}_G=\begin{pmatrix}0\\0\\1\end{pmatrix}\). For face \(BCS\), use \(\overrightarrow{BC}=\begin{pmatrix}-20\\0\\0\end{pmatrix}\) and \(\overrightarrow{BS}=\begin{pmatrix}-10\\-5\\12\end{pmatrix}\). If \(\vec{n}_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\) is perpendicular to both, the dot-product equations give \(a=0\) and \(-5b+12c=0\). Choose \(\vec{n}_1=\begin{pmatrix}0\\12\\5\end{pmatrix}\). 2. For part a, \(\cos\alpha=\frac{|\vec{n}_1\cdot\vec{n}_G|}{\|\vec{n}_1\|\|\vec{n}_G\|}=\frac{5}{13}\), so \(\alpha\approx67.4^\circ\). 3. For face \(ABS\), use \(\overrightarrow{AB}=\begin{pmatrix}0\\10\\0\end{pmatrix}\) and \(\overrightarrow{AS}=\begin{pmatrix}-10\\5\\12\end{pmatrix}\). Requiring \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\) to be perpendicular to both gives \(b=0\) and \(-10a+12c=0\). Choose \(\vec{n}_2=\begin{pmatrix}6\\0\\5\end{pmatrix}\). 4. For part b, \(\cos\beta=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{25}{13\sqrt{61}}\), so \(\beta\approx75.7^\circ\).

a) \(67.4^\circ\) b) \(75.7^\circ\)

- Use symmetry to describe the LED's coordinates. - It is enough to find the distance to one lateral face. - Find a normal vector by requiring it to be orthogonal to two directions in the face. - Use the fact that the LED lies inside the pyramid when simplifying the absolute value.

1. By symmetry, the LED has the form \(L(0,0,h)\), where \(0<h<12\). 2. Consider lateral face \(ABS\). Its direction vectors are \(\overrightarrow{AB}=\langle 0,-10,0\rangle\) and \(\overrightarrow{AS}=\langle -5,-5,12\rangle\). 3. A vector perpendicular to both directions is \(\mathbf{n}=\langle 12,0,5\rangle\), since its dot product with each direction vector is \(0\). 4. Using point \(A\), the face plane has equation \(12x+5z=60\). The normal vector has magnitude \(13\). 5. The distance from \(L\) to this face is \(\frac{\lvert 5h-60\rvert}{13}\). Because \(L\) is inside the pyramid, \(h<12\), so this becomes \(\frac{60-5h}{13}\). 6. Set the distance equal to \(2\): \(\frac{60-5h}{13}=2\). Thus, \(60-5h=26\), so \(h=\frac{34}{5}\). 7. Therefore, the LED is at \(L\left(0,0,\frac{34}{5}\right)\). By symmetry, it has the same distance from all four lateral faces.

\(L\left(0,0,\frac{34}{5}\right)\)

- A sphere tangent to the three coordinate planes has equal positive center coordinates. - Write the fourth face using its x-, y-, and z-intercepts. - Use the face's normal vector to compute the perpendicular distance from the center. - Set that distance equal to the sphere's radius.

1. Because the sphere is tangent to the three coordinate planes and lies in the first octant, its center is \(M=(r, r, r)\). 2. The face through \(X\), \(Y\), and \(Z\) has intercept equation \(\frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1\), or \(2x+2y+z=6\). A normal vector is \(\mathbf{n}=\langle2, 2, 1\rangle\), with magnitude \(3\). 3. The distance from \(M\) to this face must also equal \(r\). Since \(M\) lies below the face, \(r=\frac{6-(2r+2r+r)}{3}\). 4. Therefore, \(3r=6-5r\), so \(8r=6\) and \(r=\frac34\).

\(r=\frac34\)

- The vector from \(P\) to its perpendicular foot is orthogonal to the line's direction vector. - In right triangle \(APF\), which segments are the perpendicular legs? - For the cone, identify the radius and height from the triangle.

1. A general point on \(g\) is \(F(t)=(1+2t, 2+t, 2+2t)\), so \(\overrightarrow{PF}=\langle 2t-5, t, 2t-4\rangle\). 2. Require \(\overrightarrow{PF}\cdot\langle 2, 1, 2\rangle=0\): \(2(2t-5)+t+2(2t-4)=9t-18=0\), so \(t=2\) and \(F=(5, 4, 6)\). 3. The leg lengths are \(AF=\sqrt{4^2+2^2+4^2}=6\) and \(PF=\sqrt{(-1)^2+2^2+0^2}=\sqrt{5}\). Thus, the triangle's area is \(\frac{1}{2}(6)(\sqrt{5})=3\sqrt{5}\approx 6.71\). 4. The cone has radius \(r=\sqrt{5}\) and height \(h=6\), so \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(5)(6)=10\pi\approx 31.42\).

a) \(F(5, 4, 6)\) b) \(3\sqrt{5}\approx 6.71\) square units c) \(10\pi\approx 31.42\) cubic units

- Find a general point on \(h\) and impose a perpendicularity condition. - The point-to-line distance is the magnitude of \(\overrightarrow{QF}\). - In the cone, the radius is the perpendicular distance and the height lies along the axis.

1. A general point on \(h\) is \(F(s)=(4, -1+4s, 2+3s)\), so \(\overrightarrow{QF}=\langle 0, 4s-9, 3s-13\rangle\). 2. Require \(\overrightarrow{QF}\cdot\langle 0, 4, 3\rangle=0\): \(4(4s-9)+3(3s-13)=25s-75=0\), so \(s=3\). 3. Therefore, \(F=(4, 11, 11)\), and \(QF=\sqrt{0^2+3^2+(-4)^2}=5\). 4. The cone's radius is \(5\), and its height is \(BF=\sqrt{0^2+12^2+9^2}=15\). 5. Thus, \(V=\frac{1}{3}\pi(5^2)(15)=125\pi\approx 392.70\).

a) \(F(4, 11, 11)\) b) \(5\) units c) \(125\pi\approx 392.70\) cubic units

- Use two vectors in each plane to find normal vectors. - Apply the angle formula to the normal vectors. - Simplify the expression before solving for \(t\). - Use the condition \(t>2\) to choose the correct square-root branch.

1. A normal vector to \(E\) is \(\vec{n}_E=\begin{pmatrix}-1\\1\\0\end{pmatrix}\). A normal vector to \(F_t\) is \(\vec{n}_t=\begin{pmatrix}t-2\\-(t-2)\\2\end{pmatrix}\). 2. Therefore, \(\cos\alpha=\frac{|\vec{n}_E\cdot\vec{n}_t|}{\|\vec{n}_E\|\|\vec{n}_t\|}=\frac{|t-2|}{\sqrt{t^2-4t+6}}\). 3. For \(\alpha=30^\circ\), squaring \(\frac{|t-2|}{\sqrt{(t-2)^2+2}}=\frac{\sqrt{3}}{2}\) gives \((t-2)^2=6\). 4. Since \(t>2\), \(t=2+\sqrt{6}\).

a) \(\cos\alpha=\frac{|t-2|}{\sqrt{t^2-4t+6}}\) b) \(t=2+\sqrt{6}\)

- Use the line-plane angle formula for part a. - For each side face, use two in-plane directions and solve zero dot-product conditions to find a normal vector. - Express the plane angle using the dot product of normals. - Analyze the cosine expression as \(k\) grows.

1. For \(k=4\), a direction vector for \(AS_4\) is \(\begin{pmatrix}-2\\-2\\4\end{pmatrix}\). With base normal \(\begin{pmatrix}0\\0\\1\end{pmatrix}\), \(\sin\alpha=\frac{4}{\sqrt{24}}=\frac{2}{\sqrt{6}}\). Thus \(\alpha\approx54.7^\circ\). 2. For face \(ABS_k\), use \(\overrightarrow{AB}=\begin{pmatrix}-4\\0\\0\end{pmatrix}\) and \(\overrightarrow{AS_k}=\begin{pmatrix}-2\\-2\\k\end{pmatrix}\). A vector perpendicular to both satisfies \(a=0\) and \(-2b+kc=0\), so choose \(\vec{n}_1=\begin{pmatrix}0\\k\\2\end{pmatrix}\). For face \(BCS_k\), use \(\overrightarrow{BC}=\begin{pmatrix}0\\-4\\0\end{pmatrix}\) and \(\overrightarrow{BS_k}=\begin{pmatrix}2\\-2\\k\end{pmatrix}\). A perpendicular vector satisfies \(b=0\) and \(2a+kc=0\), so choose \(\vec{n}_2=\begin{pmatrix}-k\\0\\2\end{pmatrix}\). 3. Therefore, \(\cos\phi=\frac{4}{k^2+4}\). Setting \(\phi=60^\circ\) gives \(\frac{1}{2}=\frac{4}{k^2+4}\), so \(k=2\). 4. As \(k\to\infty\), \(\frac{4}{k^2+4}\to0\), so \(\phi\to90^\circ\). For every finite \(k>0\), the cosine remains positive, so the angle is always less than \(90^\circ\).

a) \(54.7^\circ\) b) \(k=2\) c) \(\cos\phi=\frac{4}{k^2+4}\to0\), so \(\phi\to90^\circ\), but \(\phi<90^\circ\) for every finite \(k>0\).

- Find a normal vector for each face. - Use the dot-product angle formula for planes. - Compare \(\cos^2\phi\) with \(\cos^245^\circ\). - Use that the requested angle is acute.

1. For \(a=4\), normal vectors to \(OPR_a\) and \(OQR_a\) may be simplified to \(\vec{n}_1=\begin{pmatrix}0\\-1\\1\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}1\\0\\-1\end{pmatrix}\). Thus \(\cos\gamma=\frac{1}{2}\), so \(\gamma=60^\circ\). 2. For part b, use \(\vec{n}_{OPR}=\begin{pmatrix}0\\-a\\4\end{pmatrix}\) and \(\vec{n}_{PQR}=\begin{pmatrix}a\\a\\-4\end{pmatrix}\). 3. The angle satisfies \(\cos^2\phi=\frac{a^2+16}{2a^2+16}=\frac{1}{2}+\frac{8}{2a^2+16}\). 4. Since \(a>0\), the second term is positive, so \(\cos^2\phi>\frac{1}{2}\). Because \(\phi\) is acute, \(\cos\phi>\frac{1}{\sqrt{2}}=\cos45^\circ\), and therefore \(\phi<45^\circ\).

a) \(60^\circ\) b) Since \(\cos^2\phi=\frac{1}{2}+\frac{8}{2a^2+16}>\frac{1}{2}\), it follows that \(\phi<45^\circ\) for every \(a>0\).

- Complete the square to find the sphere's center and radius. - Use the plane's normal vector to find the center-to-plane distance. - The sphere radius, cross-section radius, and center-to-plane distance form a right triangle. - The cross-section's center is the perpendicular projection of the sphere's center onto the plane.

1. Complete the square: \((x-1)^2+(y+2)^2+(z-3)^2=49\). Thus, the sphere has center \(M=(1, -2, 3)\) and radius \(7\). 2. The plane has normal vector \(\mathbf{n}=\langle3, 0, -4\rangle\), whose magnitude is \(5\). The distance from \(M\) to the plane is \(d=\frac{|3(1)-4(3)-12|}{5}=\frac{21}{5}\). 3. Since \(\frac{21}{5}<7\), the plane intersects the sphere in a circle. Its radius is \(\rho=\sqrt{7^2-\left(\frac{21}{5}\right)^2}=\frac{28}{5}=5.6\). 4. The center \(Z\) is the perpendicular projection of \(M\) onto the plane. Write \(Z=M+\lambda\mathbf{n}=(1+3\lambda, -2, 3-4\lambda)\). 5. Substitution into \(3x-4z=12\) gives \(25\lambda=21\), so \(\lambda=\frac{21}{25}\). Therefore, \(Z=\left(\frac{88}{25}, -2, -\frac{9}{25}\right)=(3.52, -2, -0.36)\).

The plane intersects the sphere in a circle. \(\rho=\frac{28}{5}=5.6\) \(Z=\left(\frac{88}{25}, -2, -\frac{9}{25}\right)=(3.52, -2, -0.36)\)

- Divide each plane equation by the magnitude of its normal vector. - Use consistent signs before adding and subtracting. - Plane normals are perpendicular when their dot product is \(0\). - Points on an angle bisector are equidistant from the original planes.

1. Both plane normals have magnitude \(3\). Choose normalized equations \(E_1: \frac{2}{3}x+\frac{2}{3}y-\frac{1}{3}z-2=0\) and \(E_2: -\frac{1}{3}x+\frac{2}{3}y-\frac{2}{3}z-1=0\). 2. Adding gives \(\frac{1}{3}x+\frac{4}{3}y-z-3=0\), so \(W_1: x+4y-3z-9=0\). 3. Subtracting gives \(x+\frac{1}{3}z-1=0\), so \(W_2: 3x+z-3=0\). 4. Normal vectors to \(W_1\) and \(W_2\) are \(\langle 1,4,-3\rangle\) and \(\langle 3,0,1\rangle\). Their dot product is \(3-3=0\), so the planes are perpendicular. 5. Substituting \(P\) into \(W_2\) gives \(3-3=0\), so \(P\in W_2\). 6. The distances are \(d(P,E_1)=\frac{\lvert -3-6\rvert}{3}=3\) and \(d(P,E_2)=\frac{\lvert 2(3)+3\rvert}{3}=3\). The distances are equal, as expected for a point on an angle-bisector plane.

a) \(E_1: \frac{2}{3}x+\frac{2}{3}y-\frac{1}{3}z-2=0\); \(E_2: -\frac{1}{3}x+\frac{2}{3}y-\frac{2}{3}z-1=0\) b) \(W_1: x+4y-3z-9=0\); \(W_2: 3x+z-3=0\) c) \(W_1\perp W_2\) d) \(P\in W_2\), and both distances are \(3\) units.

- Use the angle between the plane normals. - Normalize both plane equations before adding and subtracting. - Find one point satisfying both plane equations. - The intersection direction is perpendicular to both plane normals. - To show a line lies in a plane, test one point and its direction vector.

1. Normal vectors are \(\mathbf{n}_1=\langle 4,-4,2\rangle\) and \(\mathbf{n}_2=\langle 4,7,4\rangle\), with magnitudes \(6\) and \(9\). 2. Thus, \(\cos\alpha=\frac{\lvert\mathbf{n}_1\cdot\mathbf{n}_2\rvert}{\lVert\mathbf{n}_1\rVert\lVert\mathbf{n}_2\rVert}=\frac{4}{54}=\frac{2}{27}\), so \(\alpha\approx85.75^\circ\). 3. Normalized equations are \(\frac{2}{3}x-\frac{2}{3}y+\frac{1}{3}z-2=0\) and \(\frac{4}{9}x+\frac{7}{9}y+\frac{4}{9}z-1=0\). 4. Adding gives \(W_1: 10x+y+7z-27=0\). Subtracting gives \(W_2: 2x-13y-z-9=0\). 5. A point on both original planes is \(P(0,-1,4)\). A direction vector perpendicular to both plane normals is \(\mathbf{d}=\langle 15,4,-22\rangle\). Therefore, \(g: (x,y,z)=(0,-1,4)+t\langle 15,4,-22\rangle\). 6. Point \(P\) satisfies \(W_1\), and \(\mathbf{d}\cdot\langle 10,1,7\rangle=150+4-154=0\). Therefore, every point of \(g\) lies in \(W_1\).

a) \(\alpha\approx85.75^\circ\) b) \(W_1: 10x+y+7z-27=0\); \(W_2: 2x-13y-z-9=0\) c) \(g: (x,y,z)=(0,-1,4)+t\langle 15,4,-22\rangle\), and \(g\subset W_1\).

- Compare the center-to-plane distance with the sphere's radius. - Use the right triangle formed by the sphere radius, plane distance, and cross-section radius. - The intersection circle's center is the projection of the sphere's center onto the plane. - A line in the plane misses the sphere when it lies outside the intersection circle.

1. The sphere has center \(M=(-1, 4, 2)\) and radius \(10\). The plane's normal vector is \(\mathbf{n}=\langle3, 0, -4\rangle\), whose magnitude is \(5\). 2. The distance from \(M\) to the plane is \(d=\frac{|3(-1)-4(2)+31|}{5}=4\). Since \(4<10\), the intersection is a circle. 3. Its radius is \(r_S=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}\approx9.17\). 4. The circle's center is the perpendicular projection of \(M\) onto \(E\). Write \(M_S=M+t\mathbf{n}\). Substitution gives \(20+25t=0\), so \(t=-\frac45\) and \(M_S=\left(-\frac{17}{5}, 4, \frac{26}{5}\right)\). 5. The vectors \(\langle0, 1, 0\rangle\) and \(\langle4, 0, 3\rangle\) both lie parallel to \(E\) and are perpendicular to each other. Let \(P=M_S+2\langle4, 0, 3\rangle=\left(\frac{23}{5}, 4, \frac{56}{5}\right)\). 6. One possible line is \(g:\mathbf{x}=P+s\langle0, 1, 0\rangle\). Its distance from \(M_S\) is \(10>2\sqrt{21}\), so it lies outside the intersection circle and does not meet the sphere.

a) \(d(M,E)=4<10\), so the intersection is a circle. b) \(r_S=2\sqrt{21}\approx9.17\) c) One possible line is \(g:\mathbf{x}=\left\langle\frac{23}{5}, 4, \frac{56}{5}\right\rangle+s\langle0, 1, 0\rangle\). Its distance from the intersection circle's center is \(10>r_S\).

- Complete the square to write the sphere in standard form. - Compare the center-to-plane distance with the radius. - The nearest point on the sphere lies along the perpendicular from the center toward the plane. - Subtract the radius from the center-to-plane distance.

1. Complete the square: \(x^2+(y-3)^2+(z+4)^2=25\). Thus, \(M=(0, 3, -4)\) and \(r=5\). 2. The plane's normal vector is \(\mathbf{n}=\langle2, 1, -2\rangle\), whose magnitude is \(3\). The distance from \(M\) to the plane is \(d=\frac{|2(0)+3-2(-4)+5|}{3}=\frac{16}{3}\). 3. Since \(\frac{16}{3}>5\), the plane does not intersect the sphere. 4. The point on the sphere closest to the plane lies from \(M\) in the direction opposite \(\mathbf{n}\), because the plane expression is positive at \(M\). Therefore, \(P=M-5\frac{\mathbf{n}}{\|\mathbf{n}\|}\). 5. This gives \(P=(0, 3, -4)-\frac53\langle2, 1, -2\rangle=\left(-\frac{10}{3}, \frac43, -\frac23\right)\). 6. The minimum distance from the sphere to the plane is \(d-r=\frac{16}{3}-5=\frac13\).

a) \(M=(0, 3, -4)\) and \(r=5\) b) The plane does not intersect the sphere because \(\frac{16}{3}>5\). c) \(P=\left(-\frac{10}{3}, \frac43, -\frac23\right)\), and the minimum distance is \(\frac13\).

- Find both centers and exact radii. - Compare the center distance with the sum and difference of the radii. - The tangent point lies on the line through the two centers. - The common tangent plane is perpendicular to that line.

1. The centers are \(M_1=(1, 2, 3)\) and \(M_2=(3, 4, 5)\). The radii are \(r_1=4\sqrt3\) and \(r_2=2\sqrt3\). 2. The center distance is \(M_1M_2=\sqrt{2^2+2^2+2^2}=2\sqrt3\). Since this equals \(|r_1-r_2|\), the spheres are internally tangent. 3. The point of tangency lies on the line of centers. Because \(r_1=2M_1M_2\), move twice the vector \(\overrightarrow{M_1M_2}=\langle2, 2, 2\rangle\) from \(M_1\): \(B=(1, 2, 3)+2\langle2, 2, 2\rangle=(5, 6, 7)\). 4. The tangent plane is perpendicular to the line of centers, so a normal vector is \(\langle1, 1, 1\rangle\). 5. The plane through \(B\) is \((x-5)+(y-6)+(z-7)=0\), or \(x+y+z-18=0\).

a) The spheres are internally tangent because \(M_1M_2=|r_1-r_2|=2\sqrt3\). b) \(B=(5, 6, 7)\) c) \(x+y+z-18=0\)

- The radius equals the center-to-plane distance. - The point of tangency is the perpendicular projection of the center onto the plane. - Use the plane's normal vector as the direction of the perpendicular line. - Under reflection, the plane is the perpendicular bisector of the segment joining the original and reflected centers.

1. The radius equals the distance from \(M\) to the tangent plane: \(r=\frac{|1-10|}{\sqrt{1^2+(-2)^2+2^2}}=\frac{9}{3}=3\). 2. The point of tangency is the perpendicular projection of \(M\) onto \(E\). Write \(B=(1, 0, 0)+t\langle1, -2, 2\rangle\). 3. Substitution into the plane equation gives \(1+9t-10=0\), so \(t=1\). Therefore, \(B=(2, -2, 2)\). 4. The reflected center lies the same distance beyond \(B\), so \(M'=M+2\overrightarrow{MB}\). 5. Since \(\overrightarrow{MB}=\langle1, -2, 2\rangle\), \(M'=(1, 0, 0)+2\langle1, -2, 2\rangle=(3, -4, 4)\).

a) \(r=3\) b) \(B=(2, -2, 2)\) c) \(M'=(3, -4, 4)\)

- Find a vector orthogonal to both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) using dot products. - Use one of the given points to determine the plane's constant. - The segment joining a point and its reflection is perpendicular to the plane. - The perpendicular foot is the midpoint of the point and its reflection.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle -2,2,0\rangle\) and \(\overrightarrow{AC}=\langle 0,2,-2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. The conditions \(-2a+2b=0\) and \(2b-2c=0\) give \(a=b=c\), so choose \(\mathbf{n}=\langle 1,1,1\rangle\). 3. Using point \(A\), the plane equation is \(x+y+z=7\). 4. The line through \(P\) perpendicular to the plane is \((x,y,z)=(5,6,8)+t\langle 1,1,1\rangle\). 5. At its intersection with the plane, \((5+t)+(6+t)+(8+t)=7\), so \(t=-4\). The perpendicular foot is \(F(1,2,4)\). 6. Since \(F\) is the midpoint of \(\overline{PP'}\), \(P'=2F-P=(-3,-2,0)\).

a) \(E: x+y+z=7\) b) \(P'=(-3,-2,0)\)

- Identify a direction vector for the line and a normal vector for the plane. - Use the dot product formula for the sine of the angle between a line and a plane. - Square the expression to remove the absolute value. - Rewrite the difference between a possible upper bound and the squared sine as a nonnegative expression.

1. A direction vector for the line is \(\mathbf{v}=\langle 1, 1, 0\rangle\), and a normal vector for the plane is \(\mathbf{n}_k=\langle k, 1, 1\rangle\). 2. For the acute angle between a line and a plane, \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}_k|}{\|\mathbf{v}\|\,\|\mathbf{n}_k\|}=\frac{|k+1|}{\sqrt{2}\sqrt{k^2+2}}\). 3. Squaring removes the absolute value: \(\sin^2\alpha=\frac{(k+1)^2}{2(k^2+2)}\). 4. Compare this expression with \(\frac{3}{4}\): \(\frac{3}{4}-\sin^2\alpha=\frac{3(k^2+2)-2(k+1)^2}{4(k^2+2)}=\frac{(k-2)^2}{4(k^2+2)}\ge0\). 5. Therefore, \(\sin^2\alpha\le\frac{3}{4}\), with equality exactly when \(k=2\). Then \(\sin\alpha=\frac{\sqrt{3}}{2}\), so \(\alpha=60^\circ\).

The angle is maximized when \(k=2\), and the maximum angle is \(60^\circ\).

- Test for parallel direction vectors and then for an intersection. - Find a vector orthogonal to both line directions. - The shortest connecting segment is perpendicular to both lines. - A tangent sphere's radius equals the distance from its center to the line.

1. The direction vectors \(\langle 1,2,2\rangle\) and \(\langle 2,-2,1\rangle\) are not scalar multiples. Equating coordinates produces an inconsistent system, so the lines are skew. 2. A vector perpendicular to both directions is \(\mathbf{n}=\langle 2,1,-2\rangle\), since both dot products are \(0\). Its magnitude is \(3\). 3. The vector between the given points is \(\mathbf{w}=\langle 3,4,6\rangle\). Thus, \(d(g,h)=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{\lvert 6+4-12\rvert}{3}=\frac{2}{3}\). 4. Let \(F_g=(2,0,1)+r\langle 1,2,2\rangle\) and \(F_h=(5,4,7)+s\langle 2,-2,1\rangle\). The shortest connecting vector \(F_h-F_g\) is perpendicular to both line directions. Solving \((F_h-F_g)\cdot\langle 1,2,2\rangle=0\) and \((F_h-F_g)\cdot\langle 2,-2,1\rangle=0\) gives \(r=\frac{23}{9}\) and \(s=-\frac{4}{9}\). 5. Therefore, \(F_g=\left(\frac{41}{9},\frac{46}{9},\frac{55}{9}\right)\) and \(F_h=\left(\frac{37}{9},\frac{44}{9},\frac{59}{9}\right)\). 6. For the tangent sphere, let \(G=(2+r,2r,1+2r)\) be a point on \(g\). The perpendicular-foot condition is \((G-M)\cdot\langle 1,2,2\rangle=0\), which gives \(9r-15=0\), so \(r=\frac{5}{3}\). Then \(G-M=\left\langle-\frac{4}{3},\frac{1}{3},\frac{1}{3}\right\rangle\), and the sphere's radius is \(\sqrt{\frac{16}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{2}\).

a) The lines are skew. b) \(\frac{2}{3}\) unit c) \(F_g=\left(\frac{41}{9},\frac{46}{9},\frac{55}{9}\right)\), \(F_h=\left(\frac{37}{9},\frac{44}{9},\frac{59}{9}\right)\) d) \(r=\sqrt{2}\) units

- The tangent radius equals the shortest distance from the center to the line. - Use a dot product to locate the point on the line closest to the center. - If the radius is greater than that distance, the line intersects the sphere twice. - Substitute the line into the sphere equation to find the intersection parameters.

1. A general point on \(g\) is \(P(t)=(2+t, 3+2t, 1+2t)\). Then \(\overrightarrow{MP(t)}=\langle t-3, 2t, 2t-3\rangle\). 2. At the closest point, this vector is perpendicular to the direction vector \(\langle1, 2, 2\rangle\). Thus, \((t-3)+2(2t)+2(2t-3)=9t-9=0\), so \(t=1\). 3. The closest point is \((3, 5, 3)\), whose distance from \(M\) is \(\sqrt{(-2)^2+2^2+(-1)^2}=3\). Therefore, the tangent sphere has radius \(r=3\). 4. For \(r=5\), substitute the line into \((x-5)^2+(y-3)^2+(z-4)^2=25\). This gives \(9t^2-18t-7=0\). 5. Solving gives \(t=-\frac13\) or \(t=\frac73\). The corresponding points are \(\left(\frac53, \frac73, \frac13\right)\) and \(\left(\frac{13}{3}, \frac{23}{3}, \frac{17}{3}\right)\). 6. Since there are two intersection points, \(g\) is a secant when \(r=5\).

a) \(r=3\) b) For \(r=5\), \(g\) is a secant. The intersection points are \(\left(\frac53, \frac73, \frac13\right)\) and \(\left(\frac{13}{3}, \frac{23}{3}, \frac{17}{3}\right)\).

- Find the perpendicular foot using a dot product. - A tangent sphere's radius is the center-to-line distance. - For part d), substitute a general point on the line into the distance formula. - The resulting quadratic can have two solutions.

1. A general point on \(h\) is \(F(k)=(1+k, 1+2k, 1+2k)\), so \(\overrightarrow{AF}=\langle k-1, 2k-4, 2k\rangle\). 2. Require \(\overrightarrow{AF}\cdot\langle 1, 2, 2\rangle=0\): \((k-1)+2(2k-4)+2(2k)=9k-9=0\), so \(k=1\) and \(F=(2, 3, 3)\). 3. The distance is \(AF=\sqrt{0^2+(-2)^2+2^2}=2\sqrt{2}\approx 2.83\). 4. The tangent sphere has radius squared \(8\), so its equation is \((x-2)^2+(y-5)^2+(z-1)^2=8\). 5. For a point on \(h\) to be \(3\) units from \(A\), require \((k-1)^2+(2k-4)^2+(2k)^2=9\). This simplifies to \(9k^2-18k+8=0\), giving \(k=\frac{2}{3}\) or \(k=\frac{4}{3}\). 6. The points are \(\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right)\) and \(\left(\frac{7}{3}, \frac{11}{3}, \frac{11}{3}\right)\).

a) \(F(2, 3, 3)\) b) \(2\sqrt{2}\approx 2.83\) units c) \((x-2)^2+(y-5)^2+(z-1)^2=8\) d) \(\left(\frac{5}{3}, \frac{7}{3}, \frac{7}{3}\right)\) and \(\left(\frac{7}{3}, \frac{11}{3}, \frac{11}{3}\right)\)

- Compare \(AB\) and \(AD\) using the distance formula. - The altitude from \(D\) to \(AB\) is perpendicular to \(AB\). - Use the parallelogram relationship to find the fourth vertex. - A rhombus's diagonals bisect each other. - Use base times height for the area.

1. \(\overrightarrow{AB}=\langle 3, 4, 0\rangle\) and \(\overrightarrow{AD}=\langle -3, 4, 0\rangle\), so \(AB=AD=5\). Therefore, triangle \(ABD\) is isosceles. 2. Line \(AB\) is \(\mathbf{x}=\langle 1, 2, 1\rangle+r\langle 3, 4, 0\rangle\). A general point is \(L=(1+3r, 2+4r, 1)\). 3. Require \(\overrightarrow{DL}\cdot\langle 3, 4, 0\rangle=0\). This gives \(3(3+3r)+4(-4+4r)=25r-7=0\), so \(r=\frac{7}{25}\). 4. Thus, \(L=\left(\frac{46}{25}, \frac{78}{25}, 1\right)\), and \(DL=\frac{24}{5}=4.8\). 5. Since a rhombus is a parallelogram, \(C=B+D-A=(1, 10, 1)\). The diagonals bisect each other, so \(M=\frac{A+C}{2}=(1, 6, 1)\). 6. Using base \(AB=5\) and height \(DL=\frac{24}{5}\), the area is \(5\left(\frac{24}{5}\right)=24\).

a) \(AB=AD=5\), so triangle \(ABD\) is isosceles. b) \(L\left(\frac{46}{25}, \frac{78}{25}, 1\right)\); distance \(\frac{24}{5}=4.8\) units c) \(C(1, 10, 1)\); \(M(1, 6, 1)\); area \(24\) square units

- Use a dot product to express perpendicularity. - The point-to-line distance is the length of the perpendicular segment. - For part c), substitute a general point on the line into the distance formula. - Expect two parameter values from the quadratic equation.

1. A general point on \(g\) is \(G(k)=(1+2k, 1+2k, k)\), so \(\overrightarrow{PG}=\langle 2k-4, 2k-3, k-4\rangle\). 2. Require \(\overrightarrow{PG}\cdot\langle 2, 2, 1\rangle=0\): \(2(2k-4)+2(2k-3)+(k-4)=9k-18=0\), so \(k=2\) and \(L=(5, 5, 2)\). 3. The distance is \(PL=\sqrt{0^2+1^2+(-2)^2}=\sqrt{5}\approx 2.24\). 4. For part c), require \((2k-4)^2+(2k-3)^2+(k-4)^2=41\). This simplifies to \(9k(k-4)=0\), so \(k=0\) or \(k=4\). 5. The two points are \(Q_1=(1, 1, 0)\) and \(Q_2=(9, 9, 4)\).

a) \(L(5, 5, 2)\) b) \(\sqrt{5}\approx 2.24\) units c) \(Q_1(1, 1, 0)\) and \(Q_2(9, 9, 4)\)

- Test each possible vertex with a dot product. - Exclude values that make two points coincide. - Solve each resulting equation separately.

1. A triangle requires \(k\ne0\). 2. A right angle at \(O\) would require \(\overrightarrow{OA}\cdot\overrightarrow{OB}=k=0\), which is degenerate. 3. At \(A\), \(\overrightarrow{AO}\cdot\overrightarrow{AB} =\begin{pmatrix}-1\\-2\\-2\end{pmatrix}\cdot \begin{pmatrix}k-1\\-2\\-2\end{pmatrix} =9-k\). Thus, \(k=9\). 4. At \(B\), \(\overrightarrow{BO}\cdot\overrightarrow{BA} =\begin{pmatrix}-k\\0\\0\end{pmatrix}\cdot \begin{pmatrix}1-k\\2\\2\end{pmatrix} =k^2-k\). The nondegenerate solution is \(k=1\). 5. Therefore, \(k=1\) or \(k=9\).

\(k=1\) or \(k=9\)

- At each vertex, use two vectors that point away from that vertex. - The sign of their dot product classifies the angle. - Use the dot-product area identity for the two side vectors from one vertex.

1. \(\overrightarrow{DE}=(2, 3, 2)\), \(\overrightarrow{DF}=(-2, 2, 3)\), and \(\overrightarrow{EF}=(-4, -1, 1)\). 2. At \(D\), \(\overrightarrow{DE}\cdot\overrightarrow{DF}=8>0\). 3. At \(E\), \(\overrightarrow{ED}\cdot\overrightarrow{EF}=9>0\). 4. At \(F\), \(\overrightarrow{FD}\cdot\overrightarrow{FE}=9>0\). Thus, all three angles are acute. 5. For the area, \(\|\overrightarrow{DE}\|^2=17\), \(\|\overrightarrow{DF}\|^2=17\), and \(\overrightarrow{DE}\cdot\overrightarrow{DF}=8\). 6. The squared parallelogram area is \(\|\overrightarrow{DE}\|^2\|\overrightarrow{DF}\|^2-(\overrightarrow{DE}\cdot\overrightarrow{DF})^2=17^2-8^2=225\). 7. The triangle's area is \(\frac12\sqrt{225}=7.5\).

a) The triangle is acute. b) \(7.5\) square units

- Rearrange the equality in part a and interpret a zero dot product. - Expand the squared magnitude in part b using the dot product. - Use the angle formula for the dot product of unit vectors.

1. Statement a is false. The equation implies \(\vec{a}\cdot(\vec{b}-\vec{c})=0\), so \(\vec{b}-\vec{c}\) may be any vector perpendicular to \(\vec{a}\). For example, \(\vec{a}=(1, 0, 0)\), \(\vec{b}=(0, 1, 0)\), and \(\vec{c}=(0, 0, 0)\) satisfy the equation but \(\vec{b}\ne\vec{c}\). 2. Statement b is true. Expanding gives \(\|\vec{u}+\vec{v}\|^2=\|\vec{u}\|^2+2\vec{u}\cdot\vec{v}+\|\vec{v}\|^2\). This equals \(\|\vec{u}\|^2+\|\vec{v}\|^2\) exactly when \(\vec{u}\cdot\vec{v}=0\). 3. Statement c is true. For unit vectors, \(\vec{e}_1\cdot\vec{e}_2=\|\vec{e}_1\|\|\vec{e}_2\|\cos\theta=\cos\theta\), which lies in \([-1,1]\).

a) False. b) True. c) True.

- Express each diagonal using the side vectors. - Expand squared magnitudes. - For the geometric proof, compare triangles formed by the diagonals.

1. The diagonal vectors are \(\overrightarrow{AC}=\vec{a}+\vec{b}\) and \(\overrightarrow{BD}=\vec{b}-\vec{a}\). 2. Equal diagonal lengths give \(\|\vec{a}+\vec{b}\|^2=\|\vec{b}-\vec{a}\|^2\). 3. Expanding yields \(\|\vec{a}\|^2+2\vec{a}\cdot\vec{b}+\|\vec{b}\|^2 =\|\vec{b}\|^2-2\vec{a}\cdot\vec{b}+\|\vec{a}\|^2\). 4. Thus, \(4\vec{a}\cdot\vec{b}=0\), so the adjacent sides are perpendicular. The reverse implication follows by reversing the argument. 5. Geometrically, a rectangle's diagonals are congruent because the right triangles formed by a diagonal have matching legs. Conversely, if a parallelogram's diagonals are congruent, triangles \(ABC\) and \(BAD\) are congruent by SSS. Their corresponding adjacent angles are equal and supplementary, so each is \(90^\circ\).

a) Equal diagonals are equivalent to \(\vec{a}\cdot\vec{b}=0\), so the parallelogram is a rectangle. b) Congruent-triangle arguments prove both directions geometrically.

- Write a general point on each line. - The shortest connecting segment is perpendicular to both line directions. - Use two dot-product equations to solve for the parameters.

1. Write \(P=(1+r,r,1)\) and \(Q=(0,1+s,4+s)\). Then \(\overrightarrow{PQ}=\langle -1-r,1+s-r,3+s\rangle\). 2. For the shortest connection, \(\overrightarrow{PQ}\) must be perpendicular to both line directions. 3. From \(\overrightarrow{PQ}\cdot\langle 1,1,0\rangle=0\), obtain \(-2r+s=0\). 4. From \(\overrightarrow{PQ}\cdot\langle 0,1,1\rangle=0\), obtain \(-r+2s=-4\). 5. Solving gives \(r=-\frac{4}{3}\) and \(s=-\frac{8}{3}\). Therefore, \(P=\left(-\frac{1}{3},-\frac{4}{3},1\right)\) and \(Q=\left(0,-\frac{5}{3},\frac{4}{3}\right)\). 6. The connecting vector is \(\left\langle\frac{1}{3},-\frac{1}{3},\frac{1}{3}\right\rangle\), so the minimum distance is \(\frac{\sqrt{3}}{3}\approx0.577\).

\(P=\left(-\frac{1}{3},-\frac{4}{3},1\right)\), \(Q=\left(0,-\frac{5}{3},\frac{4}{3}\right)\), and \(d=\frac{\sqrt{3}}{3}\approx0.577\) unit

- Find a vector perpendicular to both line directions. - Form a vector between one point on each line. - Project that vector onto the common perpendicular direction.

1. A vector perpendicular to both direction vectors \(\langle 1,1,0\rangle\) and \(\langle 0,1,1\rangle\) is \(\mathbf{n}=\langle 1,-1,1\rangle\), with magnitude \(\sqrt{3}\). 2. The vector between the given points is \(\mathbf{w}=\langle 1,-2,1\rangle\). 3. The shortest distance is the magnitude of the scalar projection of \(\mathbf{w}\) onto \(\mathbf{n}\): \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\approx2.31\).

\(\frac{4\sqrt{3}}{3}\approx2.31\) units

- Find a direction perpendicular to both laser paths. - Project the vector between the paths onto that direction. - For the closest points, require the connecting vector to be perpendicular to both paths.

1. A vector perpendicular to both direction vectors is \(\mathbf{n}=\langle 1,-1,1\rangle\), with magnitude \(\sqrt{3}\). 2. The vector between the given points is \(\langle 2,0,2\rangle\). Therefore, the distance is \(d=\frac{\lvert\langle 2,0,2\rangle\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\approx2.31\,\text{m}\). 3. Let \(P=(r,r,0)\) and \(Q=(2,s,2+s)\). Then \(\overrightarrow{PQ}=\langle 2-r,s-r,2+s\rangle\). 4. The shortest connecting vector is perpendicular to both line directions. Solving \(\overrightarrow{PQ}\cdot\langle 1,1,0\rangle=0\) and \(\overrightarrow{PQ}\cdot\langle 0,1,1\rangle=0\) gives \(r=\frac{2}{3}\) and \(s=-\frac{2}{3}\). 5. Thus, \(P=\left(\frac{2}{3},\frac{2}{3},0\right)\) and \(Q=\left(2,-\frac{2}{3},\frac{4}{3}\right)\).

a) \(\frac{4\sqrt{3}}{3}\approx2.31\,\text{m}\) b) \(P=\left(\frac{2}{3},\frac{2}{3},0\right)\), \(Q=\left(2,-\frac{2}{3},\frac{4}{3}\right)\)

- Represent the space diagonals with different sign combinations of \(a\), \(b\), and \(c\). - Set a dot product equal to zero. - For part b), compare the conditions produced by different pairs.

1. Direction vectors for two space diagonals can be written as \(\vec{d}_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\) and \(\vec{d}_2=\begin{pmatrix}-a\\b\\c\end{pmatrix}\). 2. They are perpendicular when \(\vec{d}_1\cdot\vec{d}_2=-a^2+b^2+c^2=0\). Thus, one possible condition is \(a^2=b^2+c^2\). For other pairs, the analogous conditions are \(b^2=a^2+c^2\) or \(c^2=a^2+b^2\). 3. If all four diagonals were pairwise perpendicular, these three equations would have to hold simultaneously. 4. Combining \(a^2=b^2+c^2\) and \(b^2=a^2+c^2\) gives \(2c^2=0\), so \(c=0\). 5. This contradicts the requirement that every edge length be positive. Therefore, no rectangular prism has four pairwise-perpendicular space diagonals.

a) Two space diagonals can be perpendicular when one squared edge length equals the sum of the other two, such as \(a^2=b^2+c^2\). b) No. Requiring all relevant orthogonality conditions simultaneously forces an edge length to be \(0\), which is impossible for a rectangular prism.

- Express perpendicularity with a dot product. - Treat \(\vec{u}+\vec{v}\) as one vector and use a difference-of-squares pattern. - For part b), first find \(\vec{u}+\vec{v}\).

1. The diagonals are perpendicular exactly when \(\vec{d}_1\cdot\vec{d}_2=0\). 2. Let \(\vec{p}=\vec{u}+\vec{v}\). Then \((\vec{p}+\vec{w})\cdot(\vec{p}-\vec{w}) =\|\vec{p}\|^2-\|\vec{w}\|^2\). 3. Therefore, \(\vec{d}_1\cdot\vec{d}_2=0\) if and only if \(\|\vec{u}+\vec{v}\|^2=\|\vec{w}\|^2\). 4. Since magnitudes are nonnegative, this is equivalent to \(\|\vec{w}\|=\|\vec{u}+\vec{v}\|\). 5. For the given vectors, \(\vec{u}+\vec{v}=\begin{pmatrix}4\\2\\0\end{pmatrix}\), so \(\|\vec{w}\|=\sqrt{4^2+2^2}=2\sqrt{5}\).

a) \(\vec{d}_1\perp\vec{d}_2\) if and only if \(\|\vec{w}\|=\|\vec{u}+\vec{v}\|\). b) \(\|\vec{w}\|=2\sqrt{5}\), approximately \(4.47\).

- Compare adjacent side lengths and use a dot product to check for a right angle. - For each distance, write a general point on the relevant line. - The shortest connecting vector must be orthogonal to the line's direction vector.

1. \(\overrightarrow{AB}=\langle 6,0,0\rangle\) and \(\overrightarrow{AD}=\langle 0,6,0\rangle\). They have equal magnitude \(6\) and dot product \(0\). Also, opposite side vectors are equal, so \(ABCD\) is a square with side length \(6\,\text{m}\). 2. A general point on \(AB\) is \(F(t)=(2+6t,2,0)\). The condition \(\overrightarrow{SF}\cdot\overrightarrow{AB}=0\) gives \(t=\frac{1}{2}\), so \(F=(5,2,0)\). Thus, the distance is \(SF=\sqrt{0^2+(-3)^2+(-10)^2}=\sqrt{109}\approx 10.44\,\text{m}\). 3. A general point on \(CS\) is \(G(r)=(8-3r,8-3r,10r)\). Its direction vector is \(\langle -3,-3,10\rangle\). 4. Require \(\overrightarrow{AG}\cdot\langle -3,-3,10\rangle=0\). This gives \(r=\frac{18}{59}\), so \(G=\left(\frac{418}{59},\frac{418}{59},\frac{180}{59}\right)\). 5. Therefore, \(AG=\frac{60}{\sqrt{59}}\approx 7.81\,\text{m}\).

a) Adjacent side vectors have length \(6\,\text{m}\) and dot product \(0\), and opposite sides are parallel and equal. b) \(\sqrt{109}\approx 10.44\,\text{m}\) c) \(\frac{60}{\sqrt{59}}\approx 7.81\,\text{m}\)

- Determine whether the three vectors from \(O\) can lie in one plane. - Use a dot product to find the perpendicular foot on \(QR\). - For the face area, use \(\frac{1}{2}\times\text{base}\times\text{height}\).

1. The vectors \(\overrightarrow{OP}=\langle 0,0,4\rangle\), \(\overrightarrow{OQ}=\langle 6,0,0\rangle\), and \(\overrightarrow{OR}=\langle 0,8,0\rangle\) point along three different coordinate axes. They do not lie in one plane through \(O\), so the four points are noncoplanar and form a tetrahedron. 2. A general point on \(QR\) is \(F(t)=(6-6t,8t,0)\). For the perpendicular from \(O\), require \(\overrightarrow{OF}\cdot\langle -6,8,0\rangle=0\). This gives \(t=\frac{9}{25}\), so \(F=\left(\frac{96}{25},\frac{72}{25},0\right)\). 3. Thus, the distance from \(O\) to \(QR\) is \(OF=\frac{24}{5}=4.8\). 4. The same parameter gives the perpendicular foot from \(P\) because \(P\) differs from \(O\) only in the z-coordinate, while line \(QR\) lies in the xy-plane. The altitude from \(P\) to \(QR\) is \(PF=\frac{4\sqrt{61}}{5}\). 5. Since \(QR=10\), the area is \(\frac{1}{2}(10)\left(\frac{4\sqrt{61}}{5}\right)=4\sqrt{61}\approx 31.24\) square units.

a) The three vectors from \(O\) to \(P\), \(Q\), and \(R\) lie along different coordinate axes, so the four points are noncoplanar. b) \(\frac{24}{5}=4.8\) units c) \(4\sqrt{61}\approx 31.24\) square units

- Use \(B-A\) for the direction of \(g\). - Test perpendicularity with a dot product. - Find a vector perpendicular to both line directions for the distance. - Use the Gram determinant for the triangle area.

1. \(\overrightarrow{AB}=\langle 1,2,-2\rangle\), so \(g: (x,y,z)=(2,1,3)+r\langle 1,2,-2\rangle\). 2. The direction vectors have dot product \(\langle 1,2,-2\rangle\cdot\langle 2,2,1\rangle=4\), so they are not perpendicular. 3. The direction vectors are not scalar multiples, and equating the line coordinates produces an inconsistent system. Thus, the lines are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle 6,-5,-2\rangle\), with magnitude \(\sqrt{65}\). The vector from the point on \(g\) to the point on \(h\) is \(\mathbf{w}=\langle -2,3,-2\rangle\). Therefore, \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{23}{\sqrt{65}}=\frac{23\sqrt{65}}{65}\approx2.85\). 5. For triangle \(ABC\), \(\overrightarrow{AB}=\langle 1,2,-2\rangle\) and \(\overrightarrow{AC}=\langle 0,5,-1\rangle\). Its area is \(\frac{1}{2}\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=\frac{3\sqrt{10}}{2}\approx4.74\).

a) \(g: (x,y,z)=(2,1,3)+r\langle 1,2,-2\rangle\) b) No; the dot product is \(4\). c) The lines are skew, and \(d=\frac{23\sqrt{65}}{65}\approx2.85\) units. d) \(\frac{3\sqrt{10}}{2}\approx4.74\) square units

- Use the two given points to find the direction of \(h\). - Test perpendicularity with a dot product. - Equate coordinates to check for an intersection. - Project the vector between the lines onto a common perpendicular direction.

1. \(\overrightarrow{P_1P_2}=\langle 2,1,-2\rangle\), so \(h: (x,y,z)=(1,2,2)+s\langle 2,1,-2\rangle\). 2. The direction vectors have dot product \(\langle 2,-2,1\rangle\cdot\langle 2,1,-2\rangle=0\), so they are perpendicular. 3. The direction vectors are not scalar multiples. Equating coordinates gives parameter values from the first two equations that fail the third, so the lines do not intersect. Therefore, they are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle 1,2,2\rangle\), with magnitude \(3\). The vector between the given points is \(\mathbf{w}=\langle 1,2,-3\rangle\). 5. The distance is \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{1}{3}\).

a) \(h: (x,y,z)=(1,2,2)+s\langle 2,1,-2\rangle\) b) The direction vectors are perpendicular. c) The lines are skew. d) \(\frac{1}{3}\) unit

- Compare one pair of opposite-side vectors. - Use a dot product for the rectangle condition. - Compare adjacent squared side lengths for the rhombus condition. - A square must satisfy both conditions.

1. The opposite-side vectors are \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\1\end{pmatrix}\) and \(\overrightarrow{DC}=\begin{pmatrix}2\\2\\1\end{pmatrix}\). Thus, \(ABCD\) is a parallelogram for every \(k\). 2. Also, \(\overrightarrow{BC}=\begin{pmatrix}2\\0\\k-3\end{pmatrix}\). 3. For a rectangle, adjacent sides must be perpendicular: \(\overrightarrow{AB}\cdot\overrightarrow{BC}=4+k-3=k+1\). Thus, \(k=-1\). 4. For a rhombus, adjacent side lengths must be equal. Since \(\|\overrightarrow{AB}\|^2=9\) and \(\|\overrightarrow{BC}\|^2=4+(k-3)^2\), set \(9=4+(k-3)^2\). 5. This gives \((k-3)^2=5\), so \(k=3-\sqrt{5}\) or \(k=3+\sqrt{5}\). 6. A square must be both a rectangle and a rhombus. The rectangle value \(k=-1\) is neither rhombus value, so no value of \(k\) produces a square.

a) \(\overrightarrow{AB}=\overrightarrow{DC}=\begin{pmatrix}2\\2\\1\end{pmatrix}\), so \(ABCD\) is always a parallelogram. b) \(k=-1\) c) \(k=3-\sqrt{5}\) or \(k=3+\sqrt{5}\) d) No value satisfies both the rectangle and rhombus conditions, so the parallelogram can never be a square.

- Use two adjacent vectors in the parallelogram base. - The area can be found from their lengths and dot product. - Find a vector orthogonal to both base directions. - Use the point-to-plane distance as the pyramid's height, then apply \(V=\frac{1}{3}Gh\).

1. Adjacent base vectors are \(\mathbf{u}=\overrightarrow{AB}=\langle 4,0,2\rangle\) and \(\mathbf{v}=\overrightarrow{AD}=\langle 0,4,4\rangle\). 2. Their dot products are \(\mathbf{u}\cdot\mathbf{u}=20\), \(\mathbf{v}\cdot\mathbf{v}=32\), and \(\mathbf{u}\cdot\mathbf{v}=8\). The parallelogram area is \(G=\sqrt{(\mathbf{u}\cdot\mathbf{u})(\mathbf{v}\cdot\mathbf{v})-(\mathbf{u}\cdot\mathbf{v})^2}=\sqrt{20(32)-8^2}=24\). 3. A normal vector to the base is \(\mathbf{n}=\langle 1,2,-2\rangle\), because \(\mathbf{n}\cdot\mathbf{u}=0\) and \(\mathbf{n}\cdot\mathbf{v}=0\). The base plane is \(x+2y-2z-1=0\). 4. The pyramid's height is the distance from \(S\) to the base plane: \(h=\frac{\lvert 6+2(9)-2(10)-1\rvert}{\sqrt{1^2+2^2+(-2)^2}}=1\). 5. Therefore, \(V=\frac{1}{3}Gh=\frac{1}{3}(24)(1)=8\) cubic units.

\(8\) cubic units

- Use two vectors along the base triangle. - The Gram determinant gives the area of the parallelogram formed by two vectors; divide by \(2\) for the triangle. - The base points are the coordinate-axis intercepts of its plane. - Use the point-to-plane distance as the tetrahedron's height.

1. Let \(\mathbf{u}=\overrightarrow{PQ}=\langle -2,4,0\rangle\) and \(\mathbf{v}=\overrightarrow{PR}=\langle -2,0,6\rangle\). 2. Their dot products are \(\mathbf{u}\cdot\mathbf{u}=20\), \(\mathbf{v}\cdot\mathbf{v}=40\), and \(\mathbf{u}\cdot\mathbf{v}=4\). Therefore, the triangle's area is \(G=\frac{1}{2}\sqrt{(\mathbf{u}\cdot\mathbf{u})(\mathbf{v}\cdot\mathbf{v})-(\mathbf{u}\cdot\mathbf{v})^2}=\frac{1}{2}\sqrt{20(40)-4^2}=14\). 3. The base plane has intercept form \(\frac{x}{2}+\frac{y}{4}+\frac{z}{6}=1\), so its coordinate equation is \(6x+3y+2z=12\). 4. A normal vector is \(\langle 6,3,2\rangle\), with magnitude \(7\). The height is the distance from \(S\) to the base plane: \(h=\frac{\lvert 6(5)+3(1)+2(0)-12\rvert}{7}=3\). 5. Thus, \(V=\frac{1}{3}Gh=\frac{1}{3}(14)(3)=14\) cubic units.

Base area: \(14\) square units Height: \(3\) units Volume: \(14\) cubic units

- Find a vector orthogonal to both direction vectors of the plane. - Test the line's direction vector with that normal vector. - Convert the plane to coordinate form. - Use a point on the line in the point-to-plane distance formula.

1. Let \(\mathbf{u}=\langle 0,0,1\rangle\) and \(\mathbf{v}=\langle 4,-3,0\rangle\) be the plane's direction vectors. A vector \(\mathbf{n}=\langle a,b,c\rangle\) perpendicular to both must satisfy \(c=0\) and \(4a-3b=0\). Choose \(\mathbf{n}=\langle 3,4,0\rangle\). 2. The line's direction vector is \(\mathbf{d}=\langle 4,-3,7\rangle\). Since \(\mathbf{d}\cdot\mathbf{n}=4(3)+(-3)(4)+7(0)=0\), the line is parallel to the plane or lies in it. 3. Using point \((4,0,0)\), the plane equation is \(3x+4y=12\). The point \((1,1,5)\) on \(g\) gives \(3(1)+4(1)=7\ne12\), so the line is parallel to and distinct from the plane. 4. The distance is \(d=\frac{\lvert 3(1)+4(1)-12\rvert}{\sqrt{3^2+4^2}}=1\).

The line is parallel to the plane, and the distance is \(1\) unit.

- First describe all points at distance \(2\) from plane \(F\). - A fixed distance from a plane gives two planes parallel to it. - Intersect each of those planes with \(E\). - Parameterize each intersection line.

1. A normal vector to \(F\) is \(\langle 2,1,-2\rangle\), with magnitude \(3\). 2. A point \((x,y,z)\) is \(2\) units from \(F\) when \(\frac{\lvert 2x+y-2z-6\rvert}{3}=2\). 3. Therefore, the point lies in one of the two planes \(2x+y-2z=12\) or \(2x+y-2z=0\). 4. Intersect \(x+y+z=3\) with \(2x+y-2z=12\). Subtracting the equations gives \(x-3z=9\). Let \(z=t\). Then \(x=9+3t\) and \(y=-6-4t\), so one line is \((x,y,z)=(9,-6,0)+t\langle 3,-4,1\rangle\). 5. Intersect \(x+y+z=3\) with \(2x+y-2z=0\). Subtracting gives \(x-3z=-3\). Let \(z=s\). Then \(x=-3+3s\) and \(y=6-4s\), so the other line is \((x,y,z)=(-3,6,0)+s\langle 3,-4,1\rangle\).

\((x,y,z)=(9,-6,0)+t\langle 3,-4,1\rangle\) or \((x,y,z)=(-3,6,0)+s\langle 3,-4,1\rangle\), where \(s,t\in\mathbb{R}\)

- Use the distance formula to identify two planes parallel to \(F\). - Substitute the parametric coordinates of \(E\) into each parallel-plane equation. - Each resulting linear relationship between \(\lambda\) and \(\mu\) describes a line.

1. A normal vector to \(F\) is \(\langle 2,-1,2\rangle\), with magnitude \(3\). 2. Points at distance \(4\) from \(F\) satisfy \(\frac{\lvert 2x-y+2z-9\rvert}{3}=4\). Thus, they lie in one of the parallel planes \(2x-y+2z=21\) or \(2x-y+2z=-3\). 3. A point in \(E\) has coordinates \(x=2+\lambda\), \(y=1+\mu\), and \(z=\lambda-\mu\). 4. For \(2x-y+2z=21\), substitution gives \(4\lambda-3\mu=18\). One solution is \((\lambda,\mu)=(6,2)\), giving the point \((8,3,4)\). Increasing \(\lambda\) by \(3\) and \(\mu\) by \(4\) preserves the equation, producing direction vector \(3\langle 1,0,1\rangle+4\langle 0,1,-1\rangle=\langle 3,4,-1\rangle\). 5. For \(2x-y+2z=-3\), substitution gives \(4\lambda-3\mu=-6\). One solution is \((\lambda,\mu)=(0,2)\), giving \((2,3,-2)\). The direction vector is again \(\langle 3,4,-1\rangle\).

\((x,y,z)=(8,3,4)+t\langle 3,4,-1\rangle\) or \((x,y,z)=(2,3,-2)+s\langle 3,4,-1\rangle\), where \(s,t\in\mathbb{R}\)

- Face \(OAB\) lies in the xy-plane. - Find a normal vector to \(ABC\) using dot-product orthogonality. - Apply the point-to-plane distance formula. - Use two different faces as the base in \(V=\frac{1}{3}Gh\).

1. Triangle \(OAB\) is right with legs of length \(6\), so its area is \(\frac{1}{2}(6)(6)=18\). 2. For plane \(ABC\), \(\overrightarrow{AB}=\langle -6,6,0\rangle\) and \(\overrightarrow{AC}=\langle 0,6,6\rangle\). A vector \(\langle a,b,c\rangle\) orthogonal to both satisfies \(-6a+6b=0\) and \(6b+6c=0\). Choose \(\mathbf{n}=\langle 1,1,-1\rangle\). 3. Using point \(A\), the plane equation is \(x+y-z=6\). 4. The distance from the origin is \(\frac{\lvert -6\rvert}{\sqrt{1^2+1^2+(-1)^2}}=2\sqrt{3}\approx3.46\). 5. Using base \(OAB\), the height from \(C\) is \(6\), so \(V=\frac{1}{3}(18)(6)=36\). 6. For a second method, the area of triangle \(ABC\) is \(\frac{1}{2}\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=18\sqrt{3}\). Using the height from \(O\), \(V=\frac{1}{3}(18\sqrt{3})(2\sqrt{3})=36\).

a) \(18\) square units b) \(E_{ABC}: x+y-z=6\) c) \(2\sqrt{3}\approx3.46\) units d) \(36\) cubic units by both methods

- Notice that \(A\), \(B\), and \(C\) have the same z-coordinate. - Use \(V=\frac{1}{3}Gh\). - For face \(ABD\), find the component of \(\overrightarrow{AD}\) perpendicular to \(\overrightarrow{AB}\). - Rearrange the volume formula to find the final height.

1. \(\overrightarrow{AB}=\langle 4,0,0\rangle\) and \(\overrightarrow{AC}=\langle 0,6,0\rangle\) are perpendicular, so \(A_{ABC}=\frac{1}{2}(4)(6)=12\). 2. Plane \(ABC\) is \(z=2\). Therefore, the distance from \(D(3,3,6)\) to the plane is \(6-2=4\). 3. The volume is \(V=\frac{1}{3}(12)(4)=16\). 4. For face \(ABD\), use \(\overrightarrow{AB}=\langle 4,0,0\rangle\) as the base. The component of \(\overrightarrow{AD}=\langle 2,2,4\rangle\) perpendicular to \(\overrightarrow{AB}\) is \(\langle 0,2,4\rangle\), with length \(2\sqrt{5}\). Thus, \(A_{ABD}=\frac{1}{2}(4)(2\sqrt{5})=4\sqrt{5}\approx8.94\). 5. Let \(h\) be the distance from \(C\) to plane \(ABD\). From \(16=\frac{1}{3}(4\sqrt{5})h\), \(h=\frac{12}{\sqrt{5}}=\frac{12\sqrt{5}}{5}\approx5.37\).

a) \(A_{ABC}=12\) square units; \(d(D,E_{ABC})=4\) units b) \(V=16\) cubic units c) \(A_{ABD}=4\sqrt{5}\approx8.94\) square units d) \(d(C,E_{ABD})=\frac{12\sqrt{5}}{5}\approx5.37\) units

- Use the normal vector \(\begin{pmatrix}1\\1\\k\end{pmatrix}\). - Compare it with each coordinate-axis direction. - Equal axis angles require equal sine values. - Use the condition \(k>0\) only in part 1.

1. A normal vector is \(\vec{n}=\begin{pmatrix}1\\1\\k\end{pmatrix}\). For the z-axis, \(\sin45^\circ=\frac{|k|}{\sqrt{k^2+2}}\). Squaring gives \(k^2=2\), so the positive solution is \(k=\sqrt{2}\). 2. The sine values of the angles with the x-, y-, and z-axes are \(\frac{1}{\|\vec{n}\|}\), \(\frac{1}{\|\vec{n}\|}\), and \(\frac{|k|}{\|\vec{n}\|}\). They are equal when \(|k|=1\), so \(k=1\) or \(k=-1\). 3. For either value, \(\|\vec{n}\|=\sqrt{3}\), so the common angle is \(\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx35.3^\circ\).

1. \(k=\sqrt{2}\) 2. \(k=1\) or \(k=-1\), and the common angle is approximately \(35.3^\circ\).

- Find a normal vector and use it to write a coordinate equation of the plane. - Use the sine formula for the line-plane angle. - Project one additional point of the line along the plane normal.

1. A normal vector to \(E\) is \(\vec{n}=\begin{pmatrix}0\\-1\\1\end{pmatrix}\), so a coordinate equation is \(-y+z=1\). 2. Substituting \(g\) into the plane equation gives \(-\lambda+2=1\), so \(\lambda=1\). Thus \(S=(2, 1, 2)\). 3. The line direction is \(\vec{v}=\begin{pmatrix}1\\1\\0\end{pmatrix}\). Hence \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}|}{\|\vec{v}\|\|\vec{n}\|}=\frac{1}{2}\), so \(\alpha=30^\circ\). 4. Project the position point \(A=(1, 0, 2)\) onto \(E\) along the normal. The perpendicular line is \(\vec{x}=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}0\\-1\\1\end{pmatrix}\). Substitution into \(-y+z=1\) gives \(t=-\frac{1}{2}\), so the projection is \(A'=\left(1, \frac{1}{2}, \frac{3}{2}\right)\). 5. The projected line passes through \(S\) and \(A'\). A direction vector is \(\begin{pmatrix}2\\1\\1\end{pmatrix}\), so one equation is \(h:\vec{x}=\begin{pmatrix}2\\1\\2\end{pmatrix}+\sigma\begin{pmatrix}2\\1\\1\end{pmatrix}\).

a) \(S=(2, 1, 2)\), and \(\alpha=30^\circ\) b) \(h:\vec{x}=\begin{pmatrix}2\\1\\2\end{pmatrix}+\sigma\begin{pmatrix}2\\1\\1\end{pmatrix}\)

- The intersection direction is perpendicular to both plane normals; solve two zero dot-product equations. - A line in \(E_2\) perpendicular to the intersection line must have direction perpendicular to both \(\vec{n}_2\) and the intersection direction. - Use the line-plane angle formula for \(\alpha\). - Compare exact trigonometric expressions before rounding.

1. Normal vectors are \(\vec{n}_1=\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\vec{n}_2=\begin{pmatrix}1\\-2\\2\end{pmatrix}\), each with magnitude \(3\). 2. Thus \(\cos\gamma=\frac{1}{9}\), so \(\gamma\approx83.6^\circ\). 3. A direction vector \(\vec{u}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) for the intersection line must be perpendicular to both normals. The equations \(a+2b+2c=0\) and \(a-2b+2c=0\) give \(b=0\) and \(a=-2c\). Choose \(\vec{u}=\begin{pmatrix}2\\0\\-1\end{pmatrix}\). 4. A direction vector \(\vec{v}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) for \(g\) must lie in \(E_2\), so \(\vec{v}\cdot\vec{n}_2=0\), and it must be perpendicular to \(s\), so \(\vec{v}\cdot\vec{u}=0\). These equations are \(a-2b+2c=0\) and \(2a-c=0\). Choose \(\vec{v}=\begin{pmatrix}2\\5\\4\end{pmatrix}\). 5. Then \(\sin\alpha=\frac{|\vec{v}\cdot\vec{n}_1|}{\|\vec{v}\|\|\vec{n}_1\|}=\frac{4\sqrt{5}}{9}\), so \(\alpha\approx83.6^\circ\). 6. Since \(\sin\gamma=\sqrt{1-\frac{1}{81}}=\frac{4\sqrt{5}}{9}=\sin\alpha\) and both angles are acute, \(\alpha=\gamma\).

a) \(\gamma=\cos^{-1}\left(\frac{1}{9}\right)\approx83.6^\circ\) b) One direction vector is \(\vec{u}=\begin{pmatrix}2\\0\\-1\end{pmatrix}\). c) One direction vector is \(\vec{v}=\begin{pmatrix}2\\5\\4\end{pmatrix}\), and \(\alpha\approx83.6^\circ\). Thus \(\alpha=\gamma\).

- Substitute the line into the plane equation. - Use the line direction and plane normal for the line-plane angle. - Project the line direction onto the plane for the minimum-angle line. - A line lies in a plane when its point is in the plane and its direction is perpendicular to the plane normal.

1. Point \(S\) satisfies \(2\cdot1-2\cdot0+1=3\). Substituting \(g\) into \(E\) gives \(3+4t=3\), so \(t=0\). Therefore, \(S\) is the unique intersection point. 2. A normal vector is \(\vec{n}=\begin{pmatrix}2\\-2\\1\end{pmatrix}\), and the line direction is \(\vec{v}=\begin{pmatrix}2\\1\\2\end{pmatrix}\). Thus \(\sin\alpha=\frac{4}{9}\), so \(\alpha\approx26.4^\circ\). 3. The line in the plane making the smallest angle with \(g\) has the direction of the orthogonal projection of \(\vec{v}\) onto \(E\): \(\vec{v}_{E}=\vec{v}-\frac{\vec{v}\cdot\vec{n}}{\|\vec{n}\|^2}\vec{n}=\frac{1}{9}\begin{pmatrix}10\\17\\14\end{pmatrix}\). One equation is \(h:\vec{x}=\begin{pmatrix}1\\0\\1\end{pmatrix}+\lambda\begin{pmatrix}10\\17\\14\end{pmatrix}\). The angle is the line-plane angle, approximately \(26.4^\circ\). 4. The direction \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is perpendicular to \(\vec{n}\), and \(S\in E\), so \(k\subset E\). Also, \(\cos\gamma=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}\), so \(\gamma=45^\circ\).

1. \(S=(1, 0, 1)\) 2. \(\alpha\approx26.4^\circ\) 3. One possible line is \(h:\vec{x}=\begin{pmatrix}1\\0\\1\end{pmatrix}+\lambda\begin{pmatrix}10\\17\\14\end{pmatrix}\), and the angle is approximately \(26.4^\circ\). 4. Line \(k\) lies in \(E\), and the angle between \(g\) and \(k\) is \(45^\circ\).

- Find a vector perpendicular to both spanning vectors by solving zero dot-product conditions. - Substitute the line coordinates into the plane equation. - Use the plane normal and line direction to calculate the angle. - Check whether the sine or cosine relationship gives the line-plane angle.

1. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be normal to \(E\). Perpendicularity to \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) and \(\begin{pmatrix}2\\1\\0\end{pmatrix}\) gives \(a-c=0\) and \(2a+b=0\). Choose \(\vec{n}=\begin{pmatrix}1\\-2\\1\end{pmatrix}\). Using the point \((1, 2, 1)\), the Cartesian equation is \(x-2y+z=-2\). 2. Substitute the line coordinates into the plane equation: \((5+t)-2(4+2t)+(3+t)=-2\). This gives \(-2t=-2\), so \(t=1\). Therefore, \(S=(6, 6, 4)\). 3. The line direction is \(\vec{v}=\begin{pmatrix}1\\2\\1\end{pmatrix}\). Thus \(\sin\alpha=\frac{|\vec{n}\cdot\vec{v}|}{\|\vec{n}\|\|\vec{v}\|}=\frac{2}{6}=\frac{1}{3}\), so \(\alpha\approx19.5^\circ\).

a) \(E:x-2y+z=-2\) b) \(S=(6, 6, 4)\) c) \(\alpha\approx19.5^\circ\)

- First write a Cartesian equation of the plane through the three points. - Describe the flight path as a ray. - Check that the intersection parameter is consistent with the insect's direction of travel. - Use the path direction and a normal vector to the plane to find the angle.

1. The plane through the three intercepts has equation \(x+y+z=6\), with normal vector \(\vec{n}=\begin{pmatrix}1\\1\\1\end{pmatrix}\). 2. The insect's path is \(\vec{x}=\begin{pmatrix}1\\1\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\1\end{pmatrix}\), where \(t\ge0\). 3. Substitute into the plane equation: \((1+t)+(1+2t)+(2+t)=6\). Thus \(4+4t=6\), so \(t=\frac{1}{2}\). 4. Therefore, \(S=\begin{pmatrix}1\\1\\2\end{pmatrix}+\frac{1}{2}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}\frac{3}{2}\\2\\\frac{5}{2}\end{pmatrix}\), so \(S=(1.5, 2, 2.5)\). Its coordinates are nonnegative and sum to \(6\), so it lies inside the triangular net. 5. With direction vector \(\vec{v}=\begin{pmatrix}1\\2\\1\end{pmatrix}\), \(\sin\alpha=\frac{|\vec{n}\cdot\vec{v}|}{\|\vec{n}\|\|\vec{v}\|}=\frac{4}{\sqrt{3}\sqrt{6}}=\frac{2\sqrt{2}}{3}\). Therefore, \(\alpha\approx70.5^\circ\).

a) \(S=(1.5, 2, 2.5)\) b) \(\alpha\approx70.5^\circ\)

- Set the two parametric equations equal and solve the resulting system. - Use the dot-product formula for the angle between direction vectors. - Find a vector perpendicular to both line directions by solving two zero dot-product equations. - Construct each plane normal so it is perpendicular to the corresponding line direction and to that common perpendicular vector.

1. Set the line coordinates equal. Solving the resulting system gives \(r=-1\) and \(s=0\). Both lines therefore pass through \(S=(3, 0, 0)\). 2. Let \(\vec{v}_1=\begin{pmatrix}2\\1\\2\end{pmatrix}\) and \(\vec{v}_2=\begin{pmatrix}2\\2\\1\end{pmatrix}\). Then \(\cos\alpha=\frac{|\vec{v}_1\cdot\vec{v}_2|}{\|\vec{v}_1\|\|\vec{v}_2\|}=\frac{8}{9}\), so \(\alpha\approx27.3^\circ\). 3. First find a vector \(\vec{w}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) perpendicular to both line directions. The conditions \(2a+b+2c=0\) and \(2a+2b+c=0\) give \(b=c\) and \(2a+3b=0\). Choose \(\vec{w}=\begin{pmatrix}-3\\2\\2\end{pmatrix}\). 4. For \(E_1\), choose a normal perpendicular to both \(\vec{v}_1\) and \(\vec{w}\). Solving those two dot-product equations gives \(\vec{n}_1=\begin{pmatrix}-2\\-10\\7\end{pmatrix}\). For \(E_2\), solving \(\vec{n}_2\cdot\vec{v}_2=0\) and \(\vec{n}_2\cdot\vec{w}=0\) gives \(\vec{n}_2=\begin{pmatrix}2\\-7\\10\end{pmatrix}\). 5. Using \(S\), one possible pair of planes is \(E_1:-2x-10y+7z=-6\) and \(E_2:2x-7y+10z=6\). Their normals satisfy \(\cos\alpha=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{8}{9}\), so the plane angle equals the line angle.

1. \(S=(3, 0, 0)\), with \(r=-1\) and \(s=0\) 2. \(\alpha\approx27.3^\circ\) 3. One possible pair is \(E_1:-2x-10y+7z=-6\) and \(E_2:2x-7y+10z=6\).

- Convert both planes to the same form before finding their intersection. - The angle between two planes equals the angle between their normal vectors. - A direction vector for the intersection line is perpendicular to both plane normals. - Verify that the line direction is orthogonal to both normals.

1. Let \(\vec{n}_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be normal to \(E_2\). Perpendicularity to its spanning vectors gives \(a+c=0\) and \(b+c=0\), so choose \(\vec{n}_2=\begin{pmatrix}-1\\-1\\1\end{pmatrix}\). Using the given point, \(E_2\) has Cartesian equation \(-x-y+z=-1\). 2. The point \((2, 0, 1)\) satisfies both plane equations. A direction vector for the intersection line must be perpendicular to both \(\vec{n}_1=\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and \(\vec{n}_2\). One such vector is \(\begin{pmatrix}1\\-4\\-3\end{pmatrix}\). Therefore, \(k:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+t\begin{pmatrix}1\\-4\\-3\end{pmatrix}\). 3. \(\cos\phi=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\|\vec{n}_1\|\|\vec{n}_2\|}=\frac{1}{3\sqrt{3}}\), so \(\phi\approx78.9^\circ\).

a) \(k:\vec{x}=\begin{pmatrix}2\\0\\1\end{pmatrix}+t\begin{pmatrix}1\\-4\\-3\end{pmatrix}\) b) \(\phi\approx78.9^\circ\)

- Build two directions in each plane. - Find each normal by setting its dot product with both plane directions equal to \(0\). - Perpendicular planes have orthogonal normal vectors. - The normal of \(E_3\) must be orthogonal to both existing normals.

1. a) Directions in \(E_1\) are \(\overrightarrow{AB}=\langle -2,2,0\rangle\) and \(\overrightarrow{AC}=\langle 2,4,-3\rangle\). Solving \(\vec{n}_1\cdot\overrightarrow{AB}=0\) and \(\vec{n}_1\cdot\overrightarrow{AC}=0\) gives \(\vec{n}_1=\langle 1,1,2\rangle\). 2. Using point \(A\), \(E_1:x+y+2z=10\). 3. Two directions in \(E_2\) are \(\langle 1,0,1\rangle\) and the vector between the two line points, \(\langle 0,1,1\rangle\). Solving the two orthogonality equations gives \(\vec{n}_2=\langle 1,1,-1\rangle\). 4. Using \((1, 1, 1)\), \(E_2:x+y-z=1\). 5. b) The normals satisfy \(\vec{n}_1\cdot\vec{n}_2=1+1-2=0\), so the planes are perpendicular. 6. c) Let \(\vec{n}_3=\langle a,b,c\rangle\). For \(E_3\) to be perpendicular to both planes, require \(\vec{n}_3\cdot\vec{n}_1=0\) and \(\vec{n}_3\cdot\vec{n}_2=0\). 7. The equations are \(a+b+2c=0\) and \(a+b-c=0\). Subtraction gives \(3c=0\), so \(c=0\) and \(a=-b\). Choose \(\vec{n}_3=\langle 1,-1,0\rangle\). 8. A point-normal equation is \((\vec{r}-\langle 5,0,0\rangle)\cdot\langle 1,-1,0\rangle=0\), or \(x-y=5\).

a) \(E_1:x+y+2z=10\); \(E_2:x+y-z=1\) b) The planes are perpendicular because \(\langle 1,1,2\rangle\cdot\langle 1,1,-1\rangle=0\). c) \((\vec{r}-\langle 5,0,0\rangle)\cdot\langle 1,-1,0\rangle=0\)

- Find a vector orthogonal to both direction vectors in the plane. - Use the point-to-plane distance formula. - Find the perpendicular foot from \(Q\) to the plane. - The foot is the midpoint of \(Q\) and its reflection.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle -2,4,0\rangle\) and \(\overrightarrow{AC}=\langle 0,4,-2\rangle\). 2. A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) satisfies \(-2a+4b=0\) and \(4b-2c=0\). Choose \(\mathbf{n}=\langle 2,1,2\rangle\). 3. Using point \(A\), the plane equation is \(2x+y+2z=6\). 4. The distance from \(Q\) is \(d(Q,E)=\frac{\lvert 2(5)+5+2(9)-6\rvert}{3}=9\). 5. The perpendicular foot is \(F=Q-\frac{2(5)+5+2(9)-6}{\lVert\mathbf{n}\rVert^2}\mathbf{n}=Q-3\mathbf{n}=(-1,2,3)\). 6. Since \(F\) is the midpoint of \(\overline{QQ'}\), \(Q'=2F-Q=(-7,-1,-3)\).

a) \(E: 2x+y+2z=6\) b) \(9\) units c) \(Q'=(-7,-1,-3)\)

- For part a, equate coordinates and begin with the third coordinate. - For part b, solve two zero dot-product equations to find a vector perpendicular to both line directions, then project the position-point difference onto it. - What point do all lines \(h_k\) share? - For part c, separate the direction vector of \(h_k\) into two fixed directions and find a vector perpendicular to both using dot products.

1. Equating \(g\) and \(h_k\), the third coordinate gives \(s=1\), the first gives \(r=4\), and the second then gives \(k=1\). The intersection point is \((3, 1, 4)\). 2. For \(k=0\), the direction vectors are \(\vec{u}=\begin{pmatrix}1\\-1\\0\end{pmatrix}\) and \(\vec{v}=\begin{pmatrix}2\\0\\2\end{pmatrix}\). They are not parallel, and equating the line equations gives an inconsistent system, so the lines are skew. 3. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both directions. The conditions \(\vec{n}\cdot\vec{u}=0\) and \(\vec{n}\cdot\vec{v}=0\) give \(a-b=0\) and \(2a+2c=0\). Choose \(\vec{n}=\begin{pmatrix}1\\1\\-1\end{pmatrix}\). With the position-point difference \(\vec{d}=\begin{pmatrix}2\\-5\\-2\end{pmatrix}\), the distance is \(\frac{|\vec{d}\cdot\vec{n}|}{\|\vec{n}\|}=\frac{1}{\sqrt{3}}\approx0.577\). 4. Every line \(h_k\) passes through \((1, 0, 2)\), and \(\begin{pmatrix}2\\k\\2\end{pmatrix}=2\begin{pmatrix}1\\0\\1\end{pmatrix}+k\begin{pmatrix}0\\1\\0\end{pmatrix}\). Thus all direction vectors lie in the plane spanned by these two fixed vectors. 5. A normal \(\begin{pmatrix}a\\b\\c\end{pmatrix}\) to this plane must satisfy \(a+c=0\) and \(b=0\). Choose \(\begin{pmatrix}-1\\0\\1\end{pmatrix}\). Using \((1, 0, 2)\), the plane equation is \(-x+z=1\), or \(x-z+1=0\).

a) \(k=1\), with intersection point \((3, 1, 4)\) b) The lines are skew, and their distance is \(\frac{1}{\sqrt{3}}\approx0.577\). c) A common plane is \(E:x-z+1=0\).

- Substitute the line into the plane to find the fixed intersection point. - Reflect one additional point along a line parallel to the plane normal. - Use the fixed point and reflected point to construct the image line. - Apply the point-to-plane distance formula to a general point of the line.

1. a) Substitute the line into the plane: \(2\lambda-(2+\lambda)+2=6\). Thus \(\lambda=6\), so \(D(6, 8, 1)\). 2. b) Reflect the initial point \(A(0, 2, 1)\). A perpendicular line through \(A\) has direction \(\vec{n}=\langle 2,-1,2\rangle\): \(\vec{r}(t)=\langle 0,2,1\rangle+t\vec{n}\). 3. Its intersection with \(H\) occurs when \(9t=6\), so \(t=\frac23\). The reflected point is twice as far along the perpendicular line, at \(t=\frac43\): \(A'=\left(\frac83, \frac23, \frac{11}{3}\right)\). 4. The reflected line passes through fixed point \(D\) and \(A'\). A direction vector is \(D-A'=\frac13\langle 10,22,-8\rangle\), so one equation is \(f':\vec{r}(\mu)=\langle 6,8,1\rangle+\mu\langle 5,11,-4\rangle\). 5. c) A general point on \(f\) is \((\lambda, 2+\lambda, 1)\). Its distance from \(H\) is \(\frac{|2\lambda-(2+\lambda)+2-6|}{\sqrt{2^2+(-1)^2+2^2}}=\frac{|\lambda-6|}{3}\). 6. Set this equal to \(3\): \(|\lambda-6|=9\), so \(\lambda=15\) or \(\lambda=-3\). 7. The points are \(P_1(15, 17, 1)\) and \(P_2(-3, -1, 1)\).

a) \(D(6, 8, 1)\) b) \(f':\vec{r}(\mu)=\langle 6,8,1\rangle+\mu\langle 5,11,-4\rangle\) c) \(P_1(15, 17, 1)\) and \(P_2(-3, -1, 1)\)

- Compare the second components of the two direction vectors. - Find a normal vector using two dot-product conditions. - Convert the family to a standard equation and require it to hold for every \(k\). - Perpendicular planes have orthogonal normals. - Substitute the point into the family equation.

1. a) The directions \(\langle 2,0,1\rangle\) and \(\langle k,1,0\rangle\) cannot be scalar multiples because their second components are \(0\) and \(1\). Thus they determine a plane for every \(k\). 2. Let \(\vec{n}_k=\langle a,b,c\rangle\) be a normal vector. The dot-product conditions with the two plane directions give one normal \(\vec{n}_k=\langle -1,k,2\rangle\). 3. A standard equation of \(E_k\) is \(-x+ky+2z=k+1\), or \(-x+2z-1+k(y-1)=0\). 4. b) For a point to lie in every plane, the equation must hold for every real \(k\). Therefore, \(y-1=0\) and \(-x+2z-1=0\). 5. The common line is \(g:\vec{r}(t)=\langle 1,1,1\rangle+t\langle 2,0,1\rangle\). 6. c) For \(E_k\perp E_{-1}\), their normals must be orthogonal. Since \(\vec{n}_{-1}=\langle -1,-1,2\rangle\), \(\vec{n}_k\cdot\vec{n}_{-1}=1-k+4=0\). 7. Thus \(k=5\). 8. d) Substitute \(A(5, 2, 3)\) into the standard equation: \(-5+2k+6=k+1\). 9. This gives \(k=0\), so \(A\) lies in \(E_0\).

a) The two direction vectors are linearly independent for every \(k\). b) \(g:\vec{r}(t)=\langle 1,1,1\rangle+t\langle 2,0,1\rangle\) c) \(k=5\) d) \(A\) lies in \(E_0\).

- Identify the point and direction shared by \(F_0\) and \(F_1\). - Find a normal to \(F_a\) with two dot-product conditions. - Use the angle formula for plane normals. - Perpendicular planes have orthogonal normals. - For family membership, compare both direction spaces and points.

1. a) Both \(F_0\) and \(F_1\) contain the point \((2, 0, 4)\) and the direction \(\langle 1,1,0\rangle\). Their other directions are different, so their intersection line is \(h:\vec{r}(t)=\langle 2,0,4\rangle+t\langle 1,1,0\rangle\). 2. Let \(\vec{n}_a=\langle p,q,r\rangle\) be normal to \(F_a\). The conditions \(p+q=0\) and \(aq+r=0\) give \(\vec{n}_a=\langle 1,-1,a\rangle\). 3. b) Thus \(\vec{n}_0=\langle 1,-1,0\rangle\) and \(\vec{n}_1=\langle 1,-1,1\rangle\). 4. The acute angle \(\alpha\) satisfies \(\cos\alpha=\frac{|\vec{n}_0\cdot\vec{n}_1|}{\lVert\vec{n}_0\rVert\lVert\vec{n}_1\rVert}=\frac{2}{\sqrt6}\). 5. Therefore, \(\alpha=\arccos\left(\frac{2}{\sqrt6}\right)\approx35.26^\circ\). 6. c) A normal to \(H\) is \(\vec{n}_H=\langle 1,-1,2\rangle\). Perpendicular planes have orthogonal normals, so \(\vec{n}_a\cdot\vec{n}_H=1+1+2a=0\). 7. Hence \(a=-1\). 8. d) The directions of \(E^*\) match those of \(F_2\), so only \(F_2\) could equal \(E^*\). 9. However, the initial point \((2, 2, 4)\) of \(E^*\) is not in any \(F_a\): matching x and z forces \(r=0\) and \(s=0\), which would make the y-coordinate \(0\), not \(2\). 10. Therefore, \(E^*\) does not belong to the family.

a) \(h:\vec{r}(t)=\langle 2,0,4\rangle+t\langle 1,1,0\rangle\) b) \(\alpha=\arccos\left(\frac{2}{\sqrt6}\right)\approx35.26^\circ\) c) \(a=-1\) d) \(E^*\) does not belong to the family.

- Compare the direction vectors of the two lines. - Test a point from one line on the other to confirm that the lines are distinct. - Use one line direction and a vector between the lines to span the plane. - Use a dot product to test perpendicularity, and subtract the projection of the connecting vector onto the line direction to obtain the perpendicular distance. - For the right angle at \(Q\), set the dot product of the two sides meeting at \(Q\) equal to zero.

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so the lines are parallel. Their position points have different \(y\)-coordinates, while both line directions have zero \(y\)-component, so the lines are distinct. 2. A parametric equation is \(E:\vec{x}=\begin{pmatrix}1\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\0\\-1\end{pmatrix}+t\begin{pmatrix}2\\3\\-1\end{pmatrix}\). A normal vector is \(\begin{pmatrix}1\\0\\2\end{pmatrix}\), so a coordinate equation is \(x+2z=3\). 3. \(\overrightarrow{PQ}=\begin{pmatrix}2\\3\\-1\end{pmatrix}\), and \(\overrightarrow{PQ}\cdot\begin{pmatrix}2\\0\\-1\end{pmatrix}=5\neq0\). Since the line direction \(\vec{u}=\begin{pmatrix}2\\0\\-1\end{pmatrix}\) has \(\|\vec{u}\|^2=5\), the projection of \(\overrightarrow{PQ}\) onto \(\vec{u}\) is \(\frac{5}{5}\vec{u}=\vec{u}\). The perpendicular component is \(\overrightarrow{PQ}-\vec{u}=\begin{pmatrix}0\\3\\0\end{pmatrix}\), so the distance between the parallel lines is \(3\). 4. A point on \(g\) is \(R=(1+2r, 2, 1-r)\). The right-angle condition at \(Q\) is \(\overrightarrow{QP}\cdot\overrightarrow{QR}=0\), which gives \(14-5r=0\). Thus \(r=\frac{14}{5}\).

a) The lines are distinct and parallel. b) One parametric equation is \(E:\vec{x}=\begin{pmatrix}1\\2\\1\end{pmatrix}+r\begin{pmatrix}2\\0\\-1\end{pmatrix}+t\begin{pmatrix}2\\3\\-1\end{pmatrix}\), and a coordinate equation is \(x+2z=3\). c) \(\overrightarrow{PQ}\cdot\vec{u}=5\neq0\), and the distance is \(3\). d) \(r=\frac{14}{5}\)

- Substitute the parametric plane into the standard equation. - Find a plane normal with two dot-product conditions. - Use the angle formula for plane normals. - Compare the line direction with the normal of \(H_a\), then test one point.

1. a) Substitute the parametric coordinates of \(E_2\) into \(E_1\): \((1+\lambda)+(-\lambda+\mu)=2\). 2. Thus \(\mu=1\), and the intersection line is \(g:\vec{r}(t)=\langle 1,1,3\rangle+t\langle 1,-1,1\rangle\). 3. A normal to \(E_1\) is \(\vec{n}_1=\langle 1,1,0\rangle\). 4. Let \(\vec{n}_2=\langle a,b,c\rangle\). Orthogonality to the two directions of \(E_2\) gives \(a-b+c=0\) and \(b+2c=0\). 5. One normal is \(\vec{n}_2=\langle -3,-2,1\rangle\). 6. The acute plane angle satisfies \(\cos\alpha=\frac{|\vec{n}_1\cdot\vec{n}_2|}{\lVert\vec{n}_1\rVert\lVert\vec{n}_2\rVert}=\frac{5}{\sqrt{28}}\). 7. Hence \(\alpha=\arccos\left(\frac{5}{\sqrt{28}}\right)\approx19.11^\circ\). 8. b) A normal to \(H_a\) is \(\vec{n}_{H_a}=\langle a,1,1-a\rangle\), and the direction of \(g\) is \(\vec{v}=\langle 1,-1,1\rangle\). 9. Since \(\vec{v}\cdot\vec{n}_{H_a}=a-1+1-a=0\), the line is parallel to or contained in every \(H_a\). 10. Test point \((1, 1, 3)\) from \(g\): \(a+1+3(1-a)=4-2a\). 11. This equals \(4\) only when \(a=0\). Therefore, \(g\subset H_0\), while \(g\) is strictly parallel to \(H_a\) for \(a\neq0\).

a) \(g:\vec{r}(t)=\langle 1,1,3\rangle+t\langle 1,-1,1\rangle\), and \(\alpha=\arccos\left(\frac{5}{\sqrt{28}}\right)\approx19.11^\circ\) b) \(g\subset H_0\). For every \(a\neq0\), \(g\) is strictly parallel to \(H_a\).

- Use the dot product of the line direction and the plane normal. - The normal of \(F\) must be orthogonal to both the line direction and the normal of \(E\). - The normal of \(H\) must be orthogonal to both plane normals. - Solve the final three-equation system.

1. a) The normal of \(E\) is \(\vec{n}_E=\langle 3,-1,2\rangle\), and the line direction is \(\vec{u}=\langle 1,1,-1\rangle\). Since \(\vec{u}\cdot\vec{n}_E=3-1-2=0\), the line is parallel to the plane. 2. The initial point is not in \(E\), because \(3(2)-2+2(2)=8\neq10\). Thus the line is strictly parallel to \(E\). 3. b) Let \(\vec{n}_F=\langle a,b,c\rangle\). Since \(F\) contains \(g\), require \(\vec{n}_F\cdot\vec{u}=0\). Since \(F\perp E\), require \(\vec{n}_F\cdot\vec{n}_E=0\). 4. Solving \(a+b-c=0\) and \(3a-b+2c=0\) gives one normal \(\vec{n}_F=\langle 1,-5,-4\rangle\). Using \((2, 2, 2)\), \(F:x-5y-4z=-16\). 5. c) A normal to \(H\) must be orthogonal to both \(\vec{n}_E\) and \(\vec{n}_F\). Solving the two dot-product equations gives \(\vec{n}_H=\langle 1,1,-1\rangle\). 6. Using \(A\), \(H:x+y-z=1\). 7. d) Row reduction of the three plane equations gives \(x=\frac{44}{21}\), \(y=\frac{32}{21}\), and \(z=\frac{55}{21}\).

a) \(g\) is strictly parallel to \(E\). b) \(F:x-5y-4z=-16\) c) \(H:x+y-z=1\) d) \(S\left(\frac{44}{21}, \frac{32}{21}, \frac{55}{21}\right)\)

- Equate corresponding line coordinates. - Find a plane normal with two dot-product conditions. - Separate the family equation into a constant part and a coefficient of \(t\). - A plane perpendicular to every family member needs a normal orthogonal to the two fixed vectors spanning the family normals.

1. a) Equating the line coordinates gives \(\lambda=-1\) and \(\mu=0\). Both lines pass through \(S(-1, 1, 1)\). 2. Let \(\vec{n}=\langle a,b,c\rangle\) be normal to the plane containing the lines. The dot-product conditions are \(2a+b-c=0\) and \(4a+b-2c=0\). 3. Solving gives one normal \(\vec{n}=\langle 1,0,2\rangle\). 4. Using \(S\), the plane equation is \(E:x+2z=1\). 5. b) Setting \(t=0\) in the family gives \(E_0:x+2z=1\), so \(E=E_0\). 6. Rewrite the family equation as \(x+2z-1+t(x+y-1)=0\). 7. A point lies in every plane when \(x+2z=1\) and \(x+y=1\). Let \(z=u\). Then \(x=1-2u\) and \(y=2u\). 8. The common line is \(\ell:\vec{r}(u)=\langle 1,0,0\rangle+u\langle -2,2,1\rangle\). 9. c) The normal of \(E^*\) must be orthogonal to every family normal \(\langle 1+t,t,2\rangle=\langle 1,0,2\rangle+t\langle 1,1,0\rangle\). 10. The vector \(\langle -2,2,1\rangle\) is orthogonal to both fixed vectors, so use it as the normal of \(E^*\). 11. Using \(P(2, 1, 5)\), the equation is \(-2x+2y+z=3\).

a) The lines intersect at \(S(-1, 1, 1)\), and \(E:x+2z=1\). b) \(E=E_0\), and the common line is \(\ell:\vec{r}(u)=\langle 1,0,0\rangle+u\langle -2,2,1\rangle\). c) \(E^*:-2x+2y+z=3\)

- A perpendicular-bisector plane passes through the segment midpoint and has a normal parallel to the segment. - A plane perpendicular to a line can use the line direction as its normal. - Use a dot product to prove the right angle. - Under a half-turn about a line, the perpendicular foot is the midpoint of a point and its image.

1. a) The midpoint of \(\overline{AB}\) is \(M(4, 2, -1)\). A normal to the perpendicular-bisector plane is parallel to \(\overrightarrow{AB}=\langle -2,4,-4\rangle\), so use \(\vec{n}_E=\langle 1,-2,2\rangle\). 2. Using \(M\), the equation is \(x-2y+2z=-2\). 3. b) A general point of \(g\) is \((2\lambda, 1+2\lambda, \lambda)\). Substitution gives \(2\lambda-2(1+2\lambda)+2\lambda=-2\), so every point of \(g\) lies in \(E\). 4. A plane perpendicular to \(g\) can use the line direction \(\langle 2,2,1\rangle\) as its normal. Using \(A\), \(F:2x+2y+z=11\). Point \(B\) also satisfies this equation. 5. c) Substitute \(g\) into \(F\): \(4\lambda+2+4\lambda+\lambda=11\), so \(\lambda=1\) and \(S(2, 3, 1)\). 6. The vectors \(\overrightarrow{SA}=\langle 3,-3,0\rangle\) and \(\overrightarrow{SB}=\langle 1,1,-4\rangle\) have dot product \(3-3=0\), so \(\angle ASB=90^\circ\). 7. d) Point \(S\) is the foot of the perpendicular from \(A\) to the rotation axis \(g\). Under a half-turn about \(g\), \(S\) is the midpoint of \(A\) and \(A'\). 8. Therefore \(A'=2S-A=\langle 4,6,2\rangle-\langle 5,0,1\rangle=(-1, 6, 1)\).

a) \(E:x-2y+2z=-2\) b) \(g\subset E\); \(F:2x+2y+z=11\) c) \(S(2, 3, 1)\), and \(\angle ASB=90^\circ\) d) \(A'(-1, 6, 1)\)

- Express the two sides from \(C\) using \(\vec{u}\) and \(\vec{v}\). - Expand squared magnitudes with dot products. - Use a dot product for the angle at \(C\). - Compare a leg with \(AB=2\|\vec{u}\|\) in the equilateral case.

1. \(\overrightarrow{CA}=\vec{u}-\vec{v}\) and \(\overrightarrow{CB}=-\vec{u}-\vec{v}\). 2. Their squared lengths are \(\|\vec{u}-\vec{v}\|^2=\|\vec{u}\|^2-2\vec{u}\cdot\vec{v}+\|\vec{v}\|^2\) and \(\|-\vec{u}-\vec{v}\|^2=\|\vec{u}\|^2+2\vec{u}\cdot\vec{v}+\|\vec{v}\|^2\). If \(\vec{u}\cdot\vec{v}=0\), these are equal. 3. For a right angle at \(C\), \((\vec{u}-\vec{v})\cdot(-\vec{u}-\vec{v}) =\|\vec{v}\|^2-\|\vec{u}\|^2=0\). Thus, \(\|\vec{v}\|=\|\vec{u}\|\). Geometrically, \(A\) and \(B\) are endpoints of a diameter of a sphere centered at \(M\), and \(C\) lies on that sphere. 4. For an equilateral triangle, \(\vec{u}\perp\vec{v}\) and \(\|\overrightarrow{CA}\|=\|\overrightarrow{AB}\|=2\|\vec{u}\|\). Thus, \(\|\vec{u}\|^2+\|\vec{v}\|^2=4\|\vec{u}\|^2\), so \(\|\vec{v}\|=\sqrt{3}\|\vec{u}\|\).

a) \(CA=CB\) b) \(\|\vec{v}\|=\|\vec{u}\|\) c) \(\|\vec{v}\|:\|\vec{u}\|=\sqrt{3}:1\)

- Use a dot product to compare the line direction with the plane normal. - Use the point-to-plane distance formula with the common initial point. - Substitute a general point of \(g_s\) into \(E\). - Rewrite the parameter-dependent coordinates using one new parameter. - Express each family direction as a combination of two fixed vectors.

1. a) A normal vector to \(E\) is \(\vec{n}_E=\langle 1,-2,2\rangle\). The direction \(\vec{u}=\langle 2,1,0\rangle\) satisfies \(\vec{u}\cdot\vec{n}_E=2-2=0\). 2. The initial point \(A(0, 0, 3)\) is not in \(E\), because the left side is \(6\), not \(9\). Thus \(g\) is strictly parallel to \(E\). 3. The distance is \(\frac{|6-9|}{\sqrt{1^2+(-2)^2+2^2}}=1\). 4. b) When \(s=0\), the direction of \(g_s\) is \(\langle 2,1,0\rangle\). Therefore, \(g=g_0\). 5. c) Substitute a point of \(g_s\) into \(E\). The result is \(6+9s\lambda=9\). 6. If \(s=0\), there is no intersection. If \(s\neq0\), then \(\lambda=\frac{1}{3s}\), so \(Q_s=\left(\frac13+\frac{2}{3s}, -\frac23+\frac{1}{3s}, \frac{11}{3}\right)\). 7. d) Let \(\mu=\frac{1}{3s}\). Then \(Q_s=\left\langle \frac13,-\frac23,\frac{11}{3}\right\rangle+\mu\langle 2,1,0\rangle\). 8. Therefore, every \(Q_s\) lies on \(m:\vec{r}(\mu)=\left\langle \frac13,-\frac23,\frac{11}{3}\right\rangle+\mu\langle 2,1,0\rangle\). 9. e) Each family direction can be written as \(\langle 2+s,1-2s,2s\rangle=\vec{u}+s\vec{n}_E\). Since all lines also pass through \(A\), they lie in the plane through \(A\) with directions \(\vec{u}\) and \(\vec{n}_E\). 10. Let \(\vec{n}_F=\langle a,b,c\rangle\). The conditions \(\vec{n}_F\cdot\vec{u}=0\) and \(\vec{n}_F\cdot\vec{n}_E=0\) give one normal \(\vec{n}_F=\langle 2,-4,-5\rangle\). 11. Using \(A\), \(F:2x-4y-5z=-15\). 12. Since \(\vec{n}_E\cdot\vec{n}_F=2+8-10=0\), the planes are perpendicular. The line \(m\) lies in both planes, so it is their intersection line.

a) \(g\) is strictly parallel to \(E\), and the distance is \(1\). b) \(g=g_0\) c) \(s\neq0\), and \(Q_s=\left(\frac13+\frac{2}{3s}, -\frac23+\frac{1}{3s}, \frac{11}{3}\right)\) d) \(m:\vec{r}(\mu)=\left\langle \frac13,-\frac23,\frac{11}{3}\right\rangle+\mu\langle 2,1,0\rangle\) e) \(F:2x-4y-5z=-15\), and \(F\perp E\) along \(m\).

- The lines share their initial point. - Find a plane normal orthogonal to both line directions. - Separate the family equation into a constant part and a coefficient of \(k\). - The farthest plane through a fixed line is perpendicular to the shortest segment from the point to that line.

1. a) Both lines pass through \(S(1, 1, 2)\), so they intersect there. 2. Let \(\vec{n}_E=\langle a,b,c\rangle\). Orthogonality to the two line directions gives \(2a+b=0\) and \(b+2c=0\). 3. One normal is \(\vec{n}_E=\langle 1,-2,1\rangle\). Using \(S\), the plane equation is \(E:x-2y+z=1\). 4. b) A normal to \(E_k\) is \(\vec{n}_k=\langle 2,1+k,2k\rangle\). 5. Since \(\vec{n}_k\cdot\vec{n}_E=2-2(1+k)+2k=0\), every \(E_k\) is perpendicular to \(E\). 6. Rewrite the family equation as \(2x+y-3+k(y+2z-3)=0\). 7. A point lies in every \(E_k\) when \(2x+y=3\) and \(y+2z=3\). Let \(z=t\). Then \(y=3-2t\) and \(x=t\). 8. The common line is \(l:\vec{r}(t)=\langle 0,3,0\rangle+t\langle 1,-2,1\rangle\). 9. c) Among all planes through \(l\), the greatest possible distance from the origin equals the distance from the origin to \(l\). The closest point \(Q\) on \(l\) satisfies \(\overrightarrow{OQ}\cdot\langle 1,-2,1\rangle=0\). 10. With \(Q=(t, 3-2t, t)\), this gives \(t-2(3-2t)+t=0\), so \(t=1\) and \(Q=(1, 1, 1)\). 11. The maximizing plane has normal parallel to \(\overrightarrow{OQ}=\langle 1,1,1\rangle\). Requiring \(\langle 2,1+k,2k\rangle\) to be parallel to this vector gives \(k_0=1\). 12. The maximum distance is \(\lVert\overrightarrow{OQ}\rVert=\sqrt3\).

a) \(S(1, 1, 2)\), and \(E:x-2y+z=1\) b) Every \(E_k\perp E\), and \(l:\vec{r}(t)=\langle 0,3,0\rangle+t\langle 1,-2,1\rangle\) c) \(k_0=1\), and the maximum distance is \(\sqrt3\).