- Compare the given equality with the dot-product angle formula. - Use the fact that both magnitudes are nonzero. - Interpret the resulting cosine value geometrically.

1. By the dot-product angle formula, \(\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha\). 2. Compare this with the given equation: \(\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha=-\|\mathbf{u}\|\|\mathbf{v}\|\). 3. Since both vectors are nonzero, divide by their positive magnitudes to get \(\cos\alpha=-1\). 4. Therefore, \(\alpha=180^\circ\). 5. The vectors are parallel but point in opposite directions; equivalently, one is a negative scalar multiple of the other.

\(\alpha=180^\circ\). The vectors are parallel and point in opposite directions.

- A dot product combines matching components. - Multiply the first components together and the second components together. - Add the resulting products.

1. Multiply corresponding components and add: \(\mathbf{u}\cdot\mathbf{v}=1\cdot3+2\cdot(-1)=1\).

\(1\)

- Multiply corresponding components and add. - For nonzero vectors, the sign of the dot product identifies whether the angle is acute, right, or obtuse.

1. Compute the dot product: \(\mathbf{u}\cdot\mathbf{v}=2(1)+1(3)=5.\) 2. The dot product is positive, so the angle between the nonzero vectors is acute.

\(\mathbf{u}\cdot\mathbf{v}=5\); the angle is acute.

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

1. Orthogonal vectors have zero dot product, so \(\mathbf{u}\cdot\mathbf{v}=3\cdot2+k(-6)=0\). 2. Thus, \(6-6k=0\) and \(k=1\).

\(\mathbf{u}\cdot\mathbf{v}=3\cdot2+k(-6)=0\), so \(k=1\).

- A dot product combines matching components, not their lengths. - Include the negative sign in the second product.

1. Multiply corresponding components and add the results. 2. \(2\cdot1+3\cdot(-2)=2-6=-4\).

\(-4\)

- Look at the angle where the two vectors start. - How are horizontal and vertical lines related?

1. One vector is horizontal and the other is vertical. 2. Horizontal and vertical directions meet at a right angle, so the vectors are perpendicular.

Yes. They lie along perpendicular coordinate axes.

- When a vector is dotted with itself, each component is multiplied by itself. - Compare the sum with the formula for the squared magnitude.

1. \(\mathbf{a}\cdot\mathbf{a}=(-2)^2+1^2=5\). 2. The squared length is also \(\|\mathbf{a}\|^2=5\), so \(\mathbf{a}\cdot\mathbf{a}=\|\mathbf{a}\|^2\).

\(5\), which equals \(\|\mathbf{a}\|^2\).

- The dot product has the same sign as the cosine when neither vector is zero. - Which kind of angle has a negative cosine?

1. A dot product equals the product of the vector lengths and the cosine of their angle. 2. Both lengths are positive, so the negative dot product means the cosine is negative. 3. The smaller angle is therefore obtuse.

Obtuse, because a negative dot product means the cosine of the angle is negative.

- Check the dot product directly. - Is either given vector made up entirely of zero components? - Think of a right angle between two nonzero arrows.

1. Calculate \(\mathbf{u}\cdot\mathbf{v}=2\cdot1+1\cdot(-2)=0\). 2. Both vectors have nonzero components and are not zero vectors. 3. Their dot product is zero because they are perpendicular, not because either vector is zero.

\(\mathbf{u}\cdot\mathbf{v}=0\), but neither vector is zero. A zero dot product can mean two nonzero vectors are perpendicular.

- The rule for a 3D dot product is the same as in 2D, with one additional pair of coordinates. - Account for the zero middle component and the negative third component.

1. Multiply the three pairs of matching coordinates. 2. Add them: \(1\cdot3+0\cdot5+(-2)\cdot4=3-8=-5\).

\(-5\)

- A dot product with \(\mathbf{i}\) selects the signed horizontal component. - Which direction does a negative horizontal component indicate?

1. Calculate \(\mathbf{d}\cdot\mathbf{i}=(-3)\cdot1+4\cdot0=-3\). 2. The negative sign means the displacement has a \(3\)-unit component to the left.

\(-3\); the horizontal component is \(3\) units to the left.

- 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 \(\mathbf{v}=\overrightarrow{AB}=\langle3,4,0\rangle\). 2. A normal vector for the plane is \(\mathbf{n}=\langle3,0,-4\rangle\). 3. Their magnitudes are both \(5\), and \(\mathbf{v}\cdot\mathbf{n}=9\). 4. For the angle between a line and a plane, \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac{9}{25}\). 5. Therefore, \(\alpha=\sin^{-1}\left(\frac{9}{25}\right)\approx21.1^\circ\).

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

- 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 \(\mathbf{n}_1=\langle1,1,0\rangle\) and \(\mathbf{n}_2=\langle1,0,1\rangle\). 2. Their magnitudes are both \(\sqrt2\), and their dot product is \(1\). 3. The acute angle between the planes equals the acute angle between their normal vectors, so \(\cos\phi=\frac{1}{\sqrt2\sqrt2}=\frac12\). 4. Therefore, \(\phi=60^\circ\).

\(60^\circ\)

- 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 \(c=-\frac12\). Then \(a=6\left(-\frac12\right)=-3\), and the third coordinate also matches. 2. Perpendicular direction vectors have dot product \(0\). Thus, \(\langle a,2,-3\rangle\cdot\langle1,4,2\rangle=a+8-6=0\), so \(a=-2\).

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

- 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 \(\mathbf{n}=\langle2,-4,2\rangle\). 2. For the line to be parallel to the plane, its direction vector must be orthogonal to \(\mathbf{n}\): \(\langle2,a,3\rangle\cdot\langle2,-4,2\rangle=0\). 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 \(2(4)-4(1)+2(2)=8\ne5\), so the line is strictly parallel rather than contained in the plane.

\(a=\frac52\)

- How does a normal vector relate to every vector that lies in the plane? - Use the given point to form the displacement vector. - Expand the dot product to obtain the standard equation.

1. In point-normal form, a point \(P\) on the plane and a normal vector \(\mathbf{n}\) give \(\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0\). Thus, \(E:\langle3,5,-1\rangle\cdot\langle x-7,y+2,z-4\rangle=0\). 2. Expand the dot product: \(3(x-7)+5(y+2)-(z-4)=0\). 3. Simplify: \(3x+5y-z=7\).

Point-normal form: \(E:\langle3,5,-1\rangle\cdot\langle x-7,y+2,z-4\rangle=0\) 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 \(\overrightarrow{LM}=\langle 5-2,2-2,-1-2\rangle=\langle3,0,-3\rangle\). 2. Use the simpler parallel normal vector \(\mathbf{n}=\langle1,0,-1\rangle\). 3. Apply the point-normal condition at \(S\): \(\langle1,0,-1\rangle\cdot\langle x,y,z-4\rangle=0\). 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 \(\mathbf{n}_1=\langle4,0,3\rangle\) and \(\mathbf{n}_2=\langle0,5,0\rangle\). 2. The \(y\)-axis direction is \(\langle0,1,0\rangle\), and \(\mathbf{n}_1\cdot\langle0,1,0\rangle=0\). Since the origin also satisfies \(E_1\), the entire \(y\)-axis lies in \(E_1\). 3. The normal \(\mathbf{n}_2\) points in the \(y\)-direction, and \(5y=0\) is the same as \(y=0\). Thus, \(E_2\) is the \(xz\)-plane.

For \(E_1\), \(\mathbf{n}_1=\langle4,0,3\rangle\) and \(\mathbf{n}_1\cdot\langle0,1,0\rangle=0\). The origin is also in the plane, so \(E_1\) contains the entire \(y\)-axis. For \(E_2\), \(\mathbf{n}_2=\langle0,5,0\rangle\) is perpendicular to the \(xz\)-plane, and \(5y=0\) describes exactly the \(xz\)-plane.

- 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 \(\mathbf{u}=\langle1,2,2\rangle\) and \(\mathbf{v}=\langle-4,4,7\rangle\). 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 \(\mathbf{n}_1=\langle2,1,-2\rangle\) and \(\mathbf{n}_2=\langle3,-4,0\rangle\). 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, \(\alpha=\cos^{-1}\left(\frac{2}{15}\right)\approx82.3^\circ\).

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

- 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 \(\overrightarrow{DE}=\langle2,1,k-2\rangle\) and \(\overrightarrow{DF}=\langle1,2,-2\rangle\). 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 \(k=4\).

\(k=4\)

- Form two vectors that both begin at \(B\). - Find their dot product and magnitudes. - Apply the dot-product angle formula.

1. Use vectors that begin at \(B\): \(\overrightarrow{BA}=\langle-3,-2,0\rangle\) and \(\overrightarrow{BC}=\langle0,4,-3\rangle\). 2. Their dot product is \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-8\). 3. Their magnitudes are \(\|\overrightarrow{BA}\|=\sqrt{13}\) and \(\|\overrightarrow{BC}\|=5\). 4. Therefore, \(\cos\beta=\frac{-8}{5\sqrt{13}}\). 5. Thus, \(\beta=\cos^{-1}\left(\frac{-8}{5\sqrt{13}}\right)\approx116.34^\circ\).

\(\overrightarrow{BA}=\langle-3,-2,0\rangle\), \(\overrightarrow{BC}=\langle0,4,-3\rangle\); \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-8\). Hence \(\cos\beta=\frac{-8}{5\sqrt{13}}\) and \(\beta\approx116.34^\circ\).

- Multiply corresponding components and add. - Relate the sign of the dot product to the sign of the cosine of the angle. - Decide which angle type has a negative cosine.

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

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

- 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 \(k=3\).

\(k=3\)

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

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

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

- Think of vectors along different coordinate axes. - Relate a zero dot product to the angle between nonzero vectors. - Check that both example vectors are nonzero before drawing the conclusion.

1. The statement does not carry over directly to dot products. 2. For example, let \(\mathbf{u}=\langle1,0,0\rangle\) and \(\mathbf{v}=\langle0,1,0\rangle\). 3. Both vectors are nonzero, but \(\mathbf{u}\cdot\mathbf{v}=0\). 4. Two nonzero vectors have dot product \(0\) exactly when they are orthogonal.

No. For example, \(\langle1,0,0\rangle\cdot\langle0,1,0\rangle=0\), although both vectors are nonzero. They are perpendicular.

- Identify the type of object produced by each inner dot product. - Check whether the outer dot product then has two vector inputs. - An associative law requires both regroupings to be meaningful expressions of the same operation.

1. \(\mathbf{a}\cdot\mathbf{b}=1\) and \(\mathbf{b}\cdot\mathbf{c}=1\). 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. \(\mathbf{a}\cdot\mathbf{b}=1\), \(\mathbf{b}\cdot\mathbf{c}=1\) 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.

- Find the squared magnitude of each vector. - Find the dot product of the vectors. - Use the dot-product identity for the area of a parallelogram.

1. Compute the squared magnitudes: \(\|\mathbf{u}\|^2=6^2+(-2)^2=40\) and \(\|\mathbf{v}\|^2=3^2+4^2=25\). 2. Compute the dot product: \(\mathbf{u}\cdot\mathbf{v}=6\cdot3+(-2)\cdot4=10\). 3. The squared area of the parallelogram is \(\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2=40\cdot25-10^2=900\). 4. Therefore, the area is \(\sqrt{900}=30\).

\(30\) square units

- Track whether each intermediate result is a scalar or a vector. - Check that addition or subtraction combines compatible object types. - Remember that a dot product requires two vector inputs.

1. The expression is defined. \(\mathbf{a}\cdot\mathbf{b}\) is a scalar, and scaling \(\mathbf{c}\) 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 \(\mathbf{b}\cdot\mathbf{c}\) 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

- Identify the dot-product value that characterizes perpendicular vectors. - Compute the dot product as an expression in \(k\). - Solve the resulting linear equation.

1. Perpendicular vectors have dot product \(0\). 2. Compute \(\mathbf{u}\cdot\mathbf{v}=2(k+1)+3k-8=5k-6\). 3. Set the expression equal to zero: \(5k-6=0\). 4. Therefore, \(k=\frac65\).

\(\mathbf{u}\cdot\mathbf{v}=2(k+1)+3k-8=5k-6\). For perpendicular vectors, set this equal to \(0\), giving \(k=\frac65\).

- A right angle at a vertex is detected by the dot product of the two side vectors meeting there. - Start with one vertex and form both side vectors from that common initial point. - A zero dot product is enough to establish the right angle.

1. Form two side vectors meeting at \(A\): \(\overrightarrow{AB}=\langle4,2,1\rangle\) and \(\overrightarrow{AC}=\langle-1,1,2\rangle\). 2. Their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4(-1)+2(1)+1(2)=0\). 3. Therefore, \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are perpendicular, so the triangle has a right angle at \(A\).

\(\overrightarrow{AB}=\langle4,2,1\rangle\) and \(\overrightarrow{AC}=\langle-1,1,2\rangle\) have dot product \(-4+2+2=0\), so \(\triangle ABC\) has a right angle at \(A\).

- For a dot product, multiply corresponding components and add the products. - In part c, evaluate the vector sum inside the parentheses first. - You can use distributivity as a check on part c.

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

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

- Write the dot product as an expression in \(k\). - Set that expression equal to the required value \(11\). - 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 \(k=3\).

\(k=3\)

- Track whether each intermediate result is a scalar or a vector. - Check that each operation receives the type of inputs it requires. - Use \(\mathbf{a}\cdot\mathbf{a}=\|\mathbf{a}\|^2\) for the final two parts.

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

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

- Add \(\mathbf{b}\) and \(\mathbf{c}\) component by component. - Apply the component definition of the dot product to that sum. - Use real-number distributivity to separate the terms into two dot products.

1. \(\mathbf{b}+\mathbf{c}=\langle b_1+c_1,b_2+c_2,b_3+c_3\rangle\). 2. Therefore, \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})=a_1(b_1+c_1)+a_2(b_2+c_2)+a_3(b_3+c_3)\). 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 \(\mathbf{a}\cdot\mathbf{b}\) and \(\mathbf{a}\cdot\mathbf{c}\), proving the identity.

\(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})=\sum_{i=1}^{3}a_i(b_i+c_i)=\sum_{i=1}^{3}a_ib_i+\sum_{i=1}^{3}a_ic_i=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}\). This follows by distributing each component product and regrouping.

- Write the scaled vector in component form first. - Apply the component definition of the dot product. - Identify which real-number properties let you factor the common scalar \(k\).

1. \(k\mathbf{u}=\langle ku_1,ku_2,ku_3\rangle\). 2. Then \((k\mathbf{u})\cdot\mathbf{v}=(ku_1)v_1+(ku_2)v_2+(ku_3)v_3\). 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 \(k\): \(k(u_1v_1+u_2v_2+u_3v_3)=k(\mathbf{u}\cdot\mathbf{v})\).

\((k\mathbf{u})\cdot\mathbf{v}=(ku_1)v_1+(ku_2)v_2+(ku_3)v_3=k(u_1v_1+u_2v_2+u_3v_3)=k(\mathbf{u}\cdot\mathbf{v})\). The steps use associativity of real-number multiplication and the distributive property.

- Before checking arithmetic, identify the type of object a dot product should produce. - Check the sign of every component product separately. - Compare each calculation with the component definition of 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 directed segment by subtracting endpoint coordinates. - Keep \(a\) symbolic rather than substituting a numerical edge length. - Use the zero-dot-product criterion for perpendicularity.

1. The direction vectors are \(\overrightarrow{OF}=\langle a,a,a\rangle\) and \(\overrightarrow{EB}=\langle0,a,-a\rangle\). 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.

- Write two side vectors with a common initial point. - Use their magnitudes and dot product to find the area of the parallelogram they form. - The triangle has half the area of that parallelogram.

1. Use \(\overrightarrow{AB}=\langle-2,2,0\rangle\) and \(\overrightarrow{AC}=\langle-2,0,2\rangle\). 2. Both squared magnitudes are \(8\), and the dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\). 3. The squared parallelogram area is \(8\cdot8-4^2=48\). 4. The triangle has half this area, giving \(\frac12\sqrt{48}=2\sqrt3\) square units.

\(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\); area \(=\frac12\sqrt{8\cdot8-4^2}=2\sqrt3\) square units.

- Read a normal vector directly from the Cartesian equation of \(E\). - Test both spanning directions of \(F\) against that normal. - Check whether the initial point of \(F\) lies in \(E\) before deciding between identical and strictly parallel planes.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle1,1,-2\rangle\). 2. Both direction vectors of \(F\) are orthogonal to \(\mathbf{n}\): \(\langle2,0,1\rangle\cdot\mathbf{n}=2-2=0\) and \(\langle1,1,1\rangle\cdot\mathbf{n}=1+1-2=0\). 3. Therefore, \(F\) is parallel to \(E\). 4. The initial point of \(F\) does not satisfy the equation of \(E\), because \(1+2-2(1)=1\ne2\). 5. The planes are distinct and strictly parallel, so no intersection line exists.

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

- A beam striking a plane at a right angle points in a normal direction. - A displacement vector lying in the plane must be perpendicular to that normal. - Expand the resulting zero dot product to obtain standard form.

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, \(\overrightarrow{PX}=\langle x-5,y+1,z-4\rangle\) is orthogonal to \(\mathbf{v}\). Therefore, \(\langle2,2,-1\rangle\cdot\langle x-5,y+1,z-4\rangle=0\). 3. Expand: \(2(x-5)+2(y+1)-(z-4)=0\). 4. Simplifying gives \(2x+2y-z=4\).

a) \(\langle2,2,-1\rangle\cdot\langle x-5,y+1,z-4\rangle=0\) b) \(2x+2y-z=4\)

- Compute \(\mathbf{n}\cdot\mathbf{p}\) component by component. - In \(\mathbf{n}\cdot\mathbf{x}=d\), the normal-vector components become the coordinate coefficients. - Expand only after the dot-product form is established.

1. Compute \(d\): \(d=\langle1,3,-2\rangle\cdot\langle5,-2,4\rangle=1(5)+3(-2)-2(4)=5-6-8=-9\). 2. Therefore, the vector equation is \(\langle1,3,-2\rangle\cdot\mathbf{x}=-9\). 3. Writing \(\mathbf{x}=\langle x,y,z\rangle\) and expanding the dot product gives \(x+3y-2z=-9\).

a) \(d=-9\), so \(\langle1,3,-2\rangle\cdot\mathbf{x}=-9\) b) \(x+3y-2z=-9\)

- Use the given normal and point to calculate the constant \(d\). - Expand the dot-product equation to obtain standard form. - Test membership by substituting the coordinates of \(S\).

1. Compute \(d=\mathbf{n}\cdot\mathbf{q}=\langle4,-2,5\rangle\cdot\langle3,0,-1\rangle=4(3)-2(0)+5(-1)=7\). 2. Thus, \(\langle4,-2,5\rangle\cdot\mathbf{x}=\langle4,-2,5\rangle\cdot\langle3,0,-1\rangle=7\). 3. Expanding gives \(4x-2y+5z=7\). 4. For \(S\), the left side is \(4(1)-2(2)+5(0)=0\). Since \(0\ne7\), \(S\) does not lie in \(F\).

a) \(\langle4,-2,5\rangle\cdot\mathbf{x}=\langle4,-2,5\rangle\cdot\langle3,0,-1\rangle=7\) b) \(4x-2y+5z=7\) c) No, \(S\notin F\).

- Evaluate each dot product component by component. - Points in one plane with a fixed normal have the same constant value of \(\mathbf{x}\cdot\mathbf{n}\). - Substitute the coordinates of \(C\) into the resulting plane equation.

1. \(\overrightarrow{OA}\cdot\mathbf{n}=\langle1,2,3\rangle\cdot\langle1,1,1\rangle=1+2+3=6\). Also, \(\overrightarrow{OB}\cdot\mathbf{n}=\langle5,0,1\rangle\cdot\langle1,1,1\rangle=5+0+1=6\). 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 direction and the plane's two spanning directions. - Verify that the two plane directions are not scalar multiples. - Test the line direction against each plane direction with a dot product.

1. The line direction is \(\mathbf{w}=\langle2,-2,1\rangle\). The plane directions are \(\mathbf{u}=\langle1,1,0\rangle\) and \(\mathbf{v}=\langle0,1,2\rangle\). 2. The plane directions are not scalar multiples, so they define two distinct directions in the plane. 3. Compute \(\mathbf{w}\cdot\mathbf{u}=2-2+0=0\). 4. Compute \(\mathbf{w}\cdot\mathbf{v}=0-2+2=0\). 5. Because the line direction is orthogonal to both nonparallel plane directions, \(g\) is perpendicular to \(E\).

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

- Evaluate the plane's dot-product expression at each point. - Membership requires the result to equal the constant \(8\). - In part b, the membership test becomes a linear equation in \(x\).

1. For \(P\), \(\langle5,-2,1\rangle\cdot\langle1,0,3\rangle=5(1)-2(0)+3=8\). 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) \(x=3\)

- Use the known point to determine the constant \(d\). - Test each candidate point with the same dot-product expression. - A point belongs to the plane only when the result equals \(d\).

1. Compute the constant using \(A\): \(d=\langle2,3,-4\rangle\cdot\langle1,2,1\rangle=2+6-4=4\). Thus, \(E:\langle2,3,-4\rangle\cdot\mathbf{x}=4\). 2. For \(B\), \(2(4)+3(0)-4(1)=4\), so \(B\in E\). 3. For \(C\), \(2(2)+3(1)-4(1)=3\ne4\), so \(C\notin E\).

a) \(E:\langle2,3,-4\rangle\cdot\mathbf{x}=4\) b) \(B\in E\), but \(C\notin E\).

- Read a normal vector from the plane equation. - Use a dot product to test whether the line direction is parallel to the plane. - Test the line's initial point in the plane equation to distinguish containment from strict parallelism.

1. A normal to \(E\) is \(\mathbf{n}=\langle1,2,-2\rangle\). Its dot product with the line direction is \(\mathbf{n}\cdot\langle2,2,3\rangle=2+4-6=0\). 2. Thus, the line is parallel to the plane's directions. 3. Its starting point does not lie in the plane: \(4+2(1)-2(1)=4\ne8\). 4. Therefore, \(g\) is strictly parallel to \(E\), and they have no common points.

\(\mathbf{n}\cdot\langle2,2,3\rangle=0\). Since \((4,1,1)\notin E\), the line is strictly parallel to the plane and 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 into the standard-form family. - Parallel planes are identical only when they share a point.

1. A normal to \(E_1\) is \(\mathbf{n}_1=\langle2,1,-2\rangle\). The directions of \(E_2\) satisfy \(\mathbf{n}_1\cdot\langle1,0,1\rangle=2-2=0\) and \(\mathbf{n}_1\cdot\langle1,-2,0\rangle=2-2=0\). 2. Thus, \(E_2\) is parallel to every plane in the family \(E_1\). 3. The initial point \((1,2,1)\) of \(E_2\) gives \(2(1)+2-2(1)=2\), so the planes are identical for \(d=2\). 4. For \(d\ne2\), the parallel planes have no shared points and are strictly parallel.

\(\mathbf{n}_1\cdot\langle1,0,1\rangle=0\) and \(\mathbf{n}_1\cdot\langle1,-2,0\rangle=0\). The planes are identical when \(d=2\). For every \(d\ne2\), they are strictly parallel.

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

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

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

- Read each vector's components from its endpoint. - Compute the dot product from corresponding components. - Recall what a zero dot product means for two nonzero vectors.

1. From the diagram, \(\mathbf{u}=\langle2,1\rangle\) and \(\mathbf{v}=\langle-1,2\rangle\). 2. Their dot product is \(\mathbf{u}\cdot\mathbf{v}=2\cdot(-1)+1\cdot2=0\). 3. Since both vectors are nonzero and the dot product is zero, they are perpendicular.

\(\mathbf{u}\cdot\mathbf{v}=0\); the vectors are perpendicular.

- Read the two vectors from their endpoints in the diagram. - The projection must point along the same line as \(\mathbf{u}\). - Compare the dot product with \(\mathbf{u}\) to the squared magnitude of \(\mathbf{u}\).

1. From the diagram, \(\mathbf{u}=\langle2,2\rangle\) and \(\mathbf{v}=\langle4,2\rangle\). 2. The vector projection is \(\operatorname{proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{v}\cdot\mathbf{u}}{\mathbf{u}\cdot\mathbf{u}}\mathbf{u}\). 3. Here \(\mathbf{v}\cdot\mathbf{u}=12\) and \(\mathbf{u}\cdot\mathbf{u}=8\), giving \(\frac{12}{8}\langle2,2\rangle=\langle3,3\rangle\).

\(\langle3,3\rangle\)

- Use the scalar-projection formula. - Compute the dot product of the two non-axis-aligned vectors. - Divide by the magnitude of the vector being projected onto.

1. Use the scalar-projection formula \(\frac{\mathbf{v}\cdot\mathbf{u}}{\|\mathbf{u}\|}\). 2. Here, \(\mathbf{v}\cdot\mathbf{u}=4(3)+3(4)=24\) and \(\|\mathbf{u}\|=5\). 3. Therefore, the scalar projection is \(\frac{24}{5}\).

\(\frac{24}{5}\).

- What value must the dot product have for perpendicular vectors? - Use the unknown component in the dot-product calculation.

1. Perpendicular vectors have dot product \(0\). 2. The equation is \(2t+4=0\), which gives \(t=-2\).

\(2t+4=0\), so \(t=-2\).

- The dot product connects two vector lengths to the angle between them. - Find the two magnitudes independently. - Check the sign of the dot product before interpreting the angle.

1. \(\mathbf{u}\cdot\mathbf{v}=2(-1)+3(4)=10\), \(\|\mathbf{u}\|=\sqrt{13}\), and \(\|\mathbf{v}\|=\sqrt{17}\). 2. \(\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}=\frac{10}{\sqrt{221}}\). 3. Thus \(\theta=\cos^{-1}(\frac{10}{\sqrt{221}})\approx47.7^\circ\).

\(\mathbf{u}\cdot\mathbf{v}=10\), \(\|\mathbf{u}\|=\sqrt{13}\), \(\|\mathbf{v}\|=\sqrt{17}\); \(\cos\theta=\frac{10}{\sqrt{221}}\), so \(\theta\approx47.7^\circ\).

- The vector projection lies along the axis you project onto. - Start with the signed scalar component in the \(\mathbf{i}\) direction. - A projection can point opposite to the chosen unit vector.

1. The vector \(\mathbf{i}\) is a unit vector, so the projection is \((\mathbf{a}\cdot\mathbf{i})\mathbf{i}\). 2. Since \(\mathbf{a}\cdot\mathbf{i}=-2\), the projection is \(-2\langle1,0\rangle=\langle-2,0\rangle\). 3. It points left, along the negative \(x\)-direction.

\(\langle-2,0\rangle\), pointing in the negative \(x\)-direction.

- The sign of the dot product tells you whether the angle is acute or obtuse. - Find the cosine before applying the inverse cosine function. - What information disappears when a negative value is replaced by its absolute value?

1. The dot product is \(\mathbf{u}\cdot\mathbf{v}=-1\), and the lengths are \(1\) and \(\sqrt2\). 2. The cosine is \(\cos\theta=\frac{-1}{\sqrt2}\), not its absolute value. 3. Hence \(\theta=135^\circ\); taking the absolute value incorrectly turns an obtuse angle into an acute one.

\(135^\circ\). The negative sign must stay because it shows the angle is obtuse.

- Read the horizontal and vertical displacement of each vector. - Multiply corresponding components and add. - What does a zero dot product mean for two nonzero vectors?

1. The components are \(\mathbf{u}=\langle2,1\rangle\) and \(\mathbf{v}=\langle-1,2\rangle\). 2. Their dot product is \(2\cdot(-1)+1\cdot2=0\). 3. Because neither vector is zero, the zero dot product confirms they are perpendicular.

Yes. \(\mathbf{u}\cdot\mathbf{v}=-2+2=0\), so they are perpendicular.

- 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 \(\langle2,-1,2\rangle\cdot\langle1,4,1\rangle=2-4+2=0\). 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 \(\langle1,-2,1\rangle\) and \(\langle2,1,0\rangle\) are not scalar multiples. Therefore, the intersecting lines determine a unique plane. 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be normal to the plane. Perpendicularity to the two line directions gives \(a-2b+c=0\) and \(2a+b=0\). One valid choice is \(\mathbf{n}=\langle-1,2,5\rangle\). 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 \(\langle-1,2,5\rangle\), and \(E:-x+2y+5z=18\).

- Write a general vector and impose a zero dot product with each spanning direction. - Any nonzero scalar multiple of a valid normal vector is also valid. - After finding the normal, the displacement from the known plane point to \(P\) must be orthogonal to it.

1. Let \(\mathbf{n}=\langle x,y,z\rangle\). Orthogonality gives \(x+2z=0\) and \(2x+y-z=0\). 2. One nonzero solution is \(\mathbf{n}=\langle-2,5,1\rangle\). 3. \(\overrightarrow{P_0P}=\langle3,1,a-3\rangle\). Since this displacement lies in the plane, \(\langle-2,5,1\rangle\cdot\langle3,1,a-3\rangle=0\). 4. Thus, \(-6+5+a-3=0\), so \(a=4\).

One normal vector is \(\mathbf{n}=\langle-2,5,1\rangle\), and \(a=4\).

- 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 \(\overrightarrow{AB}=\langle18,0,0.5\rangle\) and \(\overrightarrow{AC}=\langle0,18,0.25\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both vectors. Then \(18a+0.5c=0\) and \(18b+0.25c=0\). Choosing \(c=-72\) gives \(a=2\), \(b=1\), and \(\mathbf{n}=\langle2,1,-72\rangle\). 3. Since \(A\) is the origin, the plane equation is \(E:2x+y-72z=0\). 4. At \(x=18\), \(y=18\), \(2\cdot18+18-72z=0\), so \(z=\frac{3}{4}=0.75\). 5. Three noncollinear contact points fix a plane. A fourth contact point at a different height cannot lie in that same plane, which explains possible wobbling.

a) One normal vector is \(\mathbf{n}=\langle2,1,-72\rangle\), since its dot products with \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are both zero. The plane is \(E:2x+y-72z=0\). b) \(z=0.75\,\text{in.}\). A fourth contact point must satisfy the plane equation exactly; otherwise, not all four legs touch the same plane.

- The sought normal must have zero dot product with each independent tabletop direction. - Use the known support point only after the normal vector has been determined. - For the center height, substitute the given horizontal coordinates into the Cartesian equation.

1. \(\overrightarrow{P_1P_2}=\langle40,0,1\rangle\) and \(\overrightarrow{P_1P_3}=\langle0,40,-1\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), the conditions are \(40a+c=0\) and \(40b-c=0\). One solution is \(\mathbf{n}=\langle-1,1,40\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle0,0,30\rangle)\cdot\langle-1,1,40\rangle=0\). Expanding gives \(-x+y+40z=1200\). 4. At \(x=20\) and \(y=20\), \(-20+20+40z=1200\), so \(z=30\).

a) \(\overrightarrow{P_1P_2}=\langle40,0,1\rangle\), \(\overrightarrow{P_1P_3}=\langle0,40,-1\rangle\), and one normal is \(\mathbf{n}=\langle-1,1,40\rangle\). b) \((\mathbf{x}-\langle0,0,30\rangle)\cdot\langle-1,1,40\rangle=0\); equivalently, \(-x+y+40z=1200\) c) \(30\,\text{in.}\)

- The axis intercepts occur where two coordinates are zero; the given trace parametrizations make those values accessible directly. - Use two independent displacement vectors lying in the plane before looking for a normal. - A plane normal has zero dot product with every direction lying in that plane.

1. The common x-axis intercept is \(S_1=(5, 0, 0)\). Setting \(r=1\) gives \(S_2=(0, 2, 0)\), and setting \(s=1\) gives \(S_3=(0, 0, 10)\). 2. \(\mathbf{u}=\langle-5,2,0\rangle\) and \(\mathbf{v}=\langle-5,0,10\rangle\). For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(-5a+2b=0\) and \(-5a+10c=0\). 3. One nonzero solution is \(\mathbf{n}=\langle2,5,1\rangle\). 4. Using \(S_1\), \((\mathbf{x}-\langle5,0,0\rangle)\cdot\langle2,5,1\rangle=0\), which expands to \(2x+5y+z=10\). 5. For \(Q\), \(2(1)+5(1)+3=10\), so \(Q\) lies in the plane.

a) \(S_1=(5, 0, 0)\), \(S_2=(0, 2, 0)\), \(S_3=(0, 0, 10)\) b) One normal vector is \(\mathbf{n}=\langle2,5,1\rangle\). c) \(2x+5y+z=10\); yes, \(Q\) lies in the plane.

- Compare the lengths and dot product of the two direction vectors. - How do the parameter intervals affect the side lengths? - What parameter values lie at the center of both intervals? - The distance from parameter \(-1\) to \(1\) is \(2\).

1. Let \(\mathbf{u}=\langle2,1,0\rangle\) and \(\mathbf{v}=\langle-1,2,0\rangle\). Their dot product is \(\mathbf{u}\cdot\mathbf{v}=2(-1)+1\cdot2+0=0\), so they are perpendicular. 2. Their lengths are equal: \(\|\mathbf{u}\|=\sqrt5\) and \(\|\mathbf{v}\|=\sqrt5\). Both parameters vary across intervals of length \(2\), so the four sides are equal and adjacent sides are perpendicular. Thus, \(M\) is a square. 3. The parameter intervals are centered at \(r=0\) and \(s=0\), so the center is \((0, 0, 5)\). 4. Each side has length \(2\sqrt5\). Therefore, the area is \((2\sqrt5)^2=20\) square units.

a) The direction vectors are perpendicular and have equal length, and both parameter intervals have length \(2\); therefore, \(M\) is a square. b) \((0, 0, 5)\) c) \(20\) square units

- What evidence is needed to show that two given lines really meet? - If the directions differ, how many planes can pass through both intersecting lines? - What property must a plane's normal satisfy relative to both line directions?

1. Equating coordinates gives \(r=2s\), \(1+2r=6-s\), and \(-1+2r=2+s\). These equations give the unique values \(s=1\), \(r=2\), which satisfy all three coordinates, so \(S=(4,5,3)\). 2. The direction vectors \(\langle1,2,2\rangle\) and \(\langle2,-1,1\rangle\) are not scalar multiples. Two intersecting lines with distinct directions determine a unique plane. 3. For \(\mathbf{n}=\langle a,b,c\rangle\), perpendicularity gives \(a+2b+2c=0\) and \(2a-b+c=0\). One nonzero solution is \(\mathbf{n}=\langle4,3,-5\rangle\). 4. Using \(S\), \(4x+3y-5z=16\).

a) Equating the two lines gives \(r=2s\), \(1+2r=6-s\), and \(-1+2r=2+s\). The only solution is \(r=2\), \(s=1\), giving \(S=(4,5,3)\). Since the directions are not scalar multiples, this intersection is unique. b) The intersecting nonparallel lines determine one plane. A normal \(\mathbf{n}=\langle4,3,-5\rangle\) has zero dot product with both direction vectors, and the plane is \(E:4x+3y-5z=16\).

- Do the line directions show that these lines intersect, coincide, or stay apart? - What additional information distinguishes a single line from two distinct parallel ones? - How could you test whether your proposed plane contains both lines?

1. The direction vector of \(h\) is \(-2\) times the direction vector of \(g\), so their directions are parallel. 2. The point \((1,0,3)\) is not on \(h\): the first coordinate requires \(k=\frac{1}{2}\), while the second requires \(k=-1\). Thus, the lines are distinct and parallel. 3. Use \(\mathbf{u}=\langle2,-1,4\rangle\) and the connecting vector \(\mathbf{v}=\langle2,2,-2\rangle\). A normal \(\mathbf{n}=\langle a,b,c\rangle\) satisfies \(2a-b+4c=0\) and \(2a+2b-2c=0\). One solution is \(\mathbf{n}=\langle-1,2,1\rangle\). 4. Using \((1,0,3)\), the common plane is \(F:-x+2y+z=2\).

a) The lines are distinct and parallel: their direction vectors are scalar multiples, but \((1,0,3)\) does not lie on \(h\). b) Distinct parallel lines determine one plane. The common direction \(\langle2,-1,4\rangle\) and connecting vector \(\langle2,2,-2\rangle\) are both perpendicular to \(\mathbf{n}=\langle-1,2,1\rangle\). The plane through \((1,0,3)\) is \(F:-x+2y+z=2\).

- What must be true for a point to lie on a parametric line? - Why does it matter whether the given point is already on the line? - How could you check that the final plane contains both the line and the separate point?

1. To place \(A\) on \(h\), the first coordinate would require \(0=2+t\), or \(t=-2\). But that value gives \(y=5-2(-2)=9\neq1\). Thus, \(A\notin h\). A line and a point outside it determine a unique plane. 2. Use \(\mathbf{u}=\langle1,-2,3\rangle\) and \(\mathbf{v}=\langle-2,-4,5\rangle\). A normal \(\mathbf{n}=\langle a,b,c\rangle\) satisfies \(a-2b+3c=0\) and \(-2a-4b+5c=0\). One solution is \(\mathbf{n}=\langle2,-11,-8\rangle\). 3. Through \(A\), the plane equation is \(E:2x-11y-8z=-43\).

a) The first coordinate on \(h\) requires \(t=-2\), but that gives \(y=9\), not \(1\). Thus, \(A\notin h\). A line and a point outside it determine one plane. b) One normal vector is \(\mathbf{n}=\langle2,-11,-8\rangle\); the equation is \(E:2x-11y-8z=-43\).

- 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 \(\mathbf{n}=\langle p,q,w\rangle\) 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 \(\mathbf{n}=\langle1,-2,-2\rangle\). 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 \(\langle1,-2,-2\rangle\cdot\langle2,1,0\rangle=2-2=0\). 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 \(a=1\).

\(a=1\)

- Which two nonparallel directions are available from the three points? - How could you check that a proposed normal is perpendicular to the whole plane? - How can you verify that your equation includes all three original points?

1. Two spanning vectors are \(\overrightarrow{PQ}=\langle2,2,-1\rangle\) and \(\overrightarrow{PR}=\langle-2,4,3\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), perpendicularity to the spanning vectors gives \(2a+2b-c=0\) and \(-2a+4b+3c=0\). One solution is \(\mathbf{n}=\langle5,-2,6\rangle\). 3. Using \(P\), the point-normal equation is \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle5,-2,6\rangle=0\). 4. Expanding gives \(E:5x-2y+6z=17\).

One normal vector is \(\mathbf{n}=\langle5,-2,6\rangle\), obtained from \(2a+2b-c=0\) and \(-2a+4b+3c=0\). Point-normal form: \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle5,-2,6\rangle=0\). 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. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality to the plane directions requires \(a+2b+2c=0\) and \(3a+c=0\). 2. From \(c=-3a\), \(2b=5a\). Taking \(a=2\) gives \(\mathbf{n}=\langle2,5,-6\rangle\). 3. Point-normal form is \(\langle2,5,-6\rangle\cdot\langle x,y-4,z+2\rangle=0\). Expanding gives \(2x+5y-6z=32\). 4. Point \(A\) satisfies \(2(4)+5(6)-6(1)=32\); point \(B\) satisfies \(2(7)+5(6)-6(2)=32\). 5. A plane containing two distinct points contains their entire line, so \(AB\) lies in \(E\).

a) One normal vector is \(\mathbf{n}=\langle2,5,-6\rangle\), satisfying \(\mathbf{n}\cdot\langle1,2,2\rangle=0\) and \(\mathbf{n}\cdot\langle3,0,1\rangle=0\). Point-normal form: \(\langle2,5,-6\rangle\cdot\langle x,y-4,z+2\rangle=0\). Standard form: \(2x+5y-6z=32\). b) Yes. Both \(A\) and \(B\) satisfy \(2x+5y-6z=32\), so the line through them lies entirely 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 \(\mathbf{n}=\langle a,b,c\rangle\). The conditions \(\mathbf{n}\cdot\langle2,1,0\rangle=0\) and \(\mathbf{n}\cdot\langle0,2,-1\rangle=0\) give \(2a+b=0\) and \(2b-c=0\). 2. Taking \(a=-1\) yields \(\mathbf{n}=\langle-1,2,4\rangle\). 3. Point-normal form is \(\langle-1,2,4\rangle\cdot\langle x-5,y+1,z-3\rangle=0\). Expanding gives \(-x+2y+4z=5\). 4. For \(P\), \(-1+2+4=5\), so \(P\in E\). For \(Q\), \(-3+4+8=9\neq5\), so \(Q\notin E\). Hence \(g\) is not contained in \(E\).

a) One normal vector is \(\mathbf{n}=\langle-1,2,4\rangle\), which has zero dot product with both plane directions. 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. The point \(P\) satisfies the plane equation, but \(Q\) gives \(9\) instead of \(5\). Thus, \(g\) 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 \(\mathbf{n}=\langle a,b,c\rangle\). 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 \(\mathbf{n}=\langle2,-4,-1\rangle\). 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 normal vector to \(F\) must be orthogonal to both \(\langle0,1,0\rangle\) and \(\langle0,0,1\rangle\). Choose \(\mathbf{n}_F=\langle1,0,0\rangle\). 2. Using the point \((3,0,0)\), the standard equation is \(x=3\). 3. 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. Plane \(G\) contains the direction \(\langle1,0,0\rangle\) of the x-axis and the direction \(\overrightarrow{OP}=\langle0,1,1\rangle\). 5. A normal vector \(\mathbf{n}_G=\langle a,b,c\rangle\) must satisfy \(a=0\) and \(b+c=0\). Choose \(\mathbf{n}_G=\langle0,-1,1\rangle\). 6. Since the origin lies in \(G\), a point-normal equation is \(\mathbf{r}\cdot\langle0,-1,1\rangle=0\), or \(-y+z=0\). 7. Finally, \(\mathbf{n}_F\cdot\mathbf{n}_G=0\), 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 \(xz\)-plane. c) A point-normal equation is \(\mathbf r\cdot\langle0,-1,1\rangle=0\). Normals \(\mathbf n_F=\langle1,0,0\rangle\) and \(\mathbf n_G=\langle0,-1,1\rangle\) have dot product \(0\), proving the planes are perpendicular.

- A normal to the plane must be orthogonal to both independent displacement vectors in the plane. - Turn each orthogonality statement into a dot product equal to zero. - Point membership can be checked only after the plane equation has been established.

1. \(\overrightarrow{PQ}=\langle-2,3,-1\rangle\) and \(\overrightarrow{PR}=\langle1,-1,1\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(-2a+3b-c=0\) and \(a-b+c=0\). One solution is \(\mathbf{n}=\langle2,1,-1\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle4,0,1\rangle)\cdot\langle2,1,-1\rangle=0\). Expanding gives \(2x+y-z=7\). 4. For \(S\), \(2(1)+4-(-1)=7\), so \(S\) lies in the plane.

a) One normal vector is \(\mathbf{n}=\langle2,1,-1\rangle\). b) \((\mathbf{x}-\langle4,0,1\rangle)\cdot\langle2,1,-1\rangle=0\); equivalently, \(2x+y-z=7\) c) Yes, \(S\) lies in the plane.

- The normal must have zero dot product with each given direction. - Solve the two homogeneous equations for any convenient nonzero normal vector. - Use the shared point only after the normal has been found.

1. For \(\mathbf{n}=\langle a,b,c\rangle\), the equations \(\mathbf{n}\cdot\mathbf{u}=0\) and \(\mathbf{n}\cdot\mathbf{v}=0\) are \(2a+b=0\) and \(-b+c=0\). 2. One nonzero solution is \(\mathbf{n}=\langle1,-2,-2\rangle\). 3. A point-normal equation is \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle1,-2,-2\rangle=0\). Expanding gives \(x-2y-2z=-3\). 4. For \(T\), \(5-2(1)-2(3)=-3\), so \(T\) lies in the plane.

a) One normal vector is \(\mathbf{n}=\langle1,-2,-2\rangle\). b) \((\mathbf{x}-\langle1,0,2\rangle)\cdot\langle1,-2,-2\rangle=0\); equivalently, \(x-2y-2z=-3\) c) Yes, \(T\) lies in the plane.

- 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 \(\mathbf{k}=\langle0,0,1\rangle\). 2. A normal vector must satisfy \(\mathbf{n}\cdot\mathbf{k}=0\), so \(n_z=0\). 3. The vector \(\overrightarrow{OP}=\langle2,4,1\rangle\) also lies in the plane. Thus \(\langle n_x,n_y,0\rangle\cdot\langle2,4,1\rangle=0\), which gives \(2n_x+4n_y=0\). 4. One suitable normal vector is \(\langle2,-1,0\rangle\). 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 \(\mathbf{n}_k=\langle k,3,0\rangle\). 2. A direction vector of the \(z\)-axis is \(\mathbf{e}_z=\langle0,0,1\rangle\). Since \(\mathbf{n}_k\cdot\mathbf{e}_z=0\) for every \(k\), the direction lies in every plane. 3. The origin satisfies each plane equation, so all these planes contain the entire \(z\)-axis. 4. The common line is \(\mathbf{r}(t)=\langle0,0,0\rangle+t\langle0,0,1\rangle\). Substitution gives \(k(0)+3(0)=0\) for every \(t\) and every allowed \(k\).

a) For each \(k\), \(\mathbf{n}_k=\langle k,3,0\rangle\) and \(\mathbf{n}_k\cdot\langle0,0,1\rangle=0\). The origin also satisfies \(kx+3y=0\), so every plane contains the \(z\)-axis. b) \(\mathbf{r}(t)=\langle0,0,0\rangle+t\langle0,0,1\rangle\), where \(t\in\mathbb{R}\). Every point on this line has \(x=y=0\) and satisfies each plane equation.

- The vector from the known point in the plane to the second point gives another direction contained in the plane. - A normal vector must be perpendicular to both independent plane directions. - Express perpendicularity with two dot products equal to zero.

1. \(\mathbf{v}=\overrightarrow{GP}=\langle4,-3,0\rangle\). 2. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality gives \(a-2b+2c=0\) and \(4a-3b=0\). One solution is \(\mathbf{n}=\langle6,8,5\rangle\). 3. Using \(G\), a point-normal equation is \((\mathbf{x}-\langle0,3,1\rangle)\cdot\langle6,8,5\rangle=0\).

a) \(\mathbf{v}=\langle4,-3,0\rangle\) b) One normal vector is \(\mathbf{n}=\langle6,8,5\rangle\). c) \((\mathbf{x}-\langle0,3,1\rangle)\cdot\langle6,8,5\rangle=0\)

- 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 \(\overrightarrow{PQ}=\langle3,3,3\rangle\) and \(\overrightarrow{PR}=\langle-3,-3,-3\rangle\). Since \(\overrightarrow{PR}=-\overrightarrow{PQ}\), the points are collinear. 2. Three collinear points lie on infinitely many planes, so no unique plane is determined. 3. Choose a normal vector perpendicular to their line, such as \(\mathbf{n}=\langle1,-1,0\rangle\). Then \(\mathbf{n}\cdot\overrightarrow{PQ}=3-3+0=0\). 4. The plane has equation \(x-y=d\). Substituting \(P\) gives \(d=2\), so one suitable plane is \(x-y=2\).

a) \(\overrightarrow{PR}=-\overrightarrow{PQ}\), so the three points are collinear and do not determine a unique plane. b) One normal is \(\mathbf{n}=\langle1,-1,0\rangle\), with \(\mathbf{n}\cdot\overrightarrow{PQ}=3-3=0\). Using \(P\) gives \(x-y=2\), which contains all three points.

- 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 \(\mathbf{v}=\overrightarrow{AB}=\langle2,2,-1\rangle\). 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 \(\langle1,0,0\rangle\). One possible plane is \(F:\mathbf{x}=\langle1,-2,3\rangle+r\langle2,2,-1\rangle+s\langle1,0,0\rangle\). 3. A normal vector to \(H\) is \(\mathbf{n}_H=\langle1,0,2\rangle\). Since \(\mathbf{n}_H\cdot\mathbf{v}=0\), 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 \(F:\mathbf{x}=\langle1,-2,3\rangle+r\langle2,2,-1\rangle+s\langle1,0,0\rangle\). 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 \(\mathbf{n}=\langle a,b,0\rangle\). To contain \(h\), its normal must be perpendicular to the line direction \(\langle1,-1,0\rangle\), so \(a-b=0\). Choose \(\mathbf{n}=\langle1,1,0\rangle\). Using the point \((0, 4, 2)\) gives \(E:x+y=4\). 2. Choose \(\mathbf{n}_G=\langle1,0,0\rangle\). 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 \(\mathbf{n}_G=\langle1,0,0\rangle\).

- 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 \(\overrightarrow{RP}=\langle3,0,2\rangle\) and \(\overrightarrow{RQ}=\langle0,4,2\rangle\). Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. Then \(3a+2c=0\) and \(4b+2c=0\). One solution is \(\mathbf{n}=\langle4,3,-6\rangle\). 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. The line direction is \(\mathbf{d}=\langle4,2,-4\rangle\), and a plane normal is \(\mathbf{n}=\langle1,1,1\rangle\). 2. \(\mathbf{n}\cdot\mathbf{d}=4+2-4=2\neq0\), so the line meets the plane exactly once. 3. Parametrize the line by \(\mathbf{r}(t)=\langle2,1,5\rangle+t\langle4,2,-4\rangle\). 4. Substitution gives \((2+4t)+(1+2t)+(5-4t)=5\), hence \(8+2t=5\) and \(t=-\frac{3}{2}\). 5. Therefore, the intersection is \(S=\langle2,1,5\rangle-\frac{3}{2}\langle4,2,-4\rangle=\langle-4,-2,11\rangle\).

Plane normal: \(\mathbf{n}=\langle1,1,1\rangle\). Line direction: \(\mathbf{d}=\langle4,2,-4\rangle\). Dot product: \(\mathbf{n}\cdot\mathbf{d}=4+2-4=2\neq0\). Therefore, the line intersects the plane exactly once, at \(S=(-4,-2,11)\).

- Use the equal axis intercepts to write the plane equation. - Compare the line's direction vector with the plane's normal vector. - A line whose direction is parallel to a plane normal is perpendicular to the plane. - Substitute the line into the plane equation to locate the intersection.

1. The three intercepts give the plane equation \(x+y+z=6\), with normal vector \(\mathbf{n}=\langle1,1,1\rangle\). 2. A direction vector of the line is \(\mathbf{d}=\overrightarrow{PQ}=\langle2,2,2\rangle=2\mathbf{n}\). Therefore \(g\) is perpendicular to \(E\). 3. Parametrize the line: \(\mathbf{r}(t)=\langle1,1,1\rangle+t\langle2,2,2\rangle\). 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 is perpendicular to the plane and intersects it 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 to \(E\) is \(\mathbf{n}=\langle3,-2,1\rangle\), and the line direction is \(\mathbf{d}_k=\langle1,2,k\rangle\). 2. Their dot product is \(\mathbf{n}\cdot\mathbf{d}_k=3-4+k=k-1\). 3. When \(k=1\), the dot product is zero, so the line is parallel to or contained in the plane. The initial point satisfies \(3(2)-2(1)+6=10\), so \(g_1\) lies in \(E\). 4. When \(k\neq1\), the dot product is nonzero, so the line intersects the plane exactly once.

Plane normal: \(\mathbf{n}=\langle3,-2,1\rangle\). Dot product: \(\mathbf{n}\cdot\mathbf{d}_k=\langle3,-2,1\rangle\cdot\langle1,2,k\rangle=k-1\). For \(k=1\), \(g_1\) lies in \(E\), since its starting point also satisfies the plane equation. For \(k\neq1\), \(g_k\) intersects \(E\) at exactly one point.

- The line direction must be orthogonal to the plane normal if the whole line lies in the plane. - Use that condition to determine the coefficient in the plane normal. - Then require a known point of the line to satisfy the plane equation.

1. The line direction is \(\mathbf{d}=\langle1,-2,3\rangle\), and a plane normal is \(\mathbf{n}=\langle a,1,2\rangle\). 2. Containment requires \(\mathbf{n}\cdot\mathbf{d}=a-2+6=a+4=0\), so \(a=-4\). 3. The line's initial point \((1,2,1)\) must satisfy the plane equation. Substituting gives \((-4)(1)+2+2(1)=b\), hence \(b=0\).

\(\mathbf{n}\cdot\mathbf{d}=\langle a,1,2\rangle\cdot\langle1,-2,3\rangle=a+4\). Setting it equal to zero gives \(a=-4\). The line's initial point gives \(b=(-4)(1)+2+2(1)=0\). Thus, \(a=-4\), \(b=0\).

- 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 \(\mathbf{v}=\langle0,1,-1\rangle\), and a plane normal is \(\mathbf{n}=\langle2,-1,1\rangle\). Since \(\mathbf{v}\cdot\mathbf{n}=-2\ne0\), 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 \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac{1}{\sqrt3}\). 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. For \(\mathbf{n}=\langle a,b,c\rangle\), orthogonality to the plane directions requires \(2a+b-c=0\) and \(3b+2c=0\). One solution is \(\mathbf{n}=\langle5,-4,6\rangle\). 2. Using \((4,1,0)\), the Cartesian equation is \(5x-4y+6z=16\). 3. Choose \(g:\mathbf{r}(t)=\langle4,1,0\rangle+t\langle2,1,-1\rangle\). Its initial point is in \(E\), and \(\mathbf{n}\cdot\langle2,1,-1\rangle=10-4-6=0\). 4. Choose \(h:\mathbf{r}(t)=t\langle0,3,2\rangle\). Its direction satisfies \(\mathbf{n}\cdot\langle0,3,2\rangle=-12+12=0\), so it is parallel to \(E\). 5. The origin is not in \(E\), since \(0\neq16\). Hence \(h\) is strictly parallel to the plane.

A plane normal is \(\mathbf{n}=\langle5,-4,6\rangle\), and the plane equation is \(5x-4y+6z=16\). a) \(g:\mathbf{r}(t)=\langle4,1,0\rangle+t\langle2,1,-1\rangle\). Its direction has dot product \(10-4-6=0\) with \(\mathbf{n}\), and its initial point lies in the plane. b) \(h:\mathbf{r}(t)=t\langle0,3,2\rangle\). Its direction has dot product \(-12+12=0\) with \(\mathbf{n}\), but its initial point gives \(0\neq16\) in the plane equation. Thus, \(h\) is parallel to and disjoint from \(E\).

- 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 \(\mathbf{n}=\langle1,2,-2\rangle\). A vector is parallel to the plane when its dot product with \(\mathbf{n}\) is \(0\). 2. Choose \(\mathbf{v}=\langle2,0,1\rangle\), since \(\mathbf{n}\cdot\mathbf{v}=2-2=0\). 3. The point \(P(6, 0, 0)\) lies in \(E\). Therefore \(k:\mathbf{r}(t)=\langle6,0,0\rangle+t\langle2,0,1\rangle\) lies in the plane. 4. The origin does not lie in \(E\), since \(0\ne6\). Therefore \(m:\mathbf{r}(t)=t\langle2,0,1\rangle\) is parallel to \(E\) and disjoint from it.

a) One possible vector is \(\mathbf{v}=\langle2,0,1\rangle\). b) \(k:\mathbf{r}(t)=\langle6,0,0\rangle+t\langle2,0,1\rangle\) \(m:\mathbf{r}(t)=t\langle2,0,1\rangle\)

- Use a dot product to make the line direction orthogonal to the plane normal. - After finding the parameter, test the line's initial point in the plane equation. - Use that point test to distinguish containment from strict parallelism.

1. The plane normal is \(\mathbf{n}=\langle3,1,-2\rangle\), and the line direction is \(\mathbf{d}_a=\langle1,a,3\rangle\). 2. The dot product is \(\mathbf{n}\cdot\mathbf{d}_a=3+a-6=a-3\). Setting this to zero gives \(a=3\). 3. The starting point \((2,6,1)\) satisfies \(3(2)+6-2(1)=10\), so \(g_3\) lies entirely in \(E\), rather than being strictly parallel to it.

Plane normal: \(\mathbf{n}=\langle3,1,-2\rangle\). Line direction: \(\mathbf{d}_a=\langle1,a,3\rangle\). Their dot product is \(\mathbf{n}\cdot\mathbf{d}_a=a-3\). Thus, \(a=3\). Since the starting point gives \(3(2)+6-2(1)=10\), the line \(g_3\) 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. - Check whether all three component ratios can agree in the perpendicular case.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle4,-2,1\rangle\), and the line direction is \(\mathbf{v}=\langle a,4,2\rangle\). 2. For the line to be parallel to the plane, require \(\mathbf{v}\cdot\mathbf{n}=0\): \(4a-8+2=0\). 3. Thus \(4a=6\), so \(a=\frac32\). 4. For the line to be perpendicular to the plane, \(\mathbf{v}\) must be a scalar multiple of \(\mathbf{n}\). 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 \(\mathbf{v}\) 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 \(\mathbf{n}=\langle1,2,-2\rangle\), and the line direction is \(\mathbf{v}=\langle2,k,-4\rangle\). 2. For the line to be parallel to the plane, require \(\mathbf{v}\cdot\mathbf{n}=0\): \(2+2k+8=0\). 3. Thus \(2k+10=0\), so \(k=-5\). 4. The initial point \((2, 1, 4)\) does not lie in \(F\), because \(2+2-8=-4\ne10\). Therefore, \(h_{-5}\) is strictly parallel to \(F\). 5. For the line to be perpendicular to the plane, \(\mathbf{v}\) must be a scalar multiple of \(\mathbf{n}\). 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 normal vector is perpendicular to every direction in its plane. A line pointing along that normal is therefore perpendicular to the plane. 2. False. The zero dot product makes the line direction parallel to the plane, so the line either has no points or infinitely many points in common with it, never exactly one. 3. True. If the direction lies in the plane and one point of the line belongs to the plane, all other points on the line remain in it. The converse also holds.

1. True. A line parallel to a plane normal is perpendicular to the plane. 2. False. The zero dot product gives a parallel direction, so the line has either no intersection points or infinitely many, not exactly one. 3. True. A line with a direction parallel to the plane and one point in the plane stays entirely in the plane; a contained line also satisfies both conditions.

- 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. The line meets the plane at exactly one point when \(\mathbf{u}\cdot\mathbf{n}\neq0\). If the dot product were zero, the direction would be parallel to the plane and the line would have either no intersections or infinitely many. 2. The point-normal condition for membership in \(E\) is \((\mathbf{p}-\mathbf{a})\cdot\mathbf{n}=0\). A strictly parallel line has no points on \(E\), so every point \(P\) on \(g\) satisfies \((\mathbf{p}-\mathbf{a})\cdot\mathbf{n}\neq0\).

a) Exactly one intersection occurs when \(\mathbf{u}\cdot\mathbf{n}\neq0\), because the line direction then has a nonzero component normal to the plane and cannot remain parallel to it. b) For every \(P\) on the strictly parallel line, \((\mathbf{p}-\mathbf{a})\cdot\mathbf{n}\neq0\); equality would put \(P\) in the plane, contradicting strict parallelism.

- 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 \(\overrightarrow{AF}\cdot\langle1,0,-1\rangle=0\): \((t-1)-(-t-3)=2t+2=0\). Thus, \(t=-1\). 3. The perpendicular foot is \(F=(0,1,2)\). 4. Since \(F\) 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. - Containment also requires a point of the line to satisfy the plane equation.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle2,1,-1\rangle\), and the line direction is \(\mathbf{v}=\langle k,2,-2\rangle\). 2. A line perpendicular to the plane has a direction parallel to \(\mathbf{n}\). 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. For a line to lie in the plane, its direction must be orthogonal to the plane normal. This would require \(\mathbf{v}\cdot\mathbf{n}=2k+2+2=0\), so \(k=-2\). 5. However, the common initial point \((1, 0, 4)\) is not in \(E\), because \(2(1)+0-4-6=-8\ne0\). 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 \(\mathbf{n}=\langle1,1,0\rangle\), and every line in the family has direction vector \(\mathbf{d}=\langle1,-1,2\rangle\). 2. A line perpendicular to \(F\) must have a direction vector parallel to \(\mathbf{n}\). The vectors \(\langle1,-1,2\rangle\) and \(\langle1,1,0\rangle\) are not scalar multiples, so no line in the family is perpendicular to \(F\). 3. Compute \(\mathbf{d}\cdot\mathbf{n}=1-1+0=0\). Thus every line in the family is parallel to \(F\). 4. The initial point \((3, 1, a)\) never lies in \(F\), because \(3+1=4\ne10\), 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 \(\langle1,-1,2\rangle\) is not parallel to \(\langle1,1,0\rangle\). 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 line direction is \(\overrightarrow{AB}=\langle1,-3,0\rangle\); a plane normal is \(\mathbf{n}=\langle6,2,-3\rangle\). 2. Their dot product is \(\mathbf{n}\cdot\overrightarrow{AB}=6(1)+2(-3)-3(0)=0\), so the line is parallel to or contained in the plane. 3. At \(A\), \(6(1)+2(5)-3(2)=10\neq12\). Thus, \(g\) is parallel to and distinct from \(E\). 4. Every point on \(g\) has the same distance from \(E\), so \(d(g,E)=\frac{|6(1)+2(5)-3(2)-12|}{\sqrt{6^2+2^2+(-3)^2}}=\frac{2}{7}\).

\(\mathbf{n}\cdot\overrightarrow{AB}=\langle6,2,-3\rangle\cdot\langle1,-3,0\rangle=0\). Since \(A\) gives \(10\neq12\) in the plane equation, \(g\) is strictly parallel to \(E\). The distance is \(\frac{2}{7}\) 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 direction is \(\mathbf{u}=\langle2,2,1\rangle\) and the plane normal is \(\mathbf{n}=\langle1,-2,2\rangle\). 2. The dot product is \(\mathbf{n}\cdot\mathbf{u}=2-4+2=0\), so the line direction is parallel to the plane. 3. The point \(P=(1,1,4)\) is not in the plane because \(1-2(1)+2(4)=7\neq5\). Thus, \(h\) is strictly parallel to \(F\). 4. The distance is the distance from \(P\) to the plane: \(d=\frac{|1-2+8-5|}{\sqrt{1^2+(-2)^2+2^2}}=\frac{2}{3}\).

\(\mathbf{n}\cdot\langle2,2,1\rangle=2-4+2=0\), so the line direction is parallel to the plane. Its point \((1,1,4)\) gives \(7\neq5\) in the plane equation, hence the line is strictly parallel. The distance is \(\frac{2}{3}\) 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. A line equation is \(\mathbf{x}=\langle2,2,1\rangle+r\langle4,0,3\rangle\), so \(L=(2+4r,2,1+3r)\). 2. Perpendicularity requires \(\overrightarrow{CL}\cdot\langle4,0,3\rangle=4(4r-2)+3(3r-4)=25r-20=0\), so \(r=\frac{4}{5}\). 3. Thus, \(L=\left(\frac{26}{5},2,\frac{17}{5}\right)\). 4. The altitude has length \(CL=\sqrt{\left(\frac{6}{5}\right)^2+(-3)^2+\left(-\frac{8}{5}\right)^2}=\sqrt{13}\). 5. Since \(AB=\sqrt{4^2+0^2+3^2}=5\), the area is \(\frac{1}{2}\cdot5\cdot\sqrt{13}=\frac{5\sqrt{13}}{2}\).

a) \(L=\left(\frac{26}{5},2,\frac{17}{5}\right)\) b) \(\sqrt{13}\) units c) \(\frac{5\sqrt{13}}{2}\) square 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 \(\frac12\).

1. A normal vector to \(E\) is \(\mathbf{n}_E=\langle1,1,0\rangle\). In \(G_a\), use \(\overrightarrow{OP}=\langle1,0,0\rangle\) and \(\overrightarrow{OQ_a}=\langle0,1,a\rangle\). If \(\mathbf{n}_a=\langle x,y,z\rangle\) is perpendicular to both, then \(x=0\) and \(y+az=0\). Choose \(\mathbf{n}_a=\langle0,-a,1\rangle\). 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 \(\cos60^\circ=\frac12\) and squaring gives \(a^2=1\). 4. Thus \(a=1\) or \(a=-1\).

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

- Read the center and radius from standard form. - Use the midpoint relationship for a diameter. - In a circle, an inscribed angle that intercepts a diameter is a right angle.

1. The center is \(M(4,-1,2)\), and the radius is \(9\). 2. Since \(M\) is the midpoint of \(\overline{AB}\), \(B=2M-A=(4,-10,2)\). 3. Points \(A\), \(B\), and \(C\) lie in a plane, and that plane intersects the sphere in a circle with diameter \(AB\). By the Inscribed Angle Theorem, \(\angle ACB=90^\circ\). 4. Therefore, \(\overrightarrow{CA}\) and \(\overrightarrow{CB}\) are perpendicular, so their dot product is \(0\).

a) \(M=(4,-1,2)\), \(r=9\) b) \(B=(4,-10,2)\) c) The vectors are perpendicular because \(AB\) is a diameter, so their dot product is \(0\).

- 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 \(\mathbf{n}=\langle4,-1,8\rangle\). 2. A line perpendicular to the plane can use \(\mathbf{n}\) as its direction vector. Since \(4(4)-6+8(1)=18\), point \(P\) lies in the plane. One equation is \(g:\mathbf{r}(t)=\langle4,6,1\rangle+t\langle4,-1,8\rangle\). 3. A line parallel to \(E\) needs a direction vector orthogonal to \(\mathbf{n}\). Choose \(\mathbf{v}=\langle1,4,0\rangle\), since \(\mathbf{v}\cdot\mathbf{n}=4-4=0\). Thus one possible line is \(h:\mathbf{r}(t)=t\langle1,4,0\rangle\). The origin is not in \(E\), so this line is strictly parallel to the plane. 4. For \(k\), \(\langle2,0,-1\rangle\cdot\langle4,-1,8\rangle=8-8=0\), 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 \(g:\mathbf{r}(t)=\langle4,6,1\rangle+t\langle4,-1,8\rangle\). b) One possible equation is \(h:\mathbf{r}(t)=t\langle1,4,0\rangle\). 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 \(\mathbf{n}=\langle1,2,-2\rangle\). 2. A perpendicular line through the origin is \(g:\mathbf{r}(t)=t\langle1,2,-2\rangle\). 3. Substituting \((t,2t,-2t)\) into the plane gives \(t+2(2t)-2(-2t)=9t=12\). Thus, \(t=\frac{4}{3}\) and \(S=\left(\frac{4}{3},\frac{8}{3},-\frac{8}{3}\right)\). 4. The direction of \(\ell\) has dot product \(\langle1,1,1\rangle\cdot\langle1,2,-2\rangle=1\neq0\), so \(\ell\) is not parallel to the plane. 5. The same vectors are not scalar multiples, so \(\ell\) is not perpendicular to the plane either.

a) \(g:\mathbf{r}(t)=t\langle1,2,-2\rangle\) b) \(S=\left(\frac{4}{3},\frac{8}{3},-\frac{8}{3}\right)\) c) The plane normal is \(\langle1,2,-2\rangle\). Its dot product with the line direction \(\langle1,1,1\rangle\) is \(1\), not zero, so the line is not parallel to the plane. The vectors are not scalar multiples either, so the line is not perpendicular to the plane.

- Use \(\langle1,1,1\rangle\) 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. Choose \(H:x-y=1\), with normal \(\mathbf{n}_H=\langle1,-1,0\rangle\). 2. Since \(\langle1,1,1\rangle\cdot\mathbf{n}_H=0\), the line direction is parallel to \(H\). The origin is not in \(H\), because \(0\neq1\), so the line and plane have no points in common. 3. An equal-coordinate point is \((a,a,a)\). Substitution into \(E_d\) gives \(3a=d\), which has exactly one solution \(a=\frac{d}{3}\) for each real \(d\). 4. The unique point is \(\left(\frac{d}{3},\frac{d}{3},\frac{d}{3}\right)\).

a) Choose \(H:x-y=1\) with normal \(\mathbf{n}_H=\langle1,-1,0\rangle\). Then \(\langle1,1,1\rangle\cdot\mathbf{n}_H=1-1=0\), so \(g\) is parallel to \(H\). Since the origin is not in \(H\), they are disjoint. b) An equal-coordinate point \((a,a,a)\) belongs to \(E_d\) when \(3a=d\), giving exactly one value \(a=\frac{d}{3}\). The unique point is \(\left(\frac{d}{3},\frac{d}{3},\frac{d}{3}\right)\).

- 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 \(\mathbf{n}=\langle2,-1,2\rangle\), with magnitude \(3\). The line direction is \(\mathbf{v}_a=\langle a,1,1\rangle\), with magnitude \(\sqrt{a^2+2}\). 2. The line-plane angle formula gives \(\frac{|2a+1|}{3\sqrt{a^2+2}}=\sin45^\circ=\frac{\sqrt2}{2}\). 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 \(\overrightarrow{AB}=\langle2,-2,2\rangle\). Since the line is perpendicular to the plane, use the parallel normal vector \(\mathbf{n}=\langle1,-1,1\rangle\). 2. The plane through \(C\) satisfies \(\langle1,-1,1\rangle\cdot\langle x-5,y-3,z-3\rangle=0\). Expanding gives \(x-y+z=5\). 3. Parametrize the line as \(\mathbf{r}(t)=\langle1,2,3\rangle+t\langle2,-2,2\rangle\). 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) Substitution of the line into the plane gives \(t=\frac{1}{2}\), so \(S=(2,1,4)\). The midpoint is \(\left(\frac{1+3}{2},\frac{2+0}{2},\frac{3+5}{2}\right)=(2,1,4)\), confirming the claim.

- 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 \(\overrightarrow{PQ}=\langle2,2,1\rangle\). 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: \(\mathbf{r}(t)=\langle1,1,1\rangle+t\langle2,2,1\rangle\). 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 \(S\notin\overline{PQ}\).

- 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 \(\overrightarrow{PQ}=\langle2,3,0\rangle\), \(\overrightarrow{PR}=\langle-3,2,3\rangle\), and \(\overrightarrow{QR}=\langle-5,-1,3\rangle\). 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) \(\overrightarrow{PQ}\cdot\overrightarrow{PR}=2(-3)+3(2)+0(3)=0\). The sides meeting at \(P\) are perpendicular, so the triangle is right at \(P\).

- Use the two side vectors that begin at \(A\). - A zero dot product proves a right angle. - For part b, use two vectors that both begin at \(B\). - Apply the dot-product angle formula after reporting the product.

1. The vectors meeting at \(A\) are \(\overrightarrow{AB}=\langle3,0,4\rangle\) and \(\overrightarrow{AC}=\langle0,4,0\rangle\). 2. Their dot product is \(3\cdot0+0\cdot4+4\cdot0=0\), so the angle at \(A\) is \(90^\circ\). 3. At \(B\), use \(\overrightarrow{BA}=\langle-3,0,-4\rangle\) and \(\overrightarrow{BC}=\langle-3,4,-4\rangle\). 4. The required dot product is \(\overrightarrow{BA}\cdot\overrightarrow{BC}=9+0+16=25\). The vectors have lengths \(5\) and \(\sqrt{41}\). 5. Hence \(\cos\beta=\frac{25}{5\sqrt{41}}=\frac{5}{\sqrt{41}}\), giving \(\beta\approx38.66^\circ\).

a) \(\overrightarrow{AB}\cdot\overrightarrow{AC}=3\cdot0+0\cdot4+4\cdot0=0\), so the triangle has a right angle at \(A\). b) \(\overrightarrow{BA}\cdot\overrightarrow{BC}=(-3)(-3)+0\cdot4+(-4)(-4)=25\). Thus \(\cos\beta=\frac{25}{5\sqrt{41}}\), so \(\beta\approx38.66^\circ\).

- 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. \(\overrightarrow{AB}=\langle2,1,-2\rangle\), \(\overrightarrow{AC}=\langle-2,2,-1\rangle\), and \(\overrightarrow{BC}=\langle-4,1,1\rangle\). 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.

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

- Add the three side lengths for the perimeter. - For the angle at \(Q\), use two vectors that begin at \(Q\). - Apply the given translation to \(P\) in part c. - To prove a line is perpendicular to a plane, relate its direction to two nonparallel directions in that plane.

1. The side lengths are \(PQ=4\), \(PR=3\), and \(QR=\sqrt{(-4)^2+3^2}=5\). 2. Therefore, the base perimeter is \(4+3+5=12\). 3. At \(Q\), \(\overrightarrow{QP}=\langle-4,0,0\rangle\) and \(\overrightarrow{QR}=\langle-4,0,3\rangle\); their dot product is \(16\). 4. Thus \(\cos\beta=\frac{16}{4\cdot5}=\frac{4}{5}\), so \(\beta\approx36.87^\circ\). 5. Translating \(P\) gives \(P'=(2,7,0)\), and \(\overrightarrow{PP'}=\langle0,4,0\rangle\). 6. Both \(\overrightarrow{PP'}\cdot\overrightarrow{PQ}=0\) and \(\overrightarrow{PP'}\cdot\overrightarrow{PR}=0\). Because \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) are nonparallel vectors in the base plane, \(PP'\) is perpendicular to that plane.

a) \(12\) units b) \(\overrightarrow{QP}\cdot\overrightarrow{QR}=16\); hence \(\cos\beta=\frac{16}{20}=\frac{4}{5}\) and \(\beta\approx36.87^\circ\). c) \(P'=(2,7,0)\), with \(\overrightarrow{PP'}=\langle0,4,0\rangle\). Then \(\overrightarrow{PP'}\cdot\overrightarrow{PQ}=0\) and \(\overrightarrow{PP'}\cdot\overrightarrow{PR}=0\). These two nonparallel base directions are both perpendicular to \(PP'\), so the edge is perpendicular to the base plane.

- Expand the squared magnitude of a difference vector. - Replace the dot product with the angle formula. - Use the equal magnitudes to eliminate the unknown length.

1. Let \(L=\|\mathbf{a}\|=\|\mathbf{b}\|>0\). Then \(\|\mathbf{a}-\mathbf{b}\|=\sqrt3L\). 2. Square the magnitude: \(\|\mathbf{a}-\mathbf{b}\|^2=\|\mathbf{a}\|^2+\|\mathbf{b}\|^2-2\mathbf{a}\cdot\mathbf{b}\). 3. Using \(\mathbf{a}\cdot\mathbf{b}=L^2\cos\alpha\), obtain \(3L^2=2L^2-2L^2\cos\alpha\). 4. Since \(L>0\), divide by \(L^2\): \(3=2-2\cos\alpha\). 5. Therefore, \(\cos\alpha=-\frac12\), so \(\alpha=120^\circ\).

\(\alpha=120^\circ\)

- Represent the shorter magnitude by one variable. - Expand the squared magnitude of the sum. - Replace the dot product with the angle formula.

1. Let \(L=\|\mathbf{u}\|>0\). Then \(\|\mathbf{v}\|=2L\) and \(\|\mathbf{u}+\mathbf{v}\|=\sqrt7L\). 2. Use \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+\|\mathbf{v}\|^2+2\mathbf{u}\cdot\mathbf{v}\). 3. Since \(\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha\), substitution gives \(7L^2=L^2+4L^2+4L^2\cos\alpha\). 4. Divide by \(L^2\): \(7=5+4\cos\alpha\). 5. Thus, \(\cos\alpha=\frac12\), so \(\alpha=60^\circ\).

\(\alpha=60^\circ\)

- Write the face diagonal with a zero \(z\)-component. - Use the dot-product angle formula. - Make sure both vectors begin at the origin.

1. The face-diagonal vector is \(\mathbf{f}=\langle4,3,0\rangle\). 2. The dot product is \(\mathbf{d}\cdot\mathbf{f}=4\cdot4+3\cdot3+12\cdot0=25\). 3. The magnitudes are \(\|\mathbf{d}\|=13\) and \(\|\mathbf{f}\|=5\). 4. Therefore, \(\cos\alpha=\frac{25}{13\cdot5}=\frac{5}{13}\). 5. Thus, \(\alpha=\cos^{-1}\left(\frac{5}{13}\right)\approx67.38^\circ\).

\(\mathbf{f}=\langle4,3,0\rangle\) and \(\mathbf{d}\cdot\mathbf{f}=4\cdot4+3\cdot3+12\cdot0=25\). Thus \(\cos\alpha=\frac{5}{13}\) and \(\alpha\approx67.38^\circ\).

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

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

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. \(\|\mathbf{a}\|=\sqrt{1^2+2^2+(-2)^2}=3\) and \(\|\mathbf{b}\|=\sqrt{4^2+0^2+3^2}=5\). 2. \(\mathbf{a}\cdot\mathbf{b}=1\cdot4+2\cdot0+(-2)\cdot3=-2\). 3. \(\mathbf{a}-\mathbf{b}=\langle-3,2,-5\rangle\) and \(2\mathbf{a}+\mathbf{b}=\langle6,4,-1\rangle\). 4. Their dot product is \((-3)\cdot6+2\cdot4+(-5)\cdot(-1)=-18+8+5=-5\).

a) \(\|\mathbf{a}\|=3\), \(\|\mathbf{b}\|=5\) b) \(\mathbf{a}\cdot\mathbf{b}=-2\) 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. - Relate that trigonometric value to the angle between the 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 dot 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) Nonzero vectors have positive magnitudes. Their dot product can still be zero because \(\cos90^\circ=0\), which happens when the vectors are perpendicular.

- Translate each perpendicularity condition into a dot-product equation. - You should obtain two linear equations involving \(x\) and \(y\). - Solve one equation for one variable and substitute into the other.

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

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

- Expand the dot product using distributivity and look for terms that cancel. - For part b, ask what must be true of \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\) when the expression equals zero. - Relate equal adjacent side lengths to a standard parallelogram classification.

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

a) Distribute: \((\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot\mathbf{a}-\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{a}-\mathbf{b}\cdot\mathbf{b}=\|\mathbf{a}\|^2-\|\mathbf{b}\|^2\), since the mixed terms cancel. b) The dot product is zero exactly when \(\|\mathbf{a}\|=\|\mathbf{b}\|\). A nondegenerate parallelogram with equal adjacent sides is a rhombus; its diagonals are perpendicular.

- Start by distinguishing the unit-vector task from the projection task. - For the projection, identify which vector supplies the direction being projected onto. - After finding the parallel component, subtract it from \(\mathbf{b}\) and check the remaining component with a dot product.

1. \(\|\mathbf{a}\|=\sqrt{3^2+4^2}=5\), so \(\hat{\mathbf{a}}=\frac{\mathbf{a}}{\|\mathbf{a}\|}=\left\langle\frac35,\frac45,0\right\rangle\). 2. The projection is \(\operatorname{proj}_{\mathbf{a}}\mathbf{b}=\frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{a}\cdot\mathbf{a}}\mathbf{a}\). 3. Since \(\mathbf{a}\cdot\mathbf{b}=11\) and \(\mathbf{a}\cdot\mathbf{a}=25\), \(\mathbf{b}_{\parallel}=\frac{11}{25}\mathbf{a}=\left\langle\frac{33}{25},\frac{44}{25},0\right\rangle\). 4. The perpendicular component is \(\mathbf{b}_{\perp}=\mathbf{b}-\mathbf{b}_{\parallel}=\left\langle-\frac{8}{25},\frac{6}{25},2\right\rangle\). 5. Check: \(\mathbf{a}\cdot\mathbf{b}_{\perp}=0\), so the decomposition has the required perpendicular component.

1. \(\hat{\mathbf{a}}=\left\langle\frac35,\frac45,0\right\rangle\) 2. \(\operatorname{proj}_{\mathbf{a}}\mathbf{b}=\left\langle\frac{33}{25},\frac{44}{25},0\right\rangle\) 3. \(\mathbf{b}_{\parallel}=\left\langle\frac{33}{25},\frac{44}{25},0\right\rangle\), \(\mathbf{b}_{\perp}=\left\langle-\frac{8}{25},\frac{6}{25},2\right\rangle\)

- 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 \(\operatorname{proj}_{\mathbf{v}}\mathbf{p}=\frac{15}{9}\mathbf{v}=\frac{5}{3}\langle1,2,2\rangle=\left\langle\frac{5}{3},\frac{10}{3},\frac{10}{3}\right\rangle\). 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 \(\sqrt{\frac{16}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{2}\).

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

- A diameter determines a right angle at any other point on the sphere. - Form the two directed segments that meet at \(C\). - Translate the right angle into a dot-product equation.

1. By Thales's theorem, \(\angle ACB=90^\circ\). 2. \(\overrightarrow{CA}=\langle-x,-4,-2\rangle\) and \(\overrightarrow{CB}=\langle4-x,-2,2\rangle\). 3. Perpendicularity gives \(\overrightarrow{CA}\cdot\overrightarrow{CB}=0\), so \((-x)(4-x)+(-4)(-2)+(-2)(2)=0\). 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 squared binomial and group like types of terms. - For the perpendicular case, identify which term becomes zero.

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

1. Expanding \(\sum_{i=1}^{3}(a_i+b_i)^2\) gives \(\sum_{i=1}^{3}a_i^2+2\sum_{i=1}^{3}a_ib_i+\sum_{i=1}^{3}b_i^2=\|\mathbf{a}\|^2+2(\mathbf{a}\cdot\mathbf{b})+\|\mathbf{b}\|^2\). 2. If \(\mathbf{a}\perp\mathbf{b}\), the dot product is zero, so \(\|\mathbf{a}+\mathbf{b}\|^2=\|\mathbf{a}\|^2+\|\mathbf{b}\|^2\). This is the Pythagorean theorem.

- Compare a pair of opposite-side displacement vectors first. - Then compare the magnitudes of adjacent sides. - Use a dot product to determine whether an adjacent angle is a right angle.

1. \(\overrightarrow{AB}=\langle2,2,1\rangle\) and \(\overrightarrow{DC}=\langle2,2,1\rangle\), so the quadrilateral is a parallelogram. 2. \(\|\overrightarrow{AB}\|=3\), and \(\overrightarrow{AD}=\langle2,-1,2\rangle\) 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.

Since \(\overrightarrow{AB}=\overrightarrow{DC}=\langle2,2,1\rangle\), the quadrilateral is a parallelogram. Also \(\|\overrightarrow{AB}\|=\|\overrightarrow{AD}\|=3\), so it is a rhombus. Next, \(\overrightarrow{AB}\cdot\overrightarrow{AD}=2\cdot2+2(-1)+1\cdot2=4\ne0\). Its adjacent sides are not perpendicular, so it is not a square.

- Translate each perpendicularity condition into a zero dot product. - Solve the resulting system to describe all common perpendicular vectors with one free parameter. - Impose the magnitude condition after you have found the common perpendicular direction.

1. Let \(\mathbf{x}=\langle x,y,z\rangle\). 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 \(\mathbf{x}=z\langle2,-2,1\rangle\). 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 \(\langle12,-12,6\rangle\) and \(\langle-12,12,-6\rangle\).

\(\langle12,-12,6\rangle\) and \(\langle-12,12,-6\rangle\)

- Represent the unknown vector with three components. - Turn the two perpendicularity conditions into dot-product equations. - Use the remaining dot-product condition to determine the scale of the common perpendicular direction.

1. Let \(\mathbf{w}=\langle x,y,z\rangle\). 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 \(\mathbf{w}=\langle-4,8,12\rangle\).

\(\mathbf{w}=\langle-4,8,12\rangle\)

- Express each diagonal as a sum or difference of the side vectors. - Expand both squared magnitudes and track the mixed dot-product terms. - For the rhombus case, use the fact that both side-vector magnitudes equal \(s\).

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

a) Expand \(\|\mathbf{a}+\mathbf{b}\|^2+\|\mathbf{b}-\mathbf{a}\|^2\). The mixed terms \(2\mathbf{a}\cdot\mathbf{b}\) and \(-2\mathbf{a}\cdot\mathbf{b}\) cancel, giving \(\|\overrightarrow{AC}\|^2+\|\overrightarrow{BD}\|^2=2(\|\mathbf{a}\|^2+\|\mathbf{b}\|^2)\). b) If both sides have length \(s\), then \(\|\overrightarrow{AC}\|^2+\|\overrightarrow{BD}\|^2=4s^2\).

- Use the midpoint relationship between \(\overrightarrow{MB}\) and \(\overrightarrow{MC}\). - Rewrite the two side vectors with one common vector and one pair of opposite vectors. - Expand both squared magnitudes and look for cancellation of the mixed terms.

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

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

- Build the known \(y\)-component into the unknown vector from the start. - Translate both perpendicularity conditions into dot-product equations. - Solve the resulting two-equation system for the remaining components.

1. Write \(\mathbf{n}=\langle x,11,z\rangle\). 2. Orthogonality to \(\mathbf{u}\) gives \(4x-22+z=0\). 3. Orthogonality to \(\mathbf{v}\) 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 \(\mathbf{n}=\langle6,11,-2\rangle\).

\(\mathbf{n}=\langle6,11,-2\rangle\)

- Use the dot-product angle relationship to connect the dot product, magnitudes, and angle. - Relate the sine and cosine of the same angle. - A triangle generated by two vectors has half the corresponding parallelogram area.

1. The dot product is \(\mathbf{a}\cdot\mathbf{b}=1\cdot3+2\cdot0+(-1)\cdot4=-1\). 2. The magnitudes are \(\|\mathbf{a}\|=\sqrt6\) and \(\|\mathbf{b}\|=5\). 3. Therefore, \(\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}=-\frac{1}{5\sqrt6}\). 4. Use \(\sin^2\theta=1-\cos^2\theta\): \(\sin\theta=\sqrt{1-\frac{1}{150}}=\sqrt{\frac{149}{150}}\). 5. The triangle's area is \(A=\frac12\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta=\frac{\sqrt{149}}2\approx6.10\).

a) \(\mathbf{a}\cdot\mathbf{b}=-1\), \(\cos\theta=-\frac{1}{5\sqrt6}\) b) \(A=\frac{\sqrt{149}}2\approx6.10\) square units

- For part a, separate the fixed part of \(R_t\) from the part multiplied by \(t\). - At each specified vertex, form the two side vectors that meet there. - Use a separate zero-dot-product equation for the right-angle condition in parts b) and c).

1. Write the position vector of \(R_t\) as \(\overrightarrow{OR_t}=\langle0,0,2\rangle+t\langle1,2,0\rangle\). Thus, the locus is a line through \((0,0,2)\) with direction vector \(\langle1,2,0\rangle\). 2. For a right angle at \(P\), require \(\overrightarrow{PQ}\cdot\overrightarrow{PR_t}=0\). Here, \(\overrightarrow{PQ}=\langle-2,2,2\rangle\) and \(\overrightarrow{PR_t}=\langle t-3,2t-2,1\rangle\). 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, \(\overrightarrow{QP}=\langle2,-2,-2\rangle\) and \(\overrightarrow{QR_t}=\langle t-1,2t-4,-1\rangle\). 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 \(\overrightarrow{OR_t}=\langle0,0,2\rangle+t\langle1,2,0\rangle\). b) \(t=-2\) c) Yes. The triangle has a right angle at \(Q\) when \(t=4\).

- Apply the translation to \(C\) first. - Use two side vectors that begin at \(A\). - Relate the side-vector magnitudes and dot product to the parallelogram area. - The triangle has half the parallelogram area.

1. \(F=C+\mathbf{v}=(4,5,6)\). 2. \(\overrightarrow{AB}=\langle3,1,0\rangle\) and \(\overrightarrow{AF}=\langle2,5,6\rangle\). 3. Their squared magnitudes are \(\|\overrightarrow{AB}\|^2=10\) and \(\|\overrightarrow{AF}\|^2=65\). 4. Their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AF}=3\cdot2+1\cdot5+0\cdot6=11\). 5. The squared parallelogram area is \(10\cdot65-11^2=529\), so the parallelogram area is \(23\). 6. Therefore, the triangle's area is \(\frac{23}{2}=11.5\).

\(\mathbf{u}\cdot\mathbf{w}=3\cdot2+1\cdot5+0\cdot6=11\). The area is \(\frac12\sqrt{10\cdot65-11^2}=\frac{23}{2}\) square units.

- What does taking midpoints change about the triangle's side vectors? - Which area relation uses only side lengths and a dot product? - Does your result represent the triangle or the parallelogram formed by its side vectors?

1. The midpoints are \(M_1=(2,0,0)\), \(M_2=(4,4,2)\), and \(M_3=(0,2,4)\). 2. Two side vectors are \(\overrightarrow{M_1M_2}=\langle2,4,2\rangle\) and \(\overrightarrow{M_1M_3}=\langle-2,2,4\rangle\). 3. Their squared magnitudes are both \(24\), and their dot product is \(2(-2)+4(2)+2(4)=12\). 4. The squared area of the parallelogram formed by the vectors is \(24\cdot24-12^2=432\). 5. Therefore, the area of the triangle is \(\frac12\sqrt{432}=6\sqrt{3}\).

\(M_1=(2,0,0)\), \(M_2=(4,4,2)\), and \(M_3=(0,2,4)\). With \(\mathbf{u}=\langle2,4,2\rangle\) and \(\mathbf{v}=\langle-2,2,4\rangle\), \(\mathbf{u}\cdot\mathbf{v}=2(-2)+4(2)+2(4)=12\). The area is \(\frac12\sqrt{24\cdot24-12^2}=6\sqrt{3}\) square units.

- Compute the dot product inside each set of parentheses first. - Use each resulting scalar to scale the vector that remains. - Compare the results and track which operation produced a scalar and which produced a vector.

1. \(\mathbf{e}_1\cdot\mathbf{e}_1=1\), so \((\mathbf{e}_1\cdot\mathbf{e}_1)\mathbf{e}_2=1\mathbf{e}_2=\langle0,1,0\rangle\). 2. \(\mathbf{e}_1\cdot\mathbf{e}_2=0\), so \((\mathbf{e}_1\cdot\mathbf{e}_2)\mathbf{e}_1=0\mathbf{e}_1=\mathbf{0}\). 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.

\((\mathbf{e}_1\cdot\mathbf{e}_1)\mathbf{e}_2=\langle0,1,0\rangle\), while \((\mathbf{e}_1\cdot\mathbf{e}_2)\mathbf{e}_1=\mathbf{0}\). The dot product produces a scalar, which then scales a vector. The two expressions use different pairs in the dot product, so they need not agree; dot product and scalar multiplication cannot be regrouped as one associative operation.

- First determine whether one given vector is a scalar multiple of the other. - Write one dot-product equation for a vector perpendicular to their common direction. - Interpret the solution set of that one linear equation geometrically in three dimensions.

1. Since \(\mathbf{v}=-\frac{3}{2}\mathbf{u}\), the two given vectors are parallel and point in opposite directions. 2. A vector \(\mathbf{w}=\langle w_1,w_2,w_3\rangle\) is perpendicular to \(\mathbf{u}\) 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 \(\mathbf{w}=\langle-2,1,0\rangle\). 4. Because \(\mathbf{v}\) is a scalar multiple of \(\mathbf{u}\), every vector perpendicular to \(\mathbf{u}\) is also perpendicular to \(\mathbf{v}\). 5. The vectors perpendicular to the line in the direction of \(\mathbf{u}\) form a plane through the origin. That plane contains infinitely many vector directions.

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

- Use the relationship between opposite sides of a parallelogram to find \(D\). - Use the two adjacent side vectors at \(A\). - The dot product can be used both in a vector area identity and in the angle formula. - Be sure to distinguish the parallelogram area from a triangle area.

1. In a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}=\langle1,2,1\rangle\), so \(D=A+\overrightarrow{BC}=(2,3,3)\). 2. The adjacent side vectors are \(\overrightarrow{AB}=\langle3,1,0\rangle\) and \(\overrightarrow{AD}=\langle1,2,1\rangle\). Their dot product is \(5\), and their squared magnitudes are \(10\) and \(6\). 3. The squared parallelogram area is \(10\cdot6-5^2=35\), so the exact area is \(\sqrt{35}\) square units. 4. For the angle, \(\cos\alpha=\frac{5}{\sqrt{10}\sqrt{6}}=\frac{5}{\sqrt{60}}\), giving \(\alpha\approx49.80^\circ\).

1. \(D(2,3,3)\) 2. \(\overrightarrow{AB}\cdot\overrightarrow{AD}=3(1)+1(2)+0(1)=5\). The area is \(\sqrt{10\cdot6-5^2}=\sqrt{35}\) square units. 3. Using the same dot product, \(\cos\alpha=\frac{5}{\sqrt{60}}\) and \(\alpha\approx49.80^\circ\).

- Form each directed side by subtracting initial-point coordinates from terminal-point coordinates. - Compare the two pairs of opposite-side vectors before using angle information. - Use a dot product of adjacent sides to test the rectangle condition.

1. Subtract endpoint coordinates: \(\overrightarrow{AB}=\langle3,1,2\rangle\) and \(\overrightarrow{DC}=\langle3,1,2\rangle\). 2. Similarly, \(\overrightarrow{AD}=\langle1,3,0\rangle\) and \(\overrightarrow{BC}=\langle1,3,0\rangle\). 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) \(\overrightarrow{AB}=\langle3,1,2\rangle\), \(\overrightarrow{DC}=\langle3,1,2\rangle\), \(\overrightarrow{AD}=\langle1,3,0\rangle\), and \(\overrightarrow{BC}=\langle1,3,0\rangle\). b) \(ABCD\) is a parallelogram, but it is not a rectangle because \(\overrightarrow{AB}\cdot\overrightarrow{AD}=6\ne0\).

- Form two side vectors from the same vertex and check their angle relationship. - Use the side lengths together with that angle relationship to determine the triangle's area. - For the normal vector, impose a zero dot product with each side vector before using the magnitude and sign conditions.

1. \(\overrightarrow{AB}=\langle4,0,3\rangle\) and \(\overrightarrow{AC}=\langle0,2,0\rangle\). 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 \(\frac12\cdot5\cdot2=5\). 4. Let \(\mathbf{n}=\langle x,y,z\rangle\). Perpendicularity gives \(4x+3z=0\) and \(2y=0\). 5. Thus, \(\mathbf{n}=t\langle-3,0,4\rangle\). The direction vector has magnitude \(5\), so \(|t|=3\) for magnitude \(15\). 6. A positive \(z\)-component requires \(t=3\), giving \(\mathbf{n}=\langle-9,0,12\rangle\).

a) \(5\) square units b) \(\mathbf{n}=\langle-9,0,12\rangle\)

- Represent the unknown vector with three components. - Translate both perpendicularity conditions into linear equations. - Describe the one-parameter solution set with a convenient direction vector.

1. Let \(\mathbf{x}=\langle x,y,z\rangle\). 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 \(\mathbf{x}=r\langle-7,-8,10\rangle\), where \(r\in\mathbb{R}\).

\(\mathbf{x}=r\langle-7,-8,10\rangle\), where \(r\in\mathbb{R}\)

- Translate perpendicularity into one equation involving \(x\) and \(y\). - Translate the magnitude condition into a second equation. - Solve the linear equation for one variable before substituting into the quadratic condition.

1. Perpendicularity requires \(\mathbf{u}\cdot\mathbf{v}=0\): \(x-4+2y=0\), 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 \(x=-\frac{2}{5}\).

\((x,y)=(2,1)\) or \((x,y)=\left(-\frac{2}{5},\frac{11}{5}\right)\)

- Use the perpendicularity condition to relate \(a\) and \(b\). - Use the magnitude condition as an independent equation. - After substitution, account for both signs allowed by a squared equation.

1. Perpendicularity gives \(\mathbf{n}\cdot\mathbf{w}=3a-4b=0\), so \(a=\frac43b\). 2. The magnitude condition gives \(\sqrt{a^2+b^2+12^2}=13\). After squaring, \(a^2+b^2=25\). 3. Substitute \(a=\frac43b\): \(\frac{16}{9}b^2+b^2=25\). 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.

\((a,b)=(4,3)\) or \((a,b)=(-4,-3)\)

- Compare the equality with the absolute-value form of the dot-product bound. - Interpret the equality case geometrically before introducing a scale factor. - Use the required magnitude to determine the possible scale factors.

1. By the dot-product angle formula, \(\left|\mathbf{a}\cdot\mathbf{x}\right|=\|\mathbf{a}\|\|\mathbf{x}\||\cos\theta|\). 2. The given equality requires \(|\cos\theta|=1\), so \(\mathbf{x}\) must be parallel to \(\mathbf{a}\). Write \(\mathbf{x}=k\mathbf{a}\). 3. Since \(\|\mathbf{a}\|=\sqrt{2^2+(-1)^2+2^2}=3\), the magnitude condition gives \(9=\|\mathbf{x}\|=3|k|\). 4. Thus, \(|k|=3\), so \(k=3\) or \(k=-3\). 5. Therefore, \(\mathbf{x}=\langle6,-3,6\rangle\) or \(\mathbf{x}=\langle-6,3,-6\rangle\).

\(\mathbf{x}=\langle6,-3,6\rangle\) or \(\mathbf{x}=\langle-6,3,-6\rangle\)

- Determine the type of each parenthesized result first. - Ask what operation can accept those input types and produce the requested output type. - Keep dot product and scalar multiplication distinct even though both use the symbol \(\cdot\) here.

1. In part a, the quantity in parentheses must be a vector so that it can be dotted with \(\mathbf{w}\). 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 \(\mathbf{w}\) 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) \(\cdot\)

- Place one vertex at the origin and align the cube's edges with the axes. - Represent two space diagonals using sign variations of \(a\). - Use the absolute dot product when finding the acute angle between lines.

1. Place the cube so its edges are parallel to the coordinate axes. 2. Two distinct space diagonals can be represented by direction vectors \(\mathbf{d}_1=\langle a,a,a\rangle\) and \(\mathbf{d}_2=\langle-a,a,a\rangle\). 3. Their dot product is \(\mathbf{d}_1\cdot\mathbf{d}_2=-a^2+a^2+a^2=a^2\). 4. Each vector has magnitude \(a\sqrt{3}\). 5. Therefore, \(\cos\alpha=\frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{\|\mathbf{d}_1\|\|\mathbf{d}_2\|}=\frac{a^2}{3a^2}=\frac{1}{3}\). 6. Any pair of distinct space-diagonal lines has direction vectors differing in the sign of one or two components, so the absolute dot-product calculation gives the same acute-angle cosine.

Use direction vectors \(\mathbf{d}_1=\langle a,a,a\rangle\) and \(\mathbf{d}_2=\langle-a,a,a\rangle\). Then \(|\mathbf{d}_1\cdot\mathbf{d}_2|=a^2\) and both lengths are \(a\sqrt{3}\), so \(\cos\alpha=\frac{a^2}{3a^2}=\frac13\). Other distinct pairs of space diagonals differ by one or two coordinate signs and give the same absolute dot product.

- Write two side vectors from \(A\). - Use the dot-product area identity, then divide the parallelogram area by \(2\). - Use equal opposite-side vectors to locate \(D\). - Relate the triangle's area to the base \(AC\) and its altitude.

1. The side vectors from \(A\) are \(\overrightarrow{AB}=\langle2,2,1\rangle\) and \(\overrightarrow{AC}=\langle0,4,4\rangle\). 2. Their squared magnitudes are \(9\) and \(32\), and their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AC}=12\). 3. The triangle's area is \(\frac12\sqrt{9\cdot32-12^2}=6\) square units. 4. Since \(\overrightarrow{AD}=\overrightarrow{BC}=\langle-2,2,3\rangle\), \(D=A+\overrightarrow{BC}=(-1,2,5)\). 5. Side \(AC\) has length \(\sqrt{32}=4\sqrt{2}\). From \(6=\frac12\cdot4\sqrt{2}\cdot h\), the altitude is \(h=\frac{3\sqrt{2}}{2}\).

a) \(\overrightarrow{AB}\cdot\overrightarrow{AC}=2\cdot0+2\cdot4+1\cdot4=12\). The area is \(\frac12\sqrt{9\cdot32-12^2}=6\) square units. b) \(D(-1,2,5)\) c) \(\frac{3\sqrt{2}}{2}\) units

- Translate perpendicularity into a zero-dot-product condition. - Simplify the resulting expression before solving. - Check whether the quadratic has two real roots.

1. Perpendicularity requires \(\mathbf{u}\cdot\mathbf{v}=0\). 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 \(t=1\)

- For part a, assign convenient values to two components and solve for the third. - For part b, compare the component equation with the standard form of a plane. - For part c, focus on whether the equation determines one vector or infinitely many.

1. Writing \(\mathbf{x}=\langle x_1,x_2,x_3\rangle\) gives \(2x_1+x_2-2x_3=12\). 2. Two solutions are \(\mathbf{x}_1=\langle6,0,0\rangle\) and \(\mathbf{x}_2=\langle0,12,0\rangle\). 3. The solution set is the plane \(2x_1+x_2-2x_3=12\), whose normal vector is \(\langle2,1,-2\rangle\). 4. An inverse operation would need to produce a unique \(\mathbf{x}\). 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 \(\mathbf{x}_1=\langle6,0,0\rangle\) and \(\mathbf{x}_2=\langle0,12,0\rangle\). 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 zero dot product to determine the missing component in part a. - Compare the current magnitude of \(\mathbf{x}\) with the target magnitude before choosing a scale factor. - In part d, use how a dot product behaves when one vector is multiplied by a scalar.

1. Perpendicularity requires \(\mathbf{v}\cdot\mathbf{x}=0\): \(2\cdot1-2\cdot2+x_3=0\). Thus, \(x_3=2\), and \(\mathbf{x}=\langle1,2,2\rangle\). 2. Its magnitude is \(\|\mathbf{x}\|=\sqrt{1^2+2^2+2^2}=3\). 3. To obtain magnitude \(12\) without changing direction, multiply by \(4\): \(\mathbf{y}=4\mathbf{x}=\langle4,8,8\rangle\). 4. By linearity of the dot product, \(\mathbf{v}\cdot\mathbf{y}=\mathbf{v}\cdot(4\mathbf{x})=4(\mathbf{v}\cdot\mathbf{x})=0\). Therefore, \(\mathbf{y}\perp\mathbf{v}\).

a) \(x_3=2\) b) \(\|\mathbf{x}\|=3\) c) \(\mathbf{y}=\langle4,8,8\rangle\) d) Since \(\mathbf{y}=4\mathbf{x}\), \(\mathbf{v}\cdot\mathbf{y}=4(\mathbf{v}\cdot\mathbf{x})=0\).

- Use a zero dot product for the parameter condition in part a. - In three dimensions, identify the geometric set whose normal is a fixed nonzero vector. - Compare the dimension of that set with the corresponding perpendicular set in two dimensions.

1. Perpendicularity requires \(\mathbf{a}\cdot\mathbf{b}=0\). 2. Compute \(2k+6k-20=0\). 3. Thus, \(8k=20\), so \(k=\frac52\). 4. In \(\mathbb{R}^3\), all vectors perpendicular to a fixed nonzero vector \(\mathbf{n}\) lie in the plane through the origin whose normal vector is \(\mathbf{n}\). 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) \(k=\frac52\) 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.

- Audit the dot-product arithmetic before checking the angle formula. - Compare the denominator with the exact dot-product angle formula. - Check the inverse-cosine value independently of the earlier steps.

1. The first error is arithmetic: \(0\cdot5=0\), not \(5\). Thus, \(\mathbf{u}\cdot\mathbf{v}=0\). 2. The second error is in the angle formula. The denominator is the product of the magnitudes, not their sum: \(\|\mathbf{u}\|\|\mathbf{v}\|=5\cdot5\). 3. The third error is the inverse-cosine value: \(\cos\phi=0.5\) would give \(\phi=60^\circ\), not \(30^\circ\). 4. Using the correct dot product, \(\cos\phi=\frac{0}{5\cdot5}=0\). 5. Therefore, \(\phi=90^\circ\).

Error 1: \(0\cdot5=0\), so the dot product is \(0\), not \(5\). Error 2: The vector magnitudes must be multiplied in the denominator, not added. Error 3: \(\cos^{-1}(0.5)=60^\circ\), not \(30^\circ\). The correct angle is \(\phi=90^\circ\).

- Form the two side vectors that meet at \(P\). - Use their dot product to determine the angle type before finding lengths. - Once the right angle is established, use the three side lengths for perimeter and the two legs for area.

1. The side vectors meeting at \(P\) are \(\overrightarrow{PQ}=\langle2,2,1\rangle\) and \(\overrightarrow{PR}=\langle-1,2,-2\rangle\). 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 \(\frac12\cdot3\cdot3=\frac92\).

a) \(\overrightarrow{PQ}\cdot\overrightarrow{PR}=2(-1)+2(2)+1(-2)=0\), so the triangle has a right angle at \(P\). b) \(6+3\sqrt{2}\) units c) \(\frac92\) square units

- Form the two side vectors that meet at \(E\). - Translate the right-angle requirement into a zero-dot-product equation. - For part b, squared side lengths are enough to test whether two sides are equal.

1. The vectors meeting at \(E\) are \(\overrightarrow{ED}=\langle-2,-1,k-1\rangle\) and \(\overrightarrow{EF}=\langle-2,3,1\rangle\). 2. Their dot product is \(\overrightarrow{ED}\cdot\overrightarrow{EF}=4-3+(k-1)=k\). 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) \(\overrightarrow{ED}\cdot\overrightarrow{EF}=4-3+(k-1)=k\). Setting this equal to zero gives \(k=0\). b) Yes. At \(k=4\), \(DE^2=EF^2=14\), so \(DE=EF=\sqrt{14}\) and the triangle is isosceles.

- Compare the two pairs of opposite-side vectors first. - Use a dot product to test whether adjacent sides are perpendicular. - A square requires one additional side-length condition beyond being a rectangle.

1. Compute the side vectors: \(\overrightarrow{AB}=\langle4,0,0\rangle\), \(\overrightarrow{BC}=\langle0,2,3\rangle\), \(\overrightarrow{DC}=\langle4,0,0\rangle\), and \(\overrightarrow{AD}=\langle0,2,3\rangle\). 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.

\(\overrightarrow{AB}=\overrightarrow{DC}=\langle4,0,0\rangle\) and \(\overrightarrow{AD}=\overrightarrow{BC}=\langle0,2,3\rangle\), so \(ABCD\) is a parallelogram. Also \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\), making it a rectangle. The adjacent side lengths are \(AB=4\) and \(BC=\sqrt{13}\), which differ, so it is not a square.

- Use equality of opposite-side displacement vectors to locate \(D\). - Compare adjacent side magnitudes to test the rhombus condition. - Use a dot product of adjacent sides to test the rectangle condition.

1. In a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\). 2. Since \(\overrightarrow{BC}=\langle0,3,4\rangle\), add this displacement to \(A\): \(D=(2,1,-1)+(0,3,4)=(2,4,3)\). 3. The adjacent side vectors are \(\overrightarrow{AB}=\langle3,4,0\rangle\) and \(\overrightarrow{BC}=\langle0,3,4\rangle\). 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 \(BC\). - After the dot-product test, compare the magnitudes of the two equal-side candidates.

1. The vector from \(B\) to \(C\) is \(\mathbf{w}=\mathbf{v}-\mathbf{u}=\langle-7,4,-3\rangle\). 2. The midpoint of \(BC\) has position relative to \(A\) equal to \(\mathbf{s}_a=\frac12(\mathbf{u}+\mathbf{v})=\left\langle\frac12,2,\frac32\right\rangle\). 3. Compute \(\mathbf{s}_a\cdot\mathbf{w}=\frac12(-7)+2\cdot4+\frac32(-3)=0\). Thus, the median is perpendicular to \(BC\). 4. Also, \(\|\mathbf{u}\|=5\) and \(\|\mathbf{v}\|=5\). 5. Therefore, \(AB=AC\), so the triangle is isosceles. The median to its base is also an altitude.

a) \(\mathbf{w}=\langle-7,4,-3\rangle\) and \(\mathbf{s}_a=\left\langle\frac12,2,\frac32\right\rangle\) b) \(\mathbf{s}_a\cdot\mathbf{w}=0\), 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.

- First verify that opposite sides are parallel. - Use a dot product on adjacent side vectors. - Use the absolute dot product to find the acute angle between diagonal lines. - A rectangle's circumcenter is the common midpoint of its diagonals.

1. The side vectors are \(\overrightarrow{AB}=\langle2,2,0\rangle\), \(\overrightarrow{BC}=\langle-2,2,2\rangle\), \(\overrightarrow{CD}=\langle-2,-2,0\rangle\), and \(\overrightarrow{DA}=\langle2,-2,-2\rangle\). 2. Since \(\overrightarrow{AB}=-\overrightarrow{CD}\) and \(\overrightarrow{BC}=-\overrightarrow{DA}\), the quadrilateral is a parallelogram. 3. Also, \(\overrightarrow{AB}\cdot\overrightarrow{BC}=2\cdot(-2)+2\cdot2+0\cdot2=0\). Thus, adjacent sides are perpendicular, so it is a rectangle. 4. The diagonal vectors are \(\overrightarrow{AC}=\langle0,4,2\rangle\) and \(\overrightarrow{BD}=\langle-4,0,2\rangle\). Their dot product is \(4\), and each has magnitude \(\sqrt{20}\). 5. Therefore, \(\cos\phi=\frac{|4|}{\sqrt{20}\sqrt{20}}=\frac15\), so \(\phi=\cos^{-1}\left(\frac15\right)\approx78.46^\circ\). 6. The circle's center is the common midpoint of the diagonals: \(M=(1,3,3)\). 7. Its radius is half a diagonal: \(r=\frac{\sqrt{20}}2=\sqrt5\).

a) \(\overrightarrow{AB}=-\overrightarrow{CD}\), \(\overrightarrow{BC}=-\overrightarrow{DA}\), and \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\), so \(ABCD\) is a rectangle. b) \(\phi\approx78.46^\circ\) c) \(M=(1,3,3)\) and \(r=\sqrt5\) units

- Write one zero-dot-product equation for each given vector. - Expect one free parameter because there are two independent linear conditions on three components. - Interpret the resulting scalar multiples 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 \(\mathbf{x}=t\langle1,1,-2\rangle\), where \(t\in\mathbb{R}\). 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) \(\mathbf{x}=t\langle1,1,-2\rangle\), 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 the known middle component in the unknown vector from the start. - Translate each perpendicularity condition into a linear equation. - Solve the resulting two-equation system for the remaining components.

1. Let \(\mathbf{c}=\langle x,15,z\rangle\). 2. From \(\mathbf{a}\cdot\mathbf{c}=0\), \(3x-15+2z=0\), so \(3x+2z=15\). 3. From \(\mathbf{b}\cdot\mathbf{c}=0\), \(x+30-z=0\), 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 \(\mathbf{c}=\langle-9,15,21\rangle\).

\(\mathbf{c}=\langle-9,15,21\rangle\)

- A normal vector must be perpendicular to each spanning vector of the plane. - Write one dot-product equation for each spanning vector. - Solve the resulting linear system for \(w_1\) and \(w_2\).

1. A vector perpendicular to the plane must be perpendicular to both spanning vectors. 2. Therefore, \(\mathbf{w}\cdot\mathbf{u}=w_1+w_2+6=0\) and \(\mathbf{w}\cdot\mathbf{v}=2w_1-w_2+18=0\). 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, \(\mathbf{w}=\langle-8,2,6\rangle\).

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

- Use equal opposite-side vectors to find \(C\). - The angle at \(B\) is formed by \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\). - Use the dot-product area identity with two adjacent side vectors. - The diagonals of a parallelogram bisect each other.

1. \(\overrightarrow{AB}=\langle3,0,4\rangle\), so \(C=D+\overrightarrow{AB}=(6,4,5)\). 2. At \(B\), \(\overrightarrow{BA}=\langle-3,0,-4\rangle\) and \(\overrightarrow{BC}=\langle1,3,0\rangle\). Their dot product is \(-3\), so \(\cos\beta=\frac{-3}{5\sqrt{10}}\) and \(\beta\approx100.94^\circ\). 3. The adjacent vectors at \(A\) are \(\overrightarrow{AB}=\langle3,0,4\rangle\) and \(\overrightarrow{AD}=\langle1,3,0\rangle\). Their squared magnitudes are \(25\) and \(10\), and their dot product is \(3\). Thus, the area is \(\sqrt{25\cdot10-3^2}=\sqrt{241}\) square units. 4. The diagonals bisect each other, giving \(M=\frac{A+C}{2}=(4,2.5,3)\).

a) \(C(6,4,5)\) b) \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-3\); \(\beta\approx100.94^\circ\) c) \(\overrightarrow{AB}\cdot\overrightarrow{AD}=3\); area \(=\sqrt{25\cdot10-3^2}=\sqrt{241}\) square units d) \(M(4,2.5,3)\)

- What does the parallelogram condition tell you about the unknown vertex? - How could you check the unknown vertex with a second property of parallelograms? - Why might an angle between two lines be different from an oriented angle between two vectors?

1. Since \(\overrightarrow{PS}=\overrightarrow{QR}=\langle1,2,3\rangle\), \(S=P+\overrightarrow{QR}=(2,3,4)\). 2. The adjacent vectors are \(\overrightarrow{PQ}=\langle4,0,0\rangle\) and \(\overrightarrow{PS}=\langle1,2,3\rangle\). Their squared magnitudes are \(16\) and \(14\), and their dot product is \(4\). The area is \(\sqrt{16\cdot14-4^2}=4\sqrt{13}\) square units. 3. The diagonals bisect each other, so \(T=\frac{P+R}{2}=(3.5,2,2.5)\). 4. The diagonal vectors are \(\overrightarrow{PR}=\langle5,2,3\rangle\) and \(\overrightarrow{QS}=\langle-3,2,3\rangle\). Their dot product is \(-2\), and their magnitudes are \(\sqrt{38}\) and \(\sqrt{22}\). 5. Thus, \(\cos\phi=\frac{|-2|}{\sqrt{38}\sqrt{22}}\), giving \(\phi\approx86.03^\circ\).

a) \(S(2,3,4)\) b) \(\overrightarrow{PQ}\cdot\overrightarrow{PS}=4\); area \(=4\sqrt{13}\) square units c) \(T(3.5,2,2.5)\); \(\overrightarrow{PR}\cdot\overrightarrow{QS}=-2\); \(\phi\approx86.03^\circ\)

- The shared position point gives the intersection immediately. - Use the dot-product angle formula for the line directions in part a. - In part b, translate “isosceles with base \(PQ\)” into equality of the other two sides. - For the reflection, first locate 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, \(\cos\alpha=\frac89\), so \(\alpha\approx27.3^\circ\). 3. For an isosceles triangle with base \(PQ\), \(SP=SQ\). Since \(\overrightarrow{SP}=\langle9,18,18\rangle\), \(SP=27\). 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 \(s=\frac{\overrightarrow{SP}\cdot\mathbf{v}_h}{\|\mathbf{v}_h\|^2}=8\), 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) \(P'=(25,13,3)\)

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

1. Let \(\mathbf{n}=\langle a,b,c\rangle\) be normal to \(E\). The equations \(a+c=0\) and \(2a+b=0\) give one normal vector \(\mathbf{n}=\langle-1,2,1\rangle\). 2. Both direction vectors of \(F\) are orthogonal to this normal: \(\langle3,1,1\rangle\cdot\mathbf{n}=-3+2+1=0\) and \(\langle1,1,-1\rangle\cdot\mathbf{n}=-1+2-1=0\). 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 \(-0+2(-1)+1=-1\ne1\). 6. Thus, the planes are distinct and strictly parallel.

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

- Read a normal vector from the plane equation. - Use dot products with that normal to classify each line direction. - For a parallel line, test one point for containment; for a nonparallel line, substitute its parameterization into the plane equation.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle1,2,-1\rangle\). For \(g\), \(\mathbf{n}\cdot\mathbf{d}_g=2-2=0\). For \(h\), \(\mathbf{n}\cdot\mathbf{d}_h=1+2-1=2\). 2. The zero dot product makes \(g\) parallel to the plane's directions. Its initial point lies in \(E\) because \(1+2(1)-(-1)=4\), so \(g\) lies entirely in \(E\). 3. The nonzero dot product means \(h\) meets \(E\) at one point. Substitute \((3+s,1+s,2+s)\) into \(E\) to get \(3+2s=4\). 4. Hence \(s=\frac12\), giving intersection point \(S\left(\frac72,\frac32,\frac52\right)\).

Plane normal: \(\mathbf{n}=\langle1,2,-1\rangle\). Dot products: \(\mathbf{n}\cdot\mathbf{d}_g=0\) and \(\mathbf{n}\cdot\mathbf{d}_h=2\). a) Line \(g\) lies entirely in plane \(E\). b) \(S\left(\frac72,\frac32,\frac52\right)\)

- Find a vector orthogonal to both spanning directions of the plane. - Test the line direction against that normal vector. - After establishing parallelism, test one point of the line to distinguish containment from strict parallelism.

1. Let \(\mathbf{n}=\langle a,b,c\rangle\) be normal to \(E\). Orthogonality to the plane directions gives \(2a+b=0\) and \(a+c=0\). One solution is \(\mathbf{n}=\langle1,-2,-1\rangle\). 2. The line direction satisfies \(\langle3,1,1\rangle\cdot\mathbf{n}=3-2-1=0\), 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 \(2-2(2)-3=-5\ne-4\). 5. Therefore, \(g\) is strictly parallel to \(E\).

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

- Find a normal vector to the parametric plane from its two spanning directions. - Parallel planes require parallel normal vectors. - After finding \(a\), test one point to distinguish identical planes from strictly parallel planes.

1. Let \(\mathbf{n}_2=\langle p,q,r\rangle\) be normal to \(E_2\). It must satisfy \(\mathbf{n}_2\cdot\langle2,1,0\rangle=0\) and \(\mathbf{n}_2\cdot\langle3,0,2\rangle=0\). 2. Solving \(2p+q=0\) and \(3p+2r=0\) gives \(\mathbf{n}_2=\langle2,-4,-3\rangle\). Check: \(\mathbf{n}_2\cdot\langle2,1,0\rangle=4-4=0\) and \(\mathbf{n}_2\cdot\langle3,0,2\rangle=6-6=0\). 3. A normal to \(E_1\) is \(\mathbf{n}_1=\langle2,-4,a\rangle\). Parallel planes require parallel normals. The first two components fix the scale at \(1\), so \(a=-3\). 4. The point \((1,0,1)\) on \(E_2\) gives \(2-3=-1\ne8\) in \(E_1\). Thus, the planes are strictly parallel.

\(\mathbf{n}_2=\langle2,-4,-3\rangle\), with \(\mathbf{n}_2\cdot\langle2,1,0\rangle=0\) and \(\mathbf{n}_2\cdot\langle3,0,2\rangle=0\). Thus, \(a=-3\). The planes are strictly parallel because \((1,0,1)\) does not lie in \(E_1\).

- A direction lying in a plane is perpendicular to the plane's normal. - For part b, impose perpendicularity to both given vectors on an unknown vector. - A two-parameter plane equation needs one point and two independent in-plane directions.

1. \(\mathbf{n}\cdot\mathbf{u}=\langle1,0,2\rangle\cdot\langle0,1,0\rangle=0\), so \(\mathbf{u}\) is parallel to the plane. 2. Let \(\mathbf{v}=\langle a,b,c\rangle\). Orthogonality to \(\mathbf{u}\) gives \(b=0\), and orthogonality to \(\mathbf{n}\) gives \(a+2c=0\). Choose \(c=1\), so \(a=-2\). Thus, \(\mathbf{v}=\langle-2,0,1\rangle\). 3. The plane is \(\mathbf{r}(s,t)=\langle3,-2,4\rangle+s\langle0,1,0\rangle+t\langle-2,0,1\rangle\).

a) \(\mathbf{n}\cdot\mathbf{u}=0\) b) \(\mathbf{v}=\langle-2,0,1\rangle\), or any nonzero scalar multiple c) \(\mathbf{r}(s,t)=\langle3,-2,4\rangle+s\langle0,1,0\rangle+t\langle-2,0,1\rangle\)

- The coordinate coefficients form a normal vector. - All normals to the same plane lie on one line through the origin. - Choose a convenient point on the plane, then compare how scalar multiples change the equation without changing the plane.

1. A normal vector is \(\mathbf{n}=\langle2,-1,2\rangle\), so \(\langle2,-1,2\rangle\cdot\mathbf{x}=12\). 2. Every nonzero scalar multiple of \(\mathbf{n}\) is also normal to the plane. The set is \(\{k\langle2,-1,2\rangle:k\in\mathbb{R}\setminus\{0\}\}\). 3. Setting \(y=z=0\) gives \(2x=12\), so \(P(6,0,0)\) lies on the plane. One point-normal equation is \(\langle2,-1,2\rangle\cdot\langle x-6,y,z\rangle=0\). 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) \(\langle2,-1,2\rangle\cdot\mathbf{x}=12\) b) \(\{k\langle2,-1,2\rangle:k\in\mathbb{R}\setminus\{0\}\}\) c) For example, \(P(6,0,0)\) and \(\langle2,-1,2\rangle\cdot\langle x-6,y,z\rangle=0\) d) No. Nonzero scalar multiples of the equation and different points on the plane give equivalent representations.

- Use the known point to determine the constant in the dot-product plane equation. - Test a point by comparing its dot product with the same constant. - Express \(\overrightarrow{PQ}\) as the difference of two position vectors and use distributivity.

1. Compute \(d=\langle2,-1,4\rangle\cdot\langle3,2,-2\rangle=2(3)+(-1)(2)+4(-2)=-4\). 2. For \(B\), \(\langle4,-3,1\rangle\cdot\langle3,2,-2\rangle=12-6-2=4\). Since \(4\ne-4\), \(B\notin E\). 3. Because \(P,Q\in E\), \(\overrightarrow{OP}\cdot\mathbf{n}=d\) and \(\overrightarrow{OQ}\cdot\mathbf{n}=d\). Since \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\), distributivity gives \(\overrightarrow{PQ}\cdot\mathbf{n}=\overrightarrow{OQ}\cdot\mathbf{n}-\overrightarrow{OP}\cdot\mathbf{n}=d-d=0\). Therefore, every vector joining two points in the plane is orthogonal to the normal vector.

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

- Apply distributivity and scalar compatibility of the dot product to \(r\mathbf{u}+s\mathbf{v}\). - Substitute the two given zero dot products. - For part 2, compare what two collinear vectors span with what two noncollinear vectors span.

1. For any \(\mathbf{x}=r\mathbf{u}+s\mathbf{v}\), distributivity gives \(\mathbf{n}\cdot\mathbf{x}=r(\mathbf{n}\cdot\mathbf{u})+s(\mathbf{n}\cdot\mathbf{v})\). 2. Since \(\mathbf{n}\cdot\mathbf{u}=0\) and \(\mathbf{n}\cdot\mathbf{v}=0\), it follows that \(\mathbf{n}\cdot\mathbf{x}=0\) for every \(r,s\in\mathbb{R}\). Thus, \(\mathbf{n}\) is perpendicular to every direction in the plane. 3. If \(\mathbf{u}\) and \(\mathbf{v}\) 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. \(\mathbf{n}\cdot(r\mathbf{u}+s\mathbf{v})=r(\mathbf{n}\cdot\mathbf{u})+s(\mathbf{n}\cdot\mathbf{v})=0\), so \(\mathbf{n}\) 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 given point by substitution. - For part c, use \(z=0\) but keep both \(x\) and \(y\) nonzero.

1. Expanding the dot product gives \(x+3y-2z=6\). 2. For \(R\), \(1+3(1)-2(-1)=6\), so \(R\in E\). For \(S\), \(2+3(2)-2(2)=4\ne6\), 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 \(T(3,1,0)\).

- Read the plane's normal vector from its equation. - Test each line direction against the normal with a dot product. - For a parallel line, test one point; for a nonparallel line, solve for its intersection parameter.

1. A normal to \(E\) is \(\mathbf{n}=\langle2,-4,3\rangle\). Compute both direction tests: \(\mathbf{n}\cdot\mathbf{d}_g=2\cdot2-4\cdot1+3\cdot0=0\) and \(\mathbf{n}\cdot\mathbf{d}_h=2\cdot1-4\cdot1+3\cdot1=1\). 2. The direction of \(g\) is parallel to \(E\). Its initial point lies in \(E\), since \(2(1)-4(2)+3(6)=12\), so the entire line lies in \(E\). 3. The nonzero second dot product means \(h\) intersects \(E\) once. Substitution gives \(2(4+s)-4s+3(1+s)=12\), hence \(11+s=12\) and \(s=1\). 4. Therefore, \(h\) meets \(E\) at \(S(5,1,2)\).

Plane normal: \(\mathbf{n}=\langle2,-4,3\rangle\). Dot products: \(\mathbf{n}\cdot\mathbf{d}_g=0\) and \(\mathbf{n}\cdot\mathbf{d}_h=1\). 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. - Use the family point to determine which member lies in the plane. - Parameterize the line through the origin and substitute it into the plane equation.

1. The plane normal is \(\mathbf{n}=\langle1,1,2\rangle\) and the common line direction is \(\mathbf{d}=\langle2,-2,0\rangle\). Their dot product is \(\mathbf{n}\cdot\mathbf{d}=2-2+0=0\). Thus every \(g_a\) is contained in or strictly parallel to \(E\). 2. The starting point \((a,2,1)\) lies in \(E\) when \(a+2+2=6\), so \(a=2\). 3. Write \(h\) as \(\mathbf{r}(s)=s\langle1,1,1\rangle\). Substitution into the plane gives \(4s=6\), hence \(s=\frac32\). 4. The intersection is \(S\left(\frac32,\frac32,\frac32\right)\).

a) \(\mathbf{n}=\langle1,1,2\rangle\) and \(\mathbf{d}=\langle2,-2,0\rangle\), so \(\mathbf{n}\cdot\mathbf{d}=2-2+0=0\). Every \(g_a\) lies in or is strictly parallel to \(E\). b) \(a=2\) c) \(S\left(\frac32,\frac32,\frac32\right)\)

- Use a dot product between the line direction and the plane normal for part a. - Test the line's initial point to distinguish containment from strict parallelism. - For \(g_0\), substitute its coordinate functions directly into the plane equation.

1. A normal to \(E\) is \(\mathbf{n}=\langle2,1,-1\rangle\). For the line direction, \(\mathbf{n}\cdot\mathbf{d}_a=2a+2-4=2a-2\). 2. Set this equal to zero to obtain \(a=1\). 3. The common starting point \((1,1,1)\) does not satisfy \(E\), since \(2+1-1=2\ne5\). Thus, \(g_1\) is strictly parallel to \(E\). 4. On \(g_0\), the coordinates are \((1,1+2t,1+4t)\). Substitution into \(E\) gives \(2-2t=5\), so \(t=-\frac32\). 5. Hence the intersection is \(S(1,-2,-5)\).

a) \(\mathbf{n}\cdot\mathbf{d}_a=2a-2\), so \(a=1\) b) \(g_1\) is strictly parallel to \(E\). c) \(S(1,-2,-5)\)

- 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, then check whether the line's initial point lies in the plane.

1. Two plane directions are \(\overrightarrow{AB}=\langle2,-1,1\rangle\) and \(\overrightarrow{AC}=\langle1,-4,2\rangle\). 2. A normal vector orthogonal to both is \(\mathbf{n}=\langle2,-3,-7\rangle\), since \(\mathbf{n}\cdot\overrightarrow{AB}=4+3-7=0\) and \(\mathbf{n}\cdot\overrightarrow{AC}=2+12-14=0\). 3. For the line direction, \(\mathbf{n}\cdot\mathbf{d}_g=\langle2,-3,-7\rangle\cdot\langle3,-5,3\rangle=6+15-21=0\). 4. Thus, \(g\) is parallel to \(E\). Since its starting point is \(B\), which lies in \(E\), the whole line lies in the plane and there are infinitely many common points.

\(\mathbf{n}=\langle2,-3,-7\rangle\) and \(\mathbf{n}\cdot\mathbf{d}_g=0\). Since \(g\) passes through \(B\in E\) and its direction is parallel to the plane, the line lies in \(E\). There are infinitely many common points.

- 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, so compare those normals with a dot product.

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 normals \(\mathbf{n}_1=\langle1,1,1\rangle\) and \(\mathbf{n}_2=\langle-1,1,1\rangle\). 3. Thus, \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac13\). 4. Therefore, \(\alpha=\cos^{-1}\left(\frac13\right)\approx70.5^\circ\).

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

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

1. A normal vector to \(E_1\) is \(\mathbf{n}_1=\langle 4,0,3\rangle\). 2. For \(E_2\), let \(\mathbf{n}_2=\langle a,b,c\rangle\). Perpendicularity to \(\langle 1,-1,0\rangle\) and \(\langle 0,1,1\rangle\) gives \(a-b=0\) and \(b+c=0\). Choose \(\mathbf{n}_2=\langle -1,-1,1\rangle\). 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. - Compare the two normal vectors with a dot product. - The angle between planes equals the acute angle between their normals.

1. For \(E_1\), a normal \(\mathbf{n}_1=\langle a,b,c\rangle\) must satisfy \(a+b=0\) and \(b+c=0\). Choose \(\mathbf{n}_1=\langle 1,-1,1\rangle\). 2. For \(E_2\), a normal \(\mathbf{n}_2=\langle a,b,c\rangle\) must satisfy \(a+c=0\) and \(a+b+c=0\). Choose \(\mathbf{n}_2=\langle -1,0,1\rangle\). 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 \(GH\) and use \(\overrightarrow{AM}\) as the line direction. - Find a normal to the plane through \(B\), \(D\), and \(E\). - Apply the sine form of the dot-product 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 \(\mathbf{v}=\langle1,2,2\rangle\). 2. A normal vector to the plane through \(B\), \(D\), and \(E\) is \(\mathbf{n}=\langle1,1,1\rangle\). 3. The line-plane angle satisfies \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac{5}{3\sqrt3}\). 4. Therefore, \(\alpha\approx74.2^\circ\).

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

- Use two edge vectors from each face. - Find a normal to each face by imposing two zero dot-product conditions. - The acute angle between the faces equals the acute angle between their normals.

1. For face \(BCS\), use \(\overrightarrow{BC}=\langle-10,0,0\rangle\) and \(\overrightarrow{BS}=\langle-5,-5,12\rangle\). A normal \(\mathbf{n}_1=\langle a,b,c\rangle\) satisfies \(a=0\) and \(-5b+12c=0\), so choose \(\mathbf{n}_1=\langle0,12,5\rangle\). 2. For face \(CDS\), use \(\overrightarrow{CD}=\langle0,-10,0\rangle\) and \(\overrightarrow{CS}=\langle5,-5,12\rangle\). A normal satisfies \(b=0\) and \(5a+12c=0\), so choose \(\mathbf{n}_2=\langle-12,0,5\rangle\). 3. The normals have dot product \(25\) and each has magnitude \(13\). 4. Thus, \(\cos\gamma=\frac{25}{169}\), giving \(\gamma\approx81.5^\circ\).

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

- Set the other two coordinates equal to zero to find each axis intercept. - Compare the plane normal with the direction of each coordinate axis. - A missing variable in a plane equation tells you which coordinate-axis direction lies parallel to the plane.

1. On the x-axis, \(y=z=0\), giving the intercept \((4,0,0)\). On the y-axis, \(x=z=0\), giving \((0,-3,0)\). 2. For \(\mathbf{n}=\langle3,-4,0\rangle\), the axis-direction dot products are \(\mathbf{n}\cdot\mathbf{e}_x=3\), \(\mathbf{n}\cdot\mathbf{e}_y=-4\), and \(\mathbf{n}\cdot\mathbf{e}_z=0\). 3. Since \(\|\mathbf{n}\|=5\), the angle with the x-axis has \(\sin\alpha_x=\frac35\), giving \(\alpha_x\approx36.9^\circ\). For the y-axis, \(\sin\alpha_y=\frac45\), giving \(\alpha_y\approx53.1^\circ\). 4. The zero dot product with \(\mathbf{e}_z\) means the z-axis direction is parallel to the plane. Because the z-axis itself is not in \(E\), the plane is strictly parallel to the z-axis, with angle \(0^\circ\).

1. \((4,0,0)\) and \((0,-3,0)\) 2. \(\mathbf{n}\cdot\mathbf{e}_x=3\), \(\mathbf{n}\cdot\mathbf{e}_y=-4\), and \(\mathbf{n}\cdot\mathbf{e}_z=0\); x-axis angle \(\approx36.9^\circ\), y-axis angle \(\approx53.1^\circ\) 3. The plane is strictly parallel to the z-axis, with angle \(0^\circ\).

- Use the intercept form to obtain a standard plane equation. - Read the plane normal from that equation. - Compare it with normals to the two coordinate planes using the acute plane-angle formula.

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 \(\mathbf{n}=\langle 1,1,2\rangle\). 2. The \(xy\)-plane has normal \(\langle 0,0,1\rangle\). Therefore, \(\cos\alpha=\frac{2}{\sqrt{6}}\), so \(\alpha\approx35.3^\circ\). 3. The \(xz\)-plane has normal \(\langle 0,1,0\rangle\). 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. - Apply the condition \(k>0\) only after solving the squared equation.

1. A normal vector to \(E_k\) is \(\mathbf{n}_k=\langle 3,4,k\rangle\). A normal to the \(xy\)-plane is \(\mathbf{e}_z=\langle 0,0,1\rangle\). 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 the normal-based sine relationship for the angle between a line and a plane. - Compare the two resulting acute angles.

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

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

- 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 \(\mathbf{r}(t)=\langle6,2,1\rangle+t\langle2,-1,2\rangle\), where \(t\ge0\). 2. Substitution into the plane equation gives \((6+2t)+(2-t)=10\), so \(t=2\). 3. The forward intersection is \(S=\langle6,2,1\rangle+2\langle2,-1,2\rangle=\langle10,0,5\rangle\). 4. The plane normal is \(\mathbf{n}=\langle1,1,0\rangle\). Thus, \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac1{3\sqrt2}\), giving \(\alpha\approx13.6^\circ\).

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

- 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 line direction is \(\mathbf{v}=\overrightarrow{AB}=\langle2,2,1\rangle\). Thus, \(g:\mathbf{r}(t)=\langle0,1,2\rangle+t\langle2,2,1\rangle\). 2. Substitute into the plane: \(2(2t)-2(1+2t)+(2+t)=2\), giving \(t=2\). 3. Therefore, \(S=\langle0,1,2\rangle+2\langle2,2,1\rangle=\langle4,5,4\rangle\). 4. A plane normal is \(\mathbf{n}=\langle2,-2,1\rangle\). Hence \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac19\), so \(\alpha\approx6.4^\circ\).

\(S=(4,5,4)\) 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 \(\mathbf{n}_1=\langle 1,1,1\rangle\) and \(\mathbf{n}_2=\langle 2,-1,1\rangle\). 2. Let the line direction be \(\mathbf{d}=\langle a,b,c\rangle\). 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 \(\mathbf{d}=\langle 2,1,-3\rangle\). 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 \(g:\mathbf{r}(t)=\langle 3,3,0\rangle+t\langle 2,1,-3\rangle\).

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

- 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 \(\mathbf{n}_E=\langle 1,2,1\rangle\) and \(\mathbf{n}_{F_k}=\langle k,-1,2\rangle\). 3. The given vector already satisfies \(\mathbf{u}\cdot\mathbf{n}_E=5-2-3=0\). 4. Require it to be orthogonal to the second normal as well: \(\mathbf{u}\cdot\mathbf{n}_{F_k}=5k+1-6=0\). 5. Thus \(5k-5=0\), so \(k=1\).

\(k=1\)

- 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. \(\overrightarrow{AB}=\langle-3,3,0\rangle\) and \(\overrightarrow{AC}=\langle-3,0,3\rangle\) are not scalar multiples, so the points determine one plane. 2. A normal perpendicular to both directions is \(\mathbf{n}=\langle1,1,1\rangle\). Through \(A\), the plane is \(E:x+y+z=3\). 3. Substituting \(g\) gives \((1+t)+1+t=3\), so \(t=\frac12\) and \(S=\left(\frac32,1,\frac12\right)\). 4. For line direction \(\mathbf{v}=\langle1,0,1\rangle\), \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac2{\sqrt6}\), giving \(\alpha\approx54.7^\circ\).

a) The points are not collinear; \(E:x+y+z=3\) b) The line intersects the plane at \(S=\left(\frac32,1,\frac12\right)\). c) \(\alpha\approx54.7^\circ\)

- What does a comparison of the normal vectors tell you about the two planes? - If two planes meet, what do all points of their common line satisfy? - Which of the requested measurements depend on direction, and which depend on position?

1. The normals are \(\mathbf{n}_1=\langle2,2,-1\rangle\) and \(\mathbf{n}_2=\langle1,-2,2\rangle\), which are not parallel, so the planes meet in a line. 2. The point \((3,6,6)\) satisfies both planes. A direction perpendicular to both normals is \(\langle2,-5,-6\rangle\). Thus, \(s:\mathbf{r}(\lambda)=\langle3,6,6\rangle+\lambda\langle2,-5,-6\rangle\). 3. \(\cos\phi=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac49\), giving \(\phi\approx63.6^\circ\). 4. The origin's distance to \(E_1\) is \(d=\frac{|0-12|}{\sqrt{2^2+2^2+(-1)^2}}=4\).

a) The planes intersect in \(s:\mathbf{r}(\lambda)=\langle3,6,6\rangle+\lambda\langle2,-5,-6\rangle\). b) \(\phi\approx63.6^\circ\) c) \(4\) units

- 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. The normals are \(\mathbf{n}_1=\langle 1,1,1\rangle\) and \(\mathbf{n}_2=\langle 1,-1,0\rangle\). 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 \(s:\mathbf{r}(t)=\langle 0,0,3\rangle+t\langle 1,1,-2\rangle\). 3. Let \(\mathbf{n}_3=\langle a,b,c\rangle\). Perpendicularity to both planes requires \(\mathbf{n}_3\cdot\mathbf{n}_1=0\) and \(\mathbf{n}_3\cdot\mathbf{n}_2=0\). 4. Thus \(a+b+c=0\) and \(a-b=0\). Hence \(a=b\) and \(c=-2a\). Choose \(\mathbf{n}_3=\langle 1,1,-2\rangle\). 5. Using \(P\), the equation is \(x+y-2z=0\). 6. 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; \(s:\mathbf{r}(t)=\langle 0,0,3\rangle+t\langle 1,1,-2\rangle\). 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. The normals are \(\mathbf{n}_F=\langle 1,1,1\rangle\) and \(\mathbf{n}_t=\langle 1,t-1,-t\rangle\). 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 \(s:\mathbf{r}(\lambda)=\lambda\langle 1,1,1\rangle\). 5. Let \(\mathbf{n}^*=\langle a,b,c\rangle\). 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 \(\mathbf{n}^*=\langle 1,-1,0\rangle\). 7. Since the plane passes through the origin, \(E^*:x-y=0\). 8. 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 \(s:\mathbf{r}(\lambda)=\lambda\langle 1,1,1\rangle\). b) \(E^*:x-y=0\) c) \(E^*=F_0\), so \(t=0\).

- 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, \(\overrightarrow{OF}=5\mathbf{u}=\langle3,4,0\rangle\), so \(F=(3,4,0)\). 4. Since \(F\) is the midpoint of \(\overline{QQ'}\), \(\overrightarrow{OQ'}=2\overrightarrow{OF}-\overrightarrow{OQ}=2\langle3,4,0\rangle-\langle7,1,5\rangle=\langle-1,7,-5\rangle\).

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

- 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 \(M=S+\frac14(C-S)\), \(M=\langle2,2,8\rangle+\frac14\langle2,2,-8\rangle=\left(\frac52,\frac52,6\right)\). 3. Directions in \(E\) are \(\overrightarrow{KL}=\langle2,0,0\rangle\) and \(\overrightarrow{KM}=\left\langle\frac32,\frac32,2\right\rangle\). 4. Let \(\mathbf{n}=\langle a,b,c\rangle\). The equations \(\mathbf{n}\cdot\overrightarrow{KL}=0\) and \(\mathbf{n}\cdot\overrightarrow{KM}=0\) give \(a=0\) and \(\frac32b+2c=0\). Choose \(\mathbf{n}=\langle0,-4,3\rangle\). 5. Using point \(K\), the plane equation is \(-4y+3z=8\). 6. Parametrize segment \(\overline{DS}\): \(\mathbf{r}(t)=\langle0,4,0\rangle+t\langle2,-2,8\rangle\), 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 \(N=\langle0,4,0\rangle+\frac34\langle2,-2,8\rangle=\left(\frac32,\frac52,6\right)\).

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

- What makes a point on the line the closest point to \(P\)? - The two requested methods should produce the same point. What could you compare to check that? - How can you tell whether a vector points along the line or straight toward it?

1. For the orthogonality method, set \(F(t)=(1+2t,2-t,-1+2t)\). Then \(\overrightarrow{PF}=\langle2t-2,4-t,2t-5\rangle\). 2. Requiring \(\overrightarrow{PF}\cdot\langle2,-1,2\rangle=0\) gives \(2(2t-2)-(4-t)+2(2t-5)=9t-18=0\). Thus, \(t=2\) and \(F=(5,0,3)\). 3. For projection, choose \(A=(1,2,-1)\) on \(g\). Then \(\overrightarrow{AP}=\langle2,-4,5\rangle\) and \(\operatorname{proj}_{\mathbf{v}}\overrightarrow{AP}=\frac{\overrightarrow{AP}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\mathbf{v}=\frac{18}{9}\langle2,-1,2\rangle=\langle4,-2,4\rangle\). 4. Therefore, \(F=A+\operatorname{proj}_{\mathbf{v}}\overrightarrow{AP}=(5,0,3)\), confirming the first result. 5. Finally, \(\overrightarrow{PF}=\langle2,2,-1\rangle\), so \(d(P,g)=\sqrt{2^2+2^2+(-1)^2}=3\).

Orthogonality: \(\overrightarrow{PF(t)}\cdot\langle2,-1,2\rangle=9t-18=0\) gives \(t=2\) and \(F=(5,0,3)\). Projection: \(\frac{\langle2,-4,5\rangle\cdot\langle2,-1,2\rangle}{\langle2,-1,2\rangle\cdot\langle2,-1,2\rangle}\langle2,-1,2\rangle=\langle4,-2,4\rangle\). Adding this to \((1,2,-1)\) also gives \(F=(5,0,3)\). Distance: \(d(P,g)=\|\langle2,2,-1\rangle\|=3\).

- 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. Rewrite the family equation as \((x+2z-4)+t(y-z-2)=0\). 2. A point lies in every plane exactly when \(x+2z-4=0\) and \(y-z-2=0\). 3. Setting \(z=\lambda\) gives \(x=4-2\lambda\) and \(y=2+\lambda\). Thus, \(g:\mathbf{r}(\lambda)=\langle4,2,0\rangle+\lambda\langle-2,1,1\rangle\). 4. At the point on \(g\) closest to the origin, the position vector \(\langle4-2\lambda,2+\lambda,\lambda\rangle\) is perpendicular to the line direction \(\langle-2,1,1\rangle\). 5. Their dot product is \(-8+4\lambda+2+\lambda+\lambda=6\lambda-6\). Setting it equal to zero gives \(\lambda=1\) and \(P=(2,3,1)\).

a) Since \((x+2z-4)+t(y-z-2)=0\) must hold for every \(t\), the common points satisfy \(x+2z=4\) and \(y-z=2\). They form the line \(g:\mathbf{r}(\lambda)=\langle4,2,0\rangle+\lambda\langle-2,1,1\rangle\). b) The closest point satisfies \(\langle4-2\lambda,2+\lambda,\lambda\rangle\cdot\langle-2,1,1\rangle=0\), giving \(\lambda=1\). Thus, \(P=(2,3,1)\).

- Which point on the line is closest to \(A\), and how could you check it? - How does reflection change the position of a point relative to the mirror line? - If a sphere just touches a line, what is special about its center-to-line distance?

1. A general point on \(g\) is \(L(t)=(t,1+2t,1)\), so \(\overrightarrow{AL}=\langle t-3,2t-1,-5\rangle\). 2. Perpendicularity gives \(\overrightarrow{AL}\cdot\langle1,2,0\rangle=(t-3)+2(2t-1)=5t-5=0\). Thus, \(t=1\) and \(L=(1,3,1)\). 3. The distance is \(AL=\sqrt{(-2)^2+1^2+(-5)^2}=\sqrt{30}\). 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) \(\sqrt{30}\) units c) \(A''=(-1,4,-4)\) d) \((x-3)^2+(y-2)^2+(z-6)^2=30\)

- 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. The line direction is \(\overrightarrow{PQ}=\langle2,4,0\rangle\), and a general point is \(L(t)=(1+2t,1+4t,2)\). 2. The displacement from \(R\) to \(L(t)\) is \(\overrightarrow{RL}=\langle2t-5,4t,-3\rangle\). Perpendicularity requires \(2(2t-5)+4(4t)=20t-10=0\), giving \(t=\frac{1}{2}\). 3. The 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. The parallelogram's area is \(PQ\cdot RL=2\sqrt{5}\cdot\sqrt{29}=2\sqrt{145}\).

a) \(\sqrt{29}\) units b) \(2\sqrt{145}\) 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. \(\overrightarrow{AB}=\langle2,0,-2\rangle\) and \(\overrightarrow{AC}=\langle0,3,0\rangle\). Their dot product is zero, so the sides are perpendicular. 2. The side lengths are \(AB=2\sqrt{2}\) and \(AC=3\), so the area is \(\frac{1}{2}\cdot2\sqrt{2}\cdot3=3\sqrt{2}\). 3. A base normal is \(\mathbf{n}=\langle1,0,1\rangle\), and \(\overrightarrow{AS}=\langle3,0,3\rangle\). The scalar-projection method gives \(h=\frac{|\overrightarrow{AS}\cdot\mathbf{n}|}{\|\mathbf{n}\|}=\frac{6}{\sqrt{2}}=3\sqrt{2}\). 4. The base plane is \(x+z=2\). The point-to-plane distance gives \(h=\frac{|4+4-2|}{\sqrt{1^2+1^2}}=3\sqrt{2}\), agreeing with the projection method.

a) \(3\sqrt{2}\) square units b) Projection onto \(\mathbf{n}=\langle1,0,1\rangle\) gives \(h=\frac{|\langle3,0,3\rangle\cdot\langle1,0,1\rangle|}{\sqrt{2}}=3\sqrt{2}\). The base plane \(x+z=2\) gives \(h=\frac{|4+4-2|}{\sqrt{2}}=3\sqrt{2}\) by the point-to-plane formula.

- What kind of vector describes the tilt of a plane independently of its shape? - How do the horizontal base and sloping faces differ in their orientations? - How can you check that the angle you report is the acute angle between the requested faces?

1. A base normal is \(\mathbf{n}_G=\langle0,0,1\rangle\). For face \(BCS\), the vectors \(\overrightarrow{BC}=\langle-24,0,0\rangle\) and \(\overrightarrow{BS}=\langle-12,-12,16\rangle\) have a common normal \(\mathbf{n}_1=\langle0,4,3\rangle\). 2. The side-face angle with the base satisfies \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_G|}{\|\mathbf{n}_1\|\|\mathbf{n}_G\|}=\frac{3}{5}\), so \(\alpha\approx53.1^\circ\). 3. For face \(ABS\), \(\overrightarrow{AB}=\langle0,24,0\rangle\) and \(\overrightarrow{AS}=\langle-12,12,16\rangle\) have normal \(\mathbf{n}_2=\langle4,0,3\rangle\). 4. For the two side faces, \(\cos\beta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac{9}{25}\). 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 \(\mathbf{n}_G=\langle0,0,1\rangle\). For face \(BCS\), use \(\overrightarrow{BC}=\langle-20,0,0\rangle\) and \(\overrightarrow{BS}=\langle-10,-5,12\rangle\). If \(\mathbf{n}_1=\langle a,b,c\rangle\) is perpendicular to both, the dot-product equations give \(a=0\) and \(-5b+12c=0\). Choose \(\mathbf{n}_1=\langle0,12,5\rangle\). 2. For part a, \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_G|}{\|\mathbf{n}_1\|\|\mathbf{n}_G\|}=\frac5{13}\), so \(\alpha\approx67.4^\circ\). 3. For face \(ABS\), use \(\overrightarrow{AB}=\langle0,10,0\rangle\) and \(\overrightarrow{AS}=\langle-10,5,12\rangle\). Requiring \(\mathbf{n}_2=\langle a,b,c\rangle\) to be perpendicular to both gives \(b=0\) and \(-10a+12c=0\). Choose \(\mathbf{n}_2=\langle6,0,5\rangle\). 4. For part b, \(\cos\beta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac{25}{13\sqrt{61}}\), so \(\beta\approx75.7^\circ\).

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

- 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 \(\overrightarrow{PF}=\langle2t-5,t,2t-4\rangle\). 2. Perpendicularity requires \(\overrightarrow{PF}\cdot\langle2,1,2\rangle=2(2t-5)+t+2(2t-4)=9t-18=0\), so \(t=2\) and \(F=(5,4,6)\). 3. The leg lengths are \(AF=\sqrt{4^2+2^2+4^2}=6\) and \(PF=\sqrt{(-1)^2+2^2}=\sqrt{5}\). Thus, the area is \(\frac{1}{2}\cdot6\cdot\sqrt{5}=3\sqrt{5}\). 4. The cone has radius \(\sqrt{5}\) and height \(6\), so \(V=\frac{1}{3}\pi(5)(6)=10\pi\).

a) \(F=(5,4,6)\) b) \(3\sqrt{5}\) square units c) \(10\pi\) 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 \(\overrightarrow{QF}=\langle0,4s-9,3s-13\rangle\). 2. Orthogonality gives \(\overrightarrow{QF}\cdot\langle0,4,3\rangle=4(4s-9)+3(3s-13)=25s-75=0\), so \(s=3\). 3. Therefore, \(F=(4,11,11)\) and \(QF=\sqrt{3^2+(-4)^2}=5\). 4. The cone has radius \(5\) and height \(BF=\sqrt{12^2+9^2}=15\). 5. Thus, \(V=\frac{1}{3}\pi\cdot5^2\cdot15=125\pi\).

a) \(F=(4,11,11)\) b) \(5\) units c) \(125\pi\) 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 \(\mathbf{n}_E=\langle-1,1,0\rangle\). A normal vector to \(F_t\) is \(\mathbf{n}_t=\langle t-2,-(t-2),2\rangle\). 2. Therefore, \(\cos\alpha=\frac{|\mathbf{n}_E\cdot\mathbf{n}_t|}{\|\mathbf{n}_E\|\|\mathbf{n}_t\|}=\frac{|t-2|}{\sqrt{t^2-4t+6}}\). 3. For \(\alpha=30^\circ\), squaring \(\frac{|t-2|}{\sqrt{(t-2)^2+2}}=\frac{\sqrt3}{2}\) gives \((t-2)^2=6\). 4. Since \(t>2\), \(t=2+\sqrt6\).

a) \(\cos\alpha=\frac{|t-2|}{\sqrt{t^2-4t+6}}\) b) \(t=2+\sqrt6\)

- 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 \(k=4\), \(\overrightarrow{AS_4}=\langle-2,-2,4\rangle\) and a base normal is \(\langle0,0,1\rangle\). Thus, \(\sin\alpha=\frac{4}{\sqrt{24}}=\frac{2}{\sqrt{6}}\), so \(\alpha\approx54.7^\circ\). 2. Face \(ABS_k\) has directions \(\langle-4,0,0\rangle\) and \(\langle-2,-2,k\rangle\), with normal \(\mathbf{n}_1=\langle0,k,2\rangle\). Face \(BCS_k\) has directions \(\langle0,-4,0\rangle\) and \(\langle2,-2,k\rangle\), with normal \(\mathbf{n}_2=\langle-k,0,2\rangle\). 3. Therefore, \(\cos\phi=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac{4}{k^2+4}\). Setting \(\phi=60^\circ\) gives \(\frac{1}{2}=\frac{4}{k^2+4}\), hence \(k=2\). 4. As \(k\to\infty\), \(\frac{4}{k^2+4}\to0\), so \(\phi\to90^\circ\). For each finite \(k>0\), the cosine is positive, so \(\phi<90^\circ\).

a) \(54.7^\circ\) b) \(k=2\) c) \(\cos\phi=\frac{4}{k^2+4}\to0\) as \(k\to\infty\), 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 \(\cos^2 45^\circ\). - Use that the requested angle is acute.

1. For \(a=4\), normal vectors to \(OPR_a\) and \(OQR_a\) may be simplified to \(\mathbf{n}_1=\langle0,-1,1\rangle\) and \(\mathbf{n}_2=\langle1,0,-1\rangle\). Thus \(\cos\gamma=\frac12\), so \(\gamma=60^\circ\). 2. For part b, use \(\mathbf{n}_{OPR}=\langle0,-a,4\rangle\) and \(\mathbf{n}_{PQR}=\langle a,a,-4\rangle\). 3. The angle satisfies \(\cos^2\phi=\frac{a^2+16}{2a^2+16}=\frac12+\frac8{2a^2+16}\). 4. Since \(a>0\), the second term is positive, so \(\cos^2\phi>\frac12\). Because \(\phi\) is acute, \(\cos\phi>\frac1{\sqrt2}=\cos45^\circ\), and therefore \(\phi<45^\circ\).

a) \(60^\circ\) b) Since \(\cos^2\phi=\frac12+\frac8{2a^2+16}>\frac12\), it follows that \(\phi<45^\circ\) for every \(a>0\).

- 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 \(0\). - After measuring the two distances, what do you notice about them?

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 \(W_1:x+4y-3z-9=0\); subtracting gives \(W_2:3x+z-3=0\). 3. The normals \(\langle1,4,-3\rangle\) and \(\langle3,0,1\rangle\) have dot product \(3-3=0\), proving perpendicularity. 4. Substituting \(P\) into \(W_2\) gives \(3-3=0\), so \(P\in W_2\). 5. The distances are \(d(P,E_1)=\frac{|-3-6|}{3}=3\) and \(d(P,E_2)=\frac{|2(3)+3|}{3}=3\). They are equal, as expected 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) \(\langle1,4,-3\rangle\cdot\langle3,0,1\rangle=3-3=0\), so \(W_1\perp W_2\). d) Substitution gives \(3(0)+3-3=0\), so \(P\in W_2\). Also, \(d(P,E_1)=\frac{|-3-6|}{3}=3\) and \(d(P,E_2)=\frac{|6+3|}{3}=3\). Thus, \(P\) is equally distant from the original planes.

- 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. Plane normals are \(\mathbf{n}_1=\langle4,-4,2\rangle\) and \(\mathbf{n}_2=\langle4,7,4\rangle\), with magnitudes \(6\) and \(9\). 2. The angle satisfies \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac{4}{54}=\frac{2}{27}\), so \(\alpha\approx85.75^\circ\). 3. Normalized plane equations are \(\frac{2}{3}x-\frac{2}{3}y+\frac{1}{3}z-2=0\) and \(\frac{4}{9}x+\frac{7}{9}y+\frac{4}{9}z-1=0\). Adding and subtracting gives \(W_1:10x+y+7z-27=0\) and \(W_2:2x-13y-z-9=0\). 4. The point \(P=(0,-1,4)\) lies on both planes, and \(\mathbf{d}=\langle15,4,-22\rangle\) is perpendicular to both plane normals. Thus, \(g:(x,y,z)=(0,-1,4)+t\langle15,4,-22\rangle\). 5. Point \(P\) satisfies \(W_1\), and \(\mathbf{d}\cdot\langle10,1,7\rangle=150+4-154=0\), proving \(g\) 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) \(g:(x,y,z)=(0,-1,4)+t\langle15,4,-22\rangle\). The starting point gives \(0-1+28-27=0\) in \(W_1\), and the direction satisfies \(\langle15,4,-22\rangle\cdot\langle10,1,7\rangle=150+4-154=0\). Therefore, the whole line lies in \(W_1\).

- 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 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 \(A\), \(\overrightarrow{AO}\cdot\overrightarrow{AB}=\langle-1,-2,-2\rangle\cdot\langle k-1,-2,-2\rangle=9-k\). Thus, \(k=9\). 4. At \(B\), \(\overrightarrow{BO}\cdot\overrightarrow{BA}=\langle-k,0,0\rangle\cdot\langle1-k,2,2\rangle=k^2-k\). 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. \(\overrightarrow{DE}=\langle2,3,2\rangle\), \(\overrightarrow{DF}=\langle-2,2,3\rangle\), and \(\overrightarrow{EF}=\langle-4,-1,1\rangle\). 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

- For part a, subtract the two dot-product expressions before deciding whether equality of vectors follows. - For part b, expand the squared magnitude with the dot product and compare terms. - For part c, connect the dot product of unit vectors to the cosine of the angle between them.

1. Statement a is false. The equation implies \(\mathbf{a}\cdot(\mathbf{b}-\mathbf{c})=0\), so \(\mathbf{b}-\mathbf{c}\) may be any vector perpendicular to \(\mathbf{a}\). For example, \(\mathbf{a}=\langle1,0,0\rangle\), \(\mathbf{b}=\langle0,1,0\rangle\), and \(\mathbf{c}=\langle0,0,0\rangle\) satisfy the equation but \(\mathbf{b}\ne\mathbf{c}\). 2. Statement b is true. Expanding gives \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+2\mathbf{u}\cdot\mathbf{v}+\|\mathbf{v}\|^2\). This equals \(\|\mathbf{u}\|^2+\|\mathbf{v}\|^2\) exactly when \(\mathbf{u}\cdot\mathbf{v}=0\). 3. Statement c is true. For unit vectors, \(\mathbf{e}_1\cdot\mathbf{e}_2=\|\mathbf{e}_1\|\|\mathbf{e}_2\|\cos\theta=\cos\theta\), which lies in \([-1,1]\).

a) False. For example, \(\mathbf{a}=\langle1,0,0\rangle\), \(\mathbf{b}=\langle0,1,0\rangle\), and \(\mathbf{c}=\mathbf{0}\) give equal dot products but \(\mathbf{b}\ne\mathbf{c}\). b) True. Expanding adds the term \(2\mathbf{u}\cdot\mathbf{v}\), which vanishes exactly when the dot product is zero. c) True. Unit vectors have dot product \(\cos\theta\), which is between \(-1\) and \(1\).

- Find \(G\) by adding the prism's height to \(C\)'s \(z\)-coordinate. - Use coordinate-axis unit vectors as normals to the coordinate planes. - Relate a line-plane angle to the angle between the line direction and a plane normal.

1. Since \(G\) is directly above \(C\), \(G=(0,5,16)\). 2. Therefore, \(\overrightarrow{AG}=\langle-12,5,16\rangle\). 3. Its magnitude is \(\|\overrightarrow{AG}\|=\sqrt{144+25+256}=5\sqrt{17}\). 4. With normals \(\mathbf{n}_{xy}=\langle0,0,1\rangle\), \(\mathbf{n}_{yz}=\langle1,0,0\rangle\), and \(\mathbf{n}_{xz}=\langle0,1,0\rangle\), the dot products are \(16\), \(-12\), and \(5\), respectively. 5. For the \(xy\)-plane, \(\sin\alpha_{xy}=\frac{16}{\sqrt{425}}\), so \(\alpha_{xy}\approx50.91^\circ\). 6. For the \(yz\)-plane, \(\sin\alpha_{yz}=\frac{12}{\sqrt{425}}\), so \(\alpha_{yz}\approx35.60^\circ\). 7. For the \(xz\)-plane, \(\sin\alpha_{xz}=\frac{5}{\sqrt{425}}\), so \(\alpha_{xz}\approx14.04^\circ\).

a) \(\overrightarrow{AG}=\langle-12,5,16\rangle\), and \(\|\overrightarrow{AG}\|=5\sqrt{17}\) units. b) With \(\mathbf{n}_{xy}=\langle0,0,1\rangle\), \(\mathbf{n}_{yz}=\langle1,0,0\rangle\), and \(\mathbf{n}_{xz}=\langle0,1,0\rangle\), the dot products in that order are \(16\), \(-12\), and \(5\). The corresponding acute angles are \(50.91^\circ\), \(35.60^\circ\), and \(14.04^\circ\), respectively.

- Form vectors from the base vertices to the apex. - Use the base plane's normal direction for the line-plane angle. - Use the dot-product angle formula for the angle between two lateral edges.

1. The edge vector is \(\overrightarrow{AS}=\langle-4,4,12\rangle\). 2. Its magnitude is \(\|\overrightarrow{AS}\|=\sqrt{16+16+144}=4\sqrt{11}\). 3. The base is the \(xy\)-plane, with unit normal \(\mathbf{n}=\langle0,0,1\rangle\). The required dot product is \(\overrightarrow{AS}\cdot\mathbf{n}=12\). 4. If \(\beta\) is the line-plane angle, \(\sin\beta=\frac{12}{4\sqrt{11}}=\frac{3}{\sqrt{11}}\), so \(\beta\approx64.76^\circ\). 5. For the lateral edges, \(\overrightarrow{SA}=\langle4,-4,-12\rangle\) and \(\overrightarrow{SB}=\langle4,4,-12\rangle\). Their dot product is \(144\), and each length is \(4\sqrt{11}\). 6. Thus \(\cos\gamma=\frac{144}{176}=\frac{9}{11}\), giving \(\gamma\approx35.10^\circ\).

a) \(AS=4\sqrt{11}\) units b) \(\overrightarrow{AS}\cdot\langle0,0,1\rangle=12\). The angle with the base plane is \(\beta\approx64.76^\circ\). c) \(\overrightarrow{SA}\cdot\overrightarrow{SB}=144\). The angle between the lateral edges is \(\gamma\approx35.10^\circ\).

- Express both diagonals using the two side vectors. - Compare their squared magnitudes so square roots do not obscure the dot-product terms. - For the geometric proof, compare triangles that contain the two diagonals and the same pair of side lengths.

1. The diagonal vectors are \(\overrightarrow{AC}=\mathbf{a}+\mathbf{b}\) and \(\overrightarrow{BD}=\mathbf{b}-\mathbf{a}\). 2. Equal diagonal lengths give \(\|\mathbf{a}+\mathbf{b}\|^2=\|\mathbf{b}-\mathbf{a}\|^2\). 3. Expanding yields \(\|\mathbf{a}\|^2+2\mathbf{a}\cdot\mathbf{b}+\|\mathbf{b}\|^2=\|\mathbf{b}\|^2-2\mathbf{a}\cdot\mathbf{b}+\|\mathbf{a}\|^2\). 4. Thus \(4\mathbf{a}\cdot\mathbf{b}=0\), 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) The squared diagonal lengths differ by \(\|\mathbf{a}+\mathbf{b}\|^2-\|\mathbf{b}-\mathbf{a}\|^2=4\mathbf{a}\cdot\mathbf{b}\). They are equal exactly when \(\mathbf{a}\cdot\mathbf{b}=0\), so the adjacent sides are perpendicular and the parallelogram is a rectangle. b) In a rectangle, its diagonals are equal by the Pythagorean theorem applied to right triangles with the same legs. Conversely, equal diagonals make triangles \(ABC\) and \(BAD\) congruent by SSS. The matching angles at \(B\) and \(A\) are therefore equal; consecutive angles of a parallelogram are supplementary, so both are right angles.

- Which two sides meet at the vertex used in the area calculation? - How could the dot product help you find an area without using coordinates of a height? - Check whether every parameter value you obtain gives a valid nondegenerate triangle.

1. The side vectors are \(\overrightarrow{AB}=\langle4,2\rangle\) and \(\overrightarrow{AC}=\langle x,5\rangle\). 2. Their squared magnitudes are \(20\) and \(x^2+25\), while \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4x+10\). 3. The square of twice the triangle's area is \(20(x^2+25)-(4x+10)^2=4(x-10)^2\). 4. Since the triangle's area is \(13\), \(\frac12\sqrt{4(x-10)^2}=13\), so \(|x-10|=13\). 5. Thus \(x=-3\) or \(x=23\). Because \(x>0\), the required value is \(x=23\).

\(\overrightarrow{AB}\cdot\overrightarrow{AC}=4x+10\). By the dot-product area identity, \(13=\frac12\sqrt{20(x^2+25)-(4x+10)^2}=|x-10|\), so \(x=-3\) or \(x=23\). The condition \(x>0\) gives \(x=23\).

- Represent different space diagonals by changing signs on the edge-length components. - Set the dot product of one chosen pair equal to zero. - For part b, compare the orthogonality equations required by different diagonal pairs.

1. Direction vectors for two space diagonals can be written as \(\mathbf{d}_1=\langle a,b,c\rangle\) and \(\mathbf{d}_2=\langle-a,b,c\rangle\). 2. They are perpendicular when \(\mathbf{d}_1\cdot\mathbf{d}_2=-a^2+b^2+c^2=0\). 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.

- Translate perpendicularity of the two diagonals into one dot-product equation. - Treat \(\mathbf{u}+\mathbf{v}\) as a single vector and look for a difference-of-squares pattern. - For part b, evaluate the magnitude of \(\mathbf{u}+\mathbf{v}\).

1. The diagonals are perpendicular exactly when \(\mathbf{d}_1\cdot\mathbf{d}_2=0\). 2. Let \(\mathbf{p}=\mathbf{u}+\mathbf{v}\). Then \((\mathbf{p}+\mathbf{w})\cdot(\mathbf{p}-\mathbf{w})=\|\mathbf{p}\|^2-\|\mathbf{w}\|^2\). 3. Therefore, \(\mathbf{d}_1\cdot\mathbf{d}_2=0\) if and only if \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{w}\|^2\). 4. Since magnitudes are nonnegative, this is equivalent to \(\|\mathbf{w}\|=\|\mathbf{u}+\mathbf{v}\|\). 5. For the given vectors, \(\mathbf{u}+\mathbf{v}=\langle4,2,0\rangle\), so \(\|\mathbf{w}\|=\sqrt{4^2+2^2}=2\sqrt5\).

a) \(\mathbf{d}_1\cdot\mathbf{d}_2=\|\mathbf{u}+\mathbf{v}\|^2-\|\mathbf{w}\|^2\); this is zero exactly when \(\|\mathbf{w}\|=\|\mathbf{u}+\mathbf{v}\|\). b) \(\|\mathbf{w}\|=2\sqrt5\)

- Which coordinate differences describe the shape's sides? - How could one vector calculation support more than one geometric classification? - What evidence would distinguish an angle calculation from an area calculation?

1. The side vectors have lengths \(\|\overrightarrow{AB}\|=\sqrt{29}\), \(\|\overrightarrow{BC}\|=\sqrt{22}\), \(\|\overrightarrow{CD}\|=\sqrt{29}\), and \(\|\overrightarrow{DA}\|=\sqrt{22}\). Also, \(\overrightarrow{AB}=-\overrightarrow{CD}\) and \(\overrightarrow{BC}=-\overrightarrow{DA}\), so the quadrilateral is a parallelogram. 2. The diagonals are \(\overrightarrow{AC}=\langle3,3,5\rangle\) and \(\overrightarrow{BD}=\langle-7,3,1\rangle\), with lengths \(\sqrt{43}\) and \(\sqrt{59}\), respectively. 3. At \(A\), the adjacent vectors are \(\overrightarrow{AB}=\langle5,0,2\rangle\) and \(\overrightarrow{AD}=\langle-2,3,3\rangle\). Their dot product is \(-4\), so \(\cos\alpha=\frac{-4}{\sqrt{29}\sqrt{22}}\). Therefore, \(\alpha\approx99.11^\circ\). 4. By the dot-product area identity, the squared area is \(\|\overrightarrow{AB}\|^2\|\overrightarrow{AD}\|^2-(\overrightarrow{AB}\cdot\overrightarrow{AD})^2=29\cdot22-(-4)^2=622\). The area is \(\sqrt{622}\) square units.

a) Side lengths: \(\sqrt{29},\sqrt{22},\sqrt{29},\sqrt{22}\); diagonal lengths: \(\sqrt{43}\) and \(\sqrt{59}\) b) \(\overrightarrow{AB}\cdot\overrightarrow{AD}=-4\); \(\alpha\approx99.11^\circ\) c) \(\sqrt{29\cdot22-(-4)^2}=\sqrt{622}\) square units

- How could you break the pentagon into shapes whose areas you can calculate? - Which vector pairs would describe those smaller shapes without overlap? - Check that your method counts the entire region exactly once.

1. Divide the pentagon along \(AC\) and \(AD\) into triangles \(ABC\), \(ACD\), and \(ADE\). 2. For \(ABC\), \(\overrightarrow{AB}=\langle2,0,2\rangle\) and \(\overrightarrow{AC}=\langle3,2,5\rangle\). Their dot product is \(16\), so the area is \(\frac12\sqrt{8\cdot38-16^2}=2\sqrt3\). 3. For \(ACD\), \(\overrightarrow{AC}=\langle3,2,5\rangle\) and \(\overrightarrow{AD}=\langle1,3,4\rangle\). Their dot product is \(29\), so the area is \(\frac12\sqrt{38\cdot26-29^2}=\frac{7\sqrt3}{2}\). 4. For \(ADE\), \(\overrightarrow{AD}=\langle1,3,4\rangle\) and \(\overrightarrow{AE}=\langle-1,1,0\rangle\). Their dot product is \(2\), so the area is \(\frac12\sqrt{26\cdot2-2^2}=2\sqrt3\). 5. Adding gives \(2\sqrt3+\frac{7\sqrt3}{2}+2\sqrt3=\frac{15\sqrt3}{2}\) square units.

\(ABC:\ \overrightarrow{AB}\cdot\overrightarrow{AC}=16\), area \(=2\sqrt3\). \(ACD:\ \overrightarrow{AC}\cdot\overrightarrow{AD}=29\), area \(=\frac{7\sqrt3}{2}\). \(ADE:\ \overrightarrow{AD}\cdot\overrightarrow{AE}=2\), area \(=2\sqrt3\). Total area: \(\frac{15\sqrt3}{2}\) square units.

- Use the given vector relationship to locate \(D\). - Find each side length from the magnitude of its connecting vector. - Use the dot-product angle formula at \(A\). - Divide the trapezoid into two triangles and use the dot-product area identity for each.

1. \(\overrightarrow{AB}=\langle6,2,-2\rangle\), so \(\overrightarrow{DC}=\frac12\overrightarrow{AB}=\langle3,1,-1\rangle\). Thus, \(D=C-\overrightarrow{DC}=(2,4,4)\). 2. The side lengths are \(AB=\sqrt{44}=2\sqrt{11}\), \(BC=\sqrt{17}\), \(CD=\sqrt{11}\), and \(DA=\sqrt{14}\). 3. At \(A\), \(\overrightarrow{AB}\cdot\overrightarrow{AD}=8\). Hence \(\cos\alpha=\frac8{\sqrt{44}\sqrt{14}}\), giving \(\alpha\approx71.20^\circ\). 4. Split along \(BD\). The dot-product area identity yields \(\text{Area}(\triangle ABD)=\sqrt{138}\) and \(\text{Area}(\triangle BCD)=\frac{\sqrt{138}}2\). 5. Total area is \(\frac{3\sqrt{138}}2\) square units.

a) \(D(2,4,4)\) b) \(AB=2\sqrt{11}\), \(BC=\sqrt{17}\), \(CD=\sqrt{11}\), \(DA=\sqrt{14}\) c) \(\overrightarrow{AB}\cdot\overrightarrow{AD}=8\); \(\alpha\approx71.20^\circ\); area \(=\sqrt{138}+\frac{\sqrt{138}}2=\frac{3\sqrt{138}}2\) square units

- Use equal opposite-side vectors to find \(D\). - In a parallelogram, equal adjacent sides make the figure a rhombus. - The diagonals of a parallelogram bisect each other. - Use a dot product to test whether the diagonal vectors are perpendicular. - A rhombus with perpendicular diagonals has area equal to half the product of the diagonal lengths.

1. Since \(\overrightarrow{AD}=\overrightarrow{BC}=\langle0,3,4\rangle\), \(D=(2,2,7)\). 2. \(\|\overrightarrow{AB}\|=5\) and \(\|\overrightarrow{BC}\|=5\), so the parallelogram is a rhombus. 3. Its diagonals bisect each other, giving \(M=\frac{A+C}{2}=(3.5,2.5,5)\). 4. The diagonals are \(\overrightarrow{AC}=\langle3,7,4\rangle\) and \(\overrightarrow{BD}=\langle-3,-1,4\rangle\), with dot product \(-9-7+16=0\). They therefore meet at \(90^\circ\), confirming that a rhombus has perpendicular diagonals. 5. Their lengths are \(\sqrt{74}\) and \(\sqrt{26}\). Thus, the area is \(\frac12\sqrt{74}\sqrt{26}=\sqrt{481}\) square units.

a) \(D(2,2,7)\); since \(AB=BC=5\), \(ABCD\) is a rhombus. b) \(M(3.5,2.5,5)\) c) \(\overrightarrow{AC}\cdot\overrightarrow{BD}=0\); the intersection angle is \(90^\circ\), confirming that a rhombus has perpendicular diagonals. d) \(\sqrt{481}\) square units

- Compare opposite-side vectors before imposing any condition on \(k\). - Use a dot product for the rectangle condition. - Compare adjacent squared side lengths for the rhombus condition, then combine both classifications for part d.

1. \(\overrightarrow{AB}=\overrightarrow{DC}=\langle2,2,1\rangle\), so one pair of opposite sides is equal and parallel, making \(ABCD\) a parallelogram for every \(k\). 2. \(\overrightarrow{BC}=\langle2,0,k-3\rangle\). The rectangle condition is \(\overrightarrow{AB}\cdot\overrightarrow{BC}=4+k-3=k+1=0\), giving \(k=-1\). 3. For a rhombus, set the adjacent squared side lengths equal: \(\|\overrightarrow{AB}\|^2=9\) and \(\|\overrightarrow{BC}\|^2=4+(k-3)^2\). 4. Thus, \((k-3)^2=5\), so \(k=3-\sqrt5\) or \(k=3+\sqrt5\). 5. A square must be both a rectangle and a rhombus. Since \(-1\) is neither rhombus value, no \(k\) makes a square.

a) \(\overrightarrow{AB}=\overrightarrow{DC}=\langle2,2,1\rangle\), so \(ABCD\) is a parallelogram. b) \(\overrightarrow{AB}\cdot\overrightarrow{BC}=k+1\), so \(k=-1\) c) \(k=3-\sqrt5\) or \(k=3+\sqrt5\) d) No value is both \(-1\) and \(3\pm\sqrt5\), so \(ABCD\) cannot be a square.

- Use the normal vector \(\mathbf{n}=\langle1,1,k\rangle\). - Compare the normal with each coordinate-axis direction using the line-plane angle relationship. - Equal axis angles require equal absolute contributions from the normal's three components.

1. The plane normal is \(\mathbf{n}=\langle1,1,k\rangle\). For the z-axis, \(\sin45^\circ=\frac{|k|}{\sqrt{k^2+2}}\), so \(k^2=2\) and the positive solution is \(k=\sqrt2\). 2. The sines of the acute angles with the three axes are \(\frac{1}{\|\mathbf{n}\|}\), \(\frac{1}{\|\mathbf{n}\|}\), and \(\frac{|k|}{\|\mathbf{n}\|}\). 3. They are equal exactly when \(|k|=1\), so \(k=1\) or \(k=-1\). 4. In either case, \(\|\mathbf{n}\|=\sqrt3\) and the common angle is \(\sin^{-1}\left(\frac1{\sqrt3}\right)\approx35.3^\circ\).

1. \(k=\sqrt2\) 2. \(k=1\) or \(k=-1\); common angle \(\approx35.3^\circ\)

- Find a normal vector and use it to write a coordinate equation of the plane. - Use the sine form of the dot-product formula for the line-plane angle. - Project one additional point of the line along the plane normal, then use that image point with the intersection point to define the projected line.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle 0,-1,1\rangle\), 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 \(\mathbf{v}=\langle 1,1,0\rangle\). Hence \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac{1}{2}\), so \(\alpha=30^\circ\). 4. Project the point \(A=(1, 0, 2)\) onto \(E\) along the normal. The perpendicular line is \(\mathbf{x}=\langle 1,0,2\rangle+t\langle 0,-1,1\rangle\). 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 \(\langle 2,1,1\rangle\), so one equation is \(h:\mathbf{x}=\langle 2,1,2\rangle+\sigma\langle 2,1,1\rangle\).

a) \(S=(2, 1, 2)\), and \(\alpha=30^\circ\) b) \(h:\mathbf{x}=\langle 2,1,2\rangle+\sigma\langle 2,1,1\rangle\)

- The intersection direction is perpendicular to both plane normals. - A line in \(E_2\) perpendicular to the intersection line must have direction perpendicular to both \(\mathbf{n}_2\) and the intersection direction. - Compare exact trigonometric expressions before rounding the two angles.

1. The normals are \(\mathbf{n}_1=\langle1,2,2\rangle\) and \(\mathbf{n}_2=\langle1,-2,2\rangle\), both with magnitude \(3\). Thus, \(\cos\gamma=\frac19\) and \(\gamma\approx83.6^\circ\). 2. The intersection direction \(\mathbf{u}=\langle a,b,c\rangle\) satisfies \(a+2b+2c=0\) and \(a-2b+2c=0\). Hence \(b=0\) and \(a=-2c\); choose \(\mathbf{u}=\langle2,0,-1\rangle\). 3. A direction \(\mathbf{v}\) for \(g\) must satisfy \(\mathbf{v}\cdot\mathbf{n}_2=0\) and \(\mathbf{v}\cdot\mathbf{u}=0\). One choice is \(\mathbf{v}=\langle2,5,4\rangle\). 4. The line-plane formula gives \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}_1|}{\|\mathbf{v}\|\|\mathbf{n}_1\|}=\frac{4\sqrt5}{9}\), so \(\alpha\approx83.6^\circ\). 5. Since \(\sin\gamma=\sqrt{1-\frac1{81}}=\frac{4\sqrt5}{9}=\sin\alpha\) and both angles are acute, \(\alpha=\gamma\).

a) \(\gamma=\cos^{-1}\left(\frac19\right)\approx83.6^\circ\) b) One direction is \(\mathbf{u}=\langle2,0,-1\rangle\). c) One direction is \(\mathbf{v}=\langle2,5,4\rangle\); \(\alpha\approx83.6^\circ\), and exactly \(\alpha=\gamma\).

- Substitute the line into the plane equation to locate the intersection. - Use the line direction and plane normal for the line-plane angle. - Of all directions in the plane, which could make the largest dot product with the line direction?

1. Point \(S\) satisfies \(2(1)-2(0)+1=3\). Substitution of \(g\) into \(E\) gives \(3+4t=3\), so \(t=0\) and \(S\) is the unique intersection. 2. With \(\mathbf{n}=\langle2,-2,1\rangle\) and \(\mathbf{v}=\langle2,1,2\rangle\), \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac49\), giving \(\alpha\approx26.4^\circ\). 3. The smallest angle occurs along the projection of \(\mathbf{v}\) onto \(E\): \(\mathbf{v}_E=\mathbf{v}-\frac{\mathbf{v}\cdot\mathbf{n}}{\|\mathbf{n}\|^2}\mathbf{n}=\frac19\langle10,17,14\rangle\). Thus one line is \(h:\mathbf{x}=\langle1,0,1\rangle+\lambda\langle10,17,14\rangle\), making angle \(\approx26.4^\circ\) with \(g\). 4. The direction \(\langle1,1,0\rangle\) is perpendicular to \(\mathbf{n}\), and \(S\in E\), so \(k\subset E\). Also, \(\cos\gamma=\frac{3}{3\sqrt2}=\frac1{\sqrt2}\), giving \(\gamma=45^\circ\).

1. \(S=(1,0,1)\) 2. \(\alpha\approx26.4^\circ\) 3. With \(\mathbf{v}=\langle2,1,2\rangle\) and \(\mathbf{n}=\langle2,-2,1\rangle\), \(\mathbf{v}_E=\mathbf{v}-\frac{\mathbf{v}\cdot\mathbf{n}}{\|\mathbf{n}\|^2}\mathbf{n}=\frac19\langle10,17,14\rangle\). Thus \(h:\mathbf{x}=\langle1,0,1\rangle+\lambda\langle10,17,14\rangle\), with minimum angle \(\approx26.4^\circ\). 4. Line \(k\) lies in \(E\), and its angle with \(g\) is \(45^\circ\).

- Find a vector perpendicular to both spanning vectors. - Use the plane normal to obtain the line-plane angle, then compare the line direction with each spanning direction. - Two comparison directions give examples. What extra argument would cover every possible direction in the plane?

1. With directions \(\mathbf{u}=\langle1,1,0\rangle\) and \(\mathbf{v}=\langle0,1,1\rangle\), one normal is \(\mathbf{n}=\langle1,-1,1\rangle\). The plane through \(A\) has equation \(x-y+z=1\). 2. \(\sin\beta=\frac{|\mathbf{w}\cdot\mathbf{n}|}{\|\mathbf{w}\|\|\mathbf{n}\|}=\frac2{\sqrt6}\), so \(\beta\approx54.7^\circ\). 3. \(\cos\gamma_1=\frac{|\mathbf{w}\cdot\mathbf{u}|}{\|\mathbf{w}\|\|\mathbf{u}\|}=\frac12\), so \(\gamma_1=60^\circ\). Similarly, \(\cos\gamma_2=\frac12\), giving \(\gamma_2=60^\circ\). 4. Here, \(\beta<\gamma_1\) and \(\beta<\gamma_2\), which are only two examples. The projection of \(\mathbf{w}\) into \(E\) is \(\mathbf{w}_E=\mathbf{w}-\frac{\mathbf{w}\cdot\mathbf{n}}{\|\mathbf{n}\|^2}\mathbf{n}=\frac{1}{3}\langle1,2,1\rangle\). 5. For any nonzero direction \(\mathbf{q}\) in \(E\), \(|\mathbf{w}\cdot\mathbf{q}|=|\mathbf{w}_E\cdot\mathbf{q}|\le\|\mathbf{w}_E\|\|\mathbf{q}\|\). Thus \(\cos\angle(\mathbf{w},\mathbf{q})\le\frac{\|\mathbf{w}_E\|}{\|\mathbf{w}\|}=\frac{1}{\sqrt3}\), so that angle is at least \(\beta\). Equality occurs along the projected direction.

1. \(E:x-y+z=1\) 2. \(\beta\approx54.7^\circ\) 3. \(\gamma_1=60^\circ\) and \(\gamma_2=60^\circ\) 4. \(\beta<60^\circ\). More generally, \(\mathbf{w}_E=\frac{1}{3}\langle1,2,1\rangle\) is the projection of \(\mathbf{w}\) into \(E\). For every nonzero \(\mathbf{q}\) parallel to \(E\), \(|\mathbf{w}\cdot\mathbf{q}|\le\|\mathbf{w}_E\|\|\mathbf{q}\|\). Hence \(\angle(\mathbf{w},\mathbf{q})\ge\beta\), with equality along \(\mathbf{w}_E\).

- What does the plane's parametric form tell you about directions that stay within it? - What would it mean for the line's direction to be parallel to the plane? - How can you verify that your angle is measured against the plane, not against its normal?

1. A normal to \(E\) must be perpendicular to \(\langle1,0,-1\rangle\) and \(\langle2,1,0\rangle\). One normal is \(\mathbf{n}=\langle1,-2,1\rangle\), giving \(x-2y+z=-2\) through \((1,2,1)\). 2. Substituting the line yields \((5+t)-2(4+2t)+(3+t)=-2\), so \(t=1\) and \(S=(6,6,4)\). 3. The line direction is \(\mathbf{v}=\langle1,2,1\rangle\). Thus, \(\sin\alpha=\frac{|\mathbf{n}\cdot\mathbf{v}|}{\|\mathbf{n}\|\|\mathbf{v}\|}=\frac13\), giving \(\alpha\approx19.5^\circ\).

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 net's plane has equation \(x+y+z=6\), with normal \(\mathbf{n}=\langle1,1,1\rangle\). 2. The insect flies along \(\mathbf{r}(t)=\langle1,1,2\rangle+t\langle1,2,1\rangle\), where \(t\ge0\). 3. Substitution gives \((1+t)+(1+2t)+(2+t)=6\), so \(t=\frac12\). 4. Therefore, \(S=\left(\frac32,2,\frac52\right)\). Its coordinates are nonnegative and sum to \(6\), so the forward intersection is inside the triangular net. 5. With flight direction \(\mathbf{v}=\langle1,2,1\rangle\), \(\sin\alpha=\frac{|\mathbf{n}\cdot\mathbf{v}|}{\|\mathbf{n}\|\|\mathbf{v}\|}=\frac{2\sqrt2}{3}\), so \(\alpha\approx70.5^\circ\).

a) \(S=\left(\frac32,2,\frac52\right)\). The ray parameter is \(t=\frac12\ge0\), and the coordinates of \(S\) are nonnegative and sum to \(6\). Therefore, \(S\) lies on the bounded triangular net. 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 \(\mathbf{v}_1=\langle 2,1,2\rangle\) and \(\mathbf{v}_2=\langle 2,2,1\rangle\). Then \(\cos\alpha=\frac{|\mathbf{v}_1\cdot\mathbf{v}_2|}{\|\mathbf{v}_1\|\|\mathbf{v}_2\|}=\frac{8}{9}\), so \(\alpha\approx27.3^\circ\). 3. First find a vector \(\mathbf{w}=\langle a,b,c\rangle\) perpendicular to both line directions. The conditions \(2a+b+2c=0\) and \(2a+2b+c=0\) give \(b=c\) and \(2a+3b=0\). Choose \(\mathbf{w}=\langle -3,2,2\rangle\). 4. For \(E_1\), choose a normal perpendicular to both \(\mathbf{v}_1\) and \(\mathbf{w}\). Solving those two dot-product equations gives \(\mathbf{n}_1=\langle -2,-10,7\rangle\). For \(E_2\), solving \(\mathbf{n}_2\cdot\mathbf{v}_2=0\) and \(\mathbf{n}_2\cdot\mathbf{w}=0\) gives \(\mathbf{n}_2=\langle 2,-7,10\rangle\). 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 \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac{8}{9}\), 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 \(E_2:2x-7y+10z=6\).

- What is easier to compare after the two planes are written in the same form? - What must be true about a point and direction that belong to the intersection line? - How could you use each original plane to check your proposed line?

1. A normal \(\mathbf{n}_2=\langle a,b,c\rangle\) to \(E_2\) satisfies \(a+c=0\) and \(b+c=0\). Choose \(\mathbf{n}_2=\langle-1,-1,1\rangle\); the plane equation is \(-x-y+z=-1\). 2. The point \((2,0,1)\) satisfies both planes. One direction perpendicular to \(\mathbf{n}_1=\langle2,-1,2\rangle\) and \(\mathbf{n}_2\) is \(\langle1,-4,-3\rangle\). Thus, \(k:\mathbf{r}(t)=\langle2,0,1\rangle+t\langle1,-4,-3\rangle\). 3. \(\cos\phi=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac1{3\sqrt3}\), giving \(\phi\approx78.9^\circ\).

a) \(k:\mathbf{r}(t)=\langle2,0,1\rangle+t\langle1,-4,-3\rangle\) 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. Directions in \(E_1\) are \(\overrightarrow{AB}=\langle -2,2,0\rangle\) and \(\overrightarrow{AC}=\langle 2,4,-3\rangle\). Solving \(\mathbf{n}_1\cdot\overrightarrow{AB}=0\) and \(\mathbf{n}_1\cdot\overrightarrow{AC}=0\) gives \(\mathbf{n}_1=\langle 1,1,2\rangle\). 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 \(\mathbf{n}_2=\langle 1,1,-1\rangle\). 4. Using \((1, 1, 1)\), \(E_2:x+y-z=1\). 5. The normals satisfy \(\mathbf{n}_1\cdot\mathbf{n}_2=1+1-2=0\), so the planes are perpendicular. 6. Let \(\mathbf{n}_3=\langle a,b,c\rangle\). For \(E_3\) to be perpendicular to both planes, require \(\mathbf{n}_3\cdot\mathbf{n}_1=0\) and \(\mathbf{n}_3\cdot\mathbf{n}_2=0\). 7. The equations are \(a+b+2c=0\) and \(a+b-c=0\). Subtraction gives \(3c=0\), so \(c=0\) and \(a=-b\). Choose \(\mathbf{n}_3=\langle 1,-1,0\rangle\). 8. A point-normal equation is \((\mathbf{r}-\langle 5,0,0\rangle)\cdot\langle 1,-1,0\rangle=0\), 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) \((\mathbf{r}-\langle 5,0,0\rangle)\cdot\langle 1,-1,0\rangle=0\)

- What consistency check decides whether two parametric lines meet? - How can you recognize a direction perpendicular to both given lines without guessing its coordinates? - Does the parameter change a line's point, its direction, or both? - Check any claimed common plane against more than one member of the line family.

1. Equate the lines. Their z-coordinates give \(s=1\), and their x-coordinates then give \(r=4\). The y-coordinates require \(5-r=sk\), so \(k=1\). The intersection is \((3,1,4)\). 2. For \(h_0\), the z- and x-equations still require \(s=1\), \(r=4\), but the y-equation becomes \(1=0\). The directions \(\langle1,-1,0\rangle\) and \(\langle2,0,2\rangle\) are not parallel, so the nonintersecting lines are skew. 3. Let \(\mathbf{n}=\langle a,b,c\rangle\). The normal-to-both conditions are \(a-b=0\) and \(2a+2c=0\). Choose \(\mathbf{n}=\langle1,1,-1\rangle\). 4. The difference between the two starting points is \(\mathbf{d}=\langle2,-5,-2\rangle\). The skew-line distance is \(\frac{|\mathbf{d}\cdot\mathbf{n}|}{\|\mathbf{n}\|}=\frac1{\sqrt3}\). 5. Every \(h_k\) passes through \((1,0,2)\), and each direction \(\langle2,k,2\rangle=2\langle1,0,1\rangle+k\langle0,1,0\rangle\) lies in the span of two fixed directions. 6. A vector normal to both fixed directions satisfies \(a+c=0\) and \(b=0\); choose \(\mathbf{n}_E=\langle-1,0,1\rangle\). Its plane through \((1,0,2)\) has equation \(-x+z=1\), or \(x-z+1=0\).

a) \(k=1\); intersection \((3,1,4)\) b) For \(h_0\), the x- and z-equations give \(r=4\), \(s=1\), but the y-equation is \(1=0\). The lines are not parallel, so they are skew. A common perpendicular direction is \(\langle1,1,-1\rangle\); their distance is \(\frac1{\sqrt3}\). c) Every \(h_k\) lies in \(E:x-z+1=0\), with normal \(\langle-1,0,1\rangle\).

- 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. 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 \(\mathbf{n}_k=\langle a,b,c\rangle\) be a normal vector. The dot-product conditions with the two plane directions give one normal \(\mathbf{n}_k=\langle -1,k,2\rangle\). 3. A standard equation of \(E_k\) is \(-x+ky+2z=k+1\), or \(-x+2z-1+k(y-1)=0\). 4. 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 \(g:\mathbf{r}(t)=\langle 1,1,1\rangle+t\langle 2,0,1\rangle\). 6. For \(E_k\perp E_{-1}\), their normals must be orthogonal. Since \(\mathbf{n}_{-1}=\langle -1,-1,2\rangle\), \(\mathbf{n}_k\cdot\mathbf{n}_{-1}=1-k+4=0\). 7. Thus \(k=5\). 8. 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) \(g:\mathbf{r}(t)=\langle 1,1,1\rangle+t\langle 2,0,1\rangle\) c) \(k=5\) d) \(A\) lies in \(E_0\).

- What stays unchanged when the parameter in the plane family varies? - How can the normal vectors help with both the angle and perpendicularity questions? - Does the same plane direction always mean the planes are identical? - Check that any claimed parameter value meets every condition in the question.

1. Both \(F_0\) and \(F_1\) contain \((2,0,4)\) and direction \(\langle1,1,0\rangle\). Their other directions differ, so their intersection is \(h:\mathbf{r}(t)=\langle2,0,4\rangle+t\langle1,1,0\rangle\). 2. A normal \(\mathbf{n}_a=\langle p,q,r\rangle\) to \(F_a\) satisfies \(p+q=0\) and \(aq+r=0\), so one normal is \(\mathbf{n}_a=\langle1,-1,a\rangle\). 3. Thus, \(\mathbf{n}_0=\langle1,-1,0\rangle\) and \(\mathbf{n}_1=\langle1,-1,1\rangle\). Their acute angle has \(\cos\alpha=\frac2{\sqrt6}\), so \(\alpha\approx35.26^\circ\). 4. A normal to \(H\) is \(\mathbf{n}_H=\langle1,-1,2\rangle\). Perpendicularity requires \(\mathbf{n}_a\cdot\mathbf{n}_H=2+2a=0\), giving \(a=-1\). 5. The directions of \(E^*\) match those of \(F_2\), but \((2,2,4)\) is not in \(F_2\): matching its x- and z-coordinates requires \(r=s=0\), forcing y-coordinate \(0\). Thus \(E^*\) is not in the family.

a) \(h:\mathbf{r}(t)=\langle2,0,4\rangle+t\langle1,1,0\rangle\) b) \(\alpha\approx35.26^\circ\) 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 \(E:\mathbf{r}=\langle 1,2,1\rangle+r\langle 2,0,-1\rangle+t\langle 2,3,-1\rangle\). A normal vector is \(\langle 1,0,2\rangle\), so a coordinate equation is \(x+2z=3\). 3. \(\overrightarrow{PQ}=\langle 2,3,-1\rangle\), and \(\overrightarrow{PQ}\cdot\langle 2,0,-1\rangle=5\ne0\). Since the line direction \(\mathbf{u}=\langle 2,0,-1\rangle\) has \(\|\mathbf{u}\|^2=5\), the projection of \(\overrightarrow{PQ}\) onto \(\mathbf{u}\) is \(\frac{5}{5}\mathbf{u}=\mathbf{u}\). The perpendicular component is \(\overrightarrow{PQ}-\mathbf{u}=\langle 0,3,0\rangle\), 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 \(E:\mathbf{r}=\langle 1,2,1\rangle+r\langle 2,0,-1\rangle+t\langle 2,3,-1\rangle\), and a coordinate equation is \(x+2z=3\). c) \(\overrightarrow{PQ}\cdot\mathbf{u}=5\ne0\), 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. Substituting the parametric coordinates of \(E_2\) into \(E_1\) gives \((1+\lambda)+(-\lambda+\mu)=2\), so \(\mu=1\). 2. The intersection is \(g:\mathbf{r}(t)=\langle1,1,3\rangle+t\langle1,-1,1\rangle\). 3. A normal to \(E_1\) is \(\mathbf{n}_1=\langle1,1,0\rangle\). A normal \(\mathbf{n}_2\) to \(E_2\) satisfies \(a-b+c=0\) and \(b+2c=0\); choose \(\mathbf{n}_2=\langle-3,-2,1\rangle\). 4. \(\cos\alpha=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}=\frac5{\sqrt{28}}\), so \(\alpha\approx19.11^\circ\). 5. A normal to \(H_a\) is \(\mathbf{n}_{H_a}=\langle a,1,1-a\rangle\). The line direction \(\mathbf{v}=\langle1,-1,1\rangle\) satisfies \(\mathbf{v}\cdot\mathbf{n}_{H_a}=a-1+1-a=0\). 6. Testing the point \((1,1,3)\) gives \(a+1+3(1-a)=4-2a\), which equals \(4\) only if \(a=0\). Thus, \(g\subset H_0\); for \(a\ne0\) the line is strictly parallel to \(H_a\).

a) \(g:\mathbf{r}(t)=\langle1,1,3\rangle+t\langle1,-1,1\rangle\); \(\alpha\approx19.11^\circ\) b) \(g\subset H_0\). For every \(a\ne0\), \(g\) 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. The normal of \(E\) is \(\mathbf{n}_E=\langle 3,-1,2\rangle\), and the line direction is \(\mathbf{u}=\langle 1,1,-1\rangle\). Since \(\mathbf{u}\cdot\mathbf{n}_E=3-1-2=0\), the line is parallel to the plane. 2. The initial point is not in \(E\), because \(3(2)-2+2(2)=8\ne10\). Thus the line is strictly parallel to \(E\). 3. Let \(\mathbf{n}_F=\langle a,b,c\rangle\). Since \(F\) contains \(g\), require \(\mathbf{n}_F\cdot\mathbf{u}=0\). Since \(F\perp E\), require \(\mathbf{n}_F\cdot\mathbf{n}_E=0\). 4. Solving \(a+b-c=0\) and \(3a-b+2c=0\) gives one normal \(\mathbf{n}_F=\langle 1,-5,-4\rangle\). Using \((2, 2, 2)\), \(F:x-5y-4z=-16\). 5. A normal to \(H\) must be orthogonal to both \(\mathbf{n}_E\) and \(\mathbf{n}_F\). Solving the two dot-product equations gives \(\mathbf{n}_H=\langle 1,1,-1\rangle\). 6. Using \(A\), \(H:x+y-z=1\). 7. 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. Equating the line coordinates gives \(\lambda=-1\) and \(\mu=0\). Both lines pass through \(S=(-1, 1, 1)\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) 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 \(\mathbf{n}=\langle 1,0,2\rangle\). 4. Using \(S\), the plane equation is \(E:x+2z=1\). 5. 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 \(\ell:\mathbf{r}(u)=\langle 1,0,0\rangle+u\langle -2,2,1\rangle\). 9. 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 \(\ell:\mathbf{r}(u)=\langle 1,0,0\rangle+u\langle -2,2,1\rangle\). c) \(E^*:-2x+2y+z=3\)

- Express the two sides from \(C\) using \(\mathbf{u}\) and \(\mathbf{v}\). - Expand squared magnitudes with dot products. - Use a dot product for the angle at \(C\). - Compare a leg with \(AB=2\|\mathbf{u}\|\) in the equilateral case.

1. \(\overrightarrow{CA}=\mathbf{u}-\mathbf{v}\) and \(\overrightarrow{CB}=-\mathbf{u}-\mathbf{v}\). 2. Their squared lengths are \(\|\mathbf{u}-\mathbf{v}\|^2=\|\mathbf{u}\|^2-2\mathbf{u}\cdot\mathbf{v}+\|\mathbf{v}\|^2\) and \(\|-\mathbf{u}-\mathbf{v}\|^2=\|\mathbf{u}\|^2+2\mathbf{u}\cdot\mathbf{v}+\|\mathbf{v}\|^2\). If \(\mathbf{u}\cdot\mathbf{v}=0\), these are equal. 3. For a right angle at \(C\), \((\mathbf{u}-\mathbf{v})\cdot(-\mathbf{u}-\mathbf{v})=\|\mathbf{v}\|^2-\|\mathbf{u}\|^2=0\). Thus \(\|\mathbf{v}\|=\|\mathbf{u}\|\). Geometrically, \(A\), \(B\), and \(C\) lie on a sphere with center \(M\), and \(AB\) is a diameter. 4. For an equilateral triangle, \(\mathbf{u}\perp\mathbf{v}\) and \(\|\overrightarrow{CA}\|=\|\overrightarrow{AB}\|=2\|\mathbf{u}\|\). Thus \(\|\mathbf{u}\|^2+\|\mathbf{v}\|^2=4\|\mathbf{u}\|^2\), so \(\|\mathbf{v}\|=\sqrt{3}\|\mathbf{u}\|\).

a) Since \(\mathbf{u}\cdot\mathbf{v}=0\), \(CA^2=\|\mathbf{u}-\mathbf{v}\|^2=\|\mathbf{u}\|^2+\|\mathbf{v}\|^2=\|-\mathbf{u}-\mathbf{v}\|^2=CB^2\); hence \(CA=CB\). b) \(\|\mathbf{v}\|=\|\mathbf{u}\|\). Then \(A\), \(B\), and \(C\) lie on a sphere centered at \(M\), with \(AB\) as a diameter, so angle \(C\) is right. c) \(\|\mathbf{v}\|:\|\mathbf{u}\|=\sqrt{3}:1\).

- 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 normal vector to \(E\) is \(\mathbf{n}_E=\langle 1,-2,2\rangle\). The direction \(\mathbf{u}=\langle 2,1,0\rangle\) satisfies \(\mathbf{u}\cdot\mathbf{n}_E=2-2=0\). 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. When \(s=0\), the direction of \(g_s\) is \(\langle 2,1,0\rangle\). Therefore, \(g=g_0\). 5. Substitute a point of \(g_s\) into \(E\). The result is \(6+9s\lambda=9\). 6. If \(s=0\), there is no intersection. If \(s\ne0\), then \(\lambda=\frac{1}{3s}\), so \(Q_s=\left(\frac13+\frac{2}{3s},-\frac23+\frac{1}{3s},\frac{11}{3}\right)\). 7. 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 \(m:\mathbf{r}(\mu)=\left\langle\frac13,-\frac23,\frac{11}{3}\right\rangle+\mu\langle 2,1,0\rangle\). 9. Each family direction can be written as \(\langle 2+s,1-2s,2s\rangle=\mathbf{u}+s\mathbf{n}_E\). Since all lines also pass through \(A\), they lie in the plane through \(A\) with directions \(\mathbf{u}\) and \(\mathbf{n}_E\). 10. Let \(\mathbf{n}_F=\langle a,b,c\rangle\). The conditions \(\mathbf{n}_F\cdot\mathbf{u}=0\) and \(\mathbf{n}_F\cdot\mathbf{n}_E=0\) give one normal \(\mathbf{n}_F=\langle 2,-4,-5\rangle\). 11. Using \(A\), \(F:2x-4y-5z=-15\). 12. Since \(\mathbf{n}_E\cdot\mathbf{n}_F=2+8-10=0\), 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) \(s\ne0\), and \(Q_s=\left(\frac13+\frac{2}{3s},-\frac23+\frac{1}{3s},\frac{11}{3}\right)\) d) \(m:\mathbf{r}(\mu)=\left\langle\frac13,-\frac23,\frac{11}{3}\right\rangle+\mu\langle 2,1,0\rangle\) 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. Both lines pass through \(S=(1, 1, 2)\), so they intersect there. 2. Let \(\mathbf{n}_E=\langle a,b,c\rangle\). Orthogonality to the two line directions gives \(2a+b=0\) and \(b+2c=0\). 3. One normal is \(\mathbf{n}_E=\langle 1,-2,1\rangle\). Using \(S\), the plane equation is \(E:x-2y+z=1\). 4. A normal to \(E_k\) is \(\mathbf{n}_k=\langle 2,1+k,2k\rangle\). 5. Since \(\mathbf{n}_k\cdot\mathbf{n}_E=2-2(1+k)+2k=0\), 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 \(l:\mathbf{r}(t)=\langle 0,3,0\rangle+t\langle 1,-2,1\rangle\). 9. 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 \(\|\overrightarrow{OQ}\|=\sqrt3\).

a) \(S=(1, 1, 2)\), and \(E:x-2y+z=1\) b) Every \(E_k\perp E\), and \(l:\mathbf{r}(t)=\langle 0,3,0\rangle+t\langle 1,-2,1\rangle\) c) \(k_0=1\), and the maximum distance is \(\sqrt3\).