- 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.

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

1. A plane parallel to the \(xy\)-plane has a normal vector parallel to the \(z\)-axis, so use \(\mathbf{n}=\langle0,0,1\rangle\). 2. The plane is \(4\) units from the origin and lies on the positive side of the \(z\)-axis, so it passes through \(P(0, 0, 4)\). 3. Point-normal form is \(\langle0,0,1\rangle\cdot\langle x,y,z-4\rangle=0\). 4. The equivalent standard equation is \(z=4\).

Point-normal form: \(\langle0,0,1\rangle\cdot\langle x,y,z-4\rangle=0\) Standard form: \(z=4\)

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

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

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

- How do the components of a normal vector appear in a plane's standard equation? - A point lies in a plane when its coordinates satisfy the equation. - What happens to every variable term at the origin?

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

a) \(4x+7y-z=0\) b) \(d=0\), because substituting the origin into \(ax+by+cz=d\) gives \(0=d\).

- Represent three equal coordinates by one variable. - Substitute that coordinate pattern into each plane equation. - A contradiction means that no point satisfies both conditions.

1. a) Write a point with three equal coordinates as \(P(t, t, t)\). 2. Substitute into \(E\): \(5t-2t-t=8\). 3. Thus \(2t=8\), so \(t=4\) and \(P=(4, 4, 4)\). 4. b) Substitute \((t, t, t)\) into \(F_k\): \(kt+(2-k)t-2t=12\). 5. The left side simplifies to \(0\), giving the contradiction \(0=12\), regardless of \(k\) and \(t\). 6. Therefore, no plane in the family contains a point with three equal coordinates.

a) \(P(4, 4, 4)\) b) Substitution gives the contradiction \(0=12\) for every \(k\), so no such point exists.

- First determine which coordinates are fixed and which are free. - What coordinate condition places a point in a coordinate plane? - If two coordinates are fixed and one varies, what geometric object results? - Rewrite an absolute-value equation as two equations.

1. In part a), \(x\) and \(y\) are fixed while \(z\) can be any real number. The set is a line parallel to the \(z\)-axis through \((0, 4, 0)\). 2. In part b), \(x=0\) places every point in the \(yz\)-plane. The condition \(y=z\) gives the line through the origin with direction vector \(\begin{pmatrix}0\\1\\1\end{pmatrix}\). 3. In part c), \(|y|=5\) means \(y=5\) or \(y=-5\). Since \(x\) and \(z\) are unrestricted, the set consists of two planes parallel to the \(xz\)-plane, each \(5\) units from it.

a) A line parallel to the \(z\)-axis through \((0, 4, 0)\). b) The line in the \(yz\)-plane through the origin with direction vector \(\begin{pmatrix}0\\1\\1\end{pmatrix}\). c) The two planes \(y=5\) and \(y=-5\), both parallel to the \(xz\)-plane.

- Substitute a few simple values into \(x=y=z\). - If only one coordinate can vary, identify the axis direction parallel to that variation. - For part c), first use \(z=0\), then interpret \(x+y=4\) in two dimensions.

1. In part a), all three coordinates have the same value. Writing that value as \(t\) gives \((x, y, z)=(t, t, t)\), so the set is the line through the origin with direction vector \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). 2. In part b), \(x\) and \(z\) are fixed while \(y\) is free. The set is the line through \((-1, 0, 2)\) parallel to the \(y\)-axis. 3. In part c), \(z=0\) places the points in the \(xy\)-plane. Within that plane, \(x+y=4\) is the line through \((4, 0, 0)\) and \((0, 4, 0)\).

a) The line through the origin with direction vector \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). b) A line parallel to the \(y\)-axis through \((-1, 0, 2)\). c) The line in the \(xy\)-plane through \((4, 0, 0)\) and \((0, 4, 0)\).

- Identify which coordinates are fixed and which are free. - Decide whether two free coordinates produce a line or a plane. - What coordinate plane is parallel to a set with constant \(y\)?

1. The \(y\)-coordinate is fixed at \(5\), while \(x\) and \(z\) can vary independently. 2. Because two coordinates are free, the set is a plane. 3. The equation is \(y=5\), so the plane is parallel to the \(xz\)-plane. It intersects the \(y\)-axis at \((0, 5, 0)\) and is \(5\) units from the \(xz\)-plane.

The plane \(y=5\), which is parallel to the \(xz\)-plane and \(5\) units from it.

- What does the constant coordinate \(z=2\) tell you? - How are the \(x\)- and \(y\)-coordinates related? - Substitute \(k=0\), \(k=1\), and \(k=-1\) to identify the pattern.

1. Since \(z=2\) for every point, the set lies in the plane \(z=2\), which is parallel to the \(xy\)-plane. 2. The coordinates satisfy \(x=y=k\). Because there is one free parameter, the set is a line. 3. The line passes through \((0, 0, 2)\) and has direction vector \(\begin{pmatrix}1\\1\\0\end{pmatrix}\). It is parallel to the line \(y=x\) in the \(xy\)-plane.

A line in the plane \(z=2\) through \((0, 0, 2)\) with direction vector \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

- Determine which coordinates are fixed and which can vary. - One free coordinate produces a line; two free coordinates produce a plane. - A coordinate equal to \(0\) identifies a coordinate plane or axis. - Which axis contains points of the form \((x, 0, 0)\)?

1. One equation that fixes a single coordinate defines a plane parallel to the coordinate plane formed by the other two axes. 2. Two equations that fix two coordinates define a line parallel to the remaining coordinate axis. 3. Therefore: a) \(x=3.5\) is a plane parallel to the \(yz\)-plane, \(3.5\) units in the positive \(x\)-direction. b) \(y=0\) and \(z=0\) define the \(x\)-axis. c) \(x=-2\) and \(y=5\) define a line through \((-2, 5, 0)\) parallel to the \(z\)-axis. d) \(z=0\) is the \(xy\)-plane.

a) The plane \(x=3.5\), parallel to the \(yz\)-plane. b) The \(x\)-axis. c) A line through \((-2, 5, 0)\) parallel to the \(z\)-axis. d) The \(xy\)-plane.

- Which coordinate is \(0\) in each coordinate plane? - How many coordinates are \(0\) on a coordinate axis? - For a plane parallel to a coordinate plane, identify the coordinate that stays constant. - Use the coordinates of the given point to determine the fixed values.

1. A point lies in the \(xz\)-plane when its \(y\)-coordinate is \(0\). 2. A point lies on the \(z\)-axis when both its \(x\)- and \(y\)-coordinates are \(0\). 3. A plane parallel to the \(xy\)-plane has a constant \(z\)-coordinate. Passing through \(S\) makes that value \(-4\). 4. A line parallel to the \(y\)-axis has fixed \(x\)- and \(z\)-coordinates. Passing through \(A\) makes them \(3\) and \(1\).

a) \(y=0\) b) \(x=0\) and \(y=0\) c) \(z=-4\) d) \(x=3\) and \(z=1\)

- Find the magnitude of the direction vector. - The displacement from \(S\) is \(t\) times that vector. - Use both signs of the parameter.

1. The direction vector has magnitude \(\sqrt{(-4)^2+7^2+4^2}=9\). 2. A point with parameter \(t\) is \(9|t|\) units from \(S\). Thus, \(9|t|=27\), so \(t=3\) or \(t=-3\). 3. For \(t=3\), \(\mathbf{x}=\langle -11,25,10\rangle\). 4. For \(t=-3\), \(\mathbf{x}=\langle 13,-17,-14\rangle\).

\((-11,25,10)\) and \((13,-17,-14)\)

- 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.

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

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

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

- A beam striking a plane at a right angle points in a normal direction. - A displacement vector 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) \(2x+2y-z=4\)

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

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

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

- Compute \(\mathbf{n}\cdot\mathbf{p}\) 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, \(S\notin F\).

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

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

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

- Read a normal vector directly from the plane equation. - A line perpendicular to a plane uses a direction parallel to the plane's normal. - Combine that direction with the required point to write the parametric line.

1. A normal vector to plane \(E\) is \(\mathbf{n}=\langle4,-1,2\rangle\). 2. A line perpendicular to a plane has a direction vector parallel to the plane's normal vector. 3. Using \(P\) as the position point gives \(g:\mathbf{x}=\langle3,-1,4\rangle+r\langle4,-1,2\rangle\).

\(g:\mathbf{x}=\langle3,-1,4\rangle+r\langle4,-1,2\rangle\), where \(r\in\mathbb{R}\)

- 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 \(C\notin E\).

- Parametrize the line from its point and direction. - Use the line direction and plane normal to confirm a unique intersection. - Substitute the line coordinates into the plane equation and solve for the parameter.

1. Parametrize the line: \(\mathbf{r}(t)=\langle5,-2,1\rangle+t\langle2,1,-1\rangle\). 2. A normal vector to the plane is \(\mathbf{n}=\langle3,-1,2\rangle\). Since \(\mathbf{n}\cdot\mathbf{d}=6-1-2=3\ne0\), the line intersects the plane once. 3. Substitute the line coordinates into the plane equation: \(3(5+2t)-(-2+t)+2(1-t)=10\). 4. Simplifying gives \(19+3t=10\), so \(t=-3\). 5. The intersection point is \(S=(5,-2,1)-3(2,1,-1)=(-1,-5,4)\).

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

- Subtract the coordinates of \(L\) from those of \(M\) to get the ray direction. - Parametrize the ray with \(t\ge0\). - Substitute the ray coordinates into the plane equation and check that the resulting parameter lies on the ray.

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

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

- 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.

- Compare the coefficients of \(x\), \(y\), and \(z\). - Find the scale factor between each pair of normal vectors. - Apply the same scale factor to the constant term. - Decide whether the full equations are multiples or only the normal vectors are.

1. In part a, the normal vectors are \(\mathbf{n}_1=\langle4,-6,2\rangle\) and \(\mathbf{n}_2=\langle-2,3,-1\rangle\). Since \(\mathbf{n}_1=-2\mathbf{n}_2\), the planes are parallel or identical. 2. The constants have the same scale factor because \(10=-2(-5)\). Thus, the entire equation of \(E_1\) is \(-2\) times the equation of \(E_2\), so the planes are identical. 3. In part b, the normal vectors of \(E_2\) and \(E_3\) satisfy \(\mathbf{n}_2=-\mathbf{n}_3\), so these planes are parallel or identical. 4. Multiplying the equation of \(E_3\) by \(-1\) gives \(-2x+3y-z=-7\), not \(-5\). Therefore, \(E_2\) and \(E_3\) are distinct and strictly parallel. 5. Proportional normal vectors mean the planes have the same orientation. If the constants have the same proportionality factor, the equations describe the same plane; otherwise, they describe distinct parallel planes.

a) \(E_1\) and \(E_2\) are identical. b) \(E_2\) and \(E_3\) are strictly parallel. c) Proportional normals give parallel orientation; matching proportional constants give identical planes.

- Compare the normal vectors as scalar multiples. - Normalize both plane equations. - Equal left sides with equal constants describe the same plane; unequal constants describe strictly parallel planes.

1. The normal vectors are \(\mathbf{n}_1=\langle3,-6,9\rangle\) and \(\mathbf{n}_2=\langle-1,2,-3\rangle\). 2. Since \(\mathbf{n}_1=-3\mathbf{n}_2\), the planes are either identical or strictly parallel. 3. Divide the equation of \(E_1\) by \(3\): \(x-2y+3z=4\). 4. Multiply the equation of \(E_2\) by \(-1\): \(x-2y+3z=-k\). 5. The planes are identical when \(-k=4\), so \(k=-4\). 6. When \(k=0\), the normalized equations have the same left side but different right sides, \(4\) and \(0\), so the planes are strictly parallel.

a) \(k=-4\) b) For \(k=0\), the planes are strictly parallel.

- 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.

- Multiply one entire equation by a constant and compare it with the other. - Parallel planes have parallel normal vectors. - Substitute the origin into the general parallel-plane equation. - The constant controls the plane's offset, not its orientation.

1. Multiplying the equation of \(F\) by \(-2\) gives \(3x+4z=12\), which is exactly the equation of \(E\). Therefore the planes are identical. 2. A plane parallel to \(E\) can use the normal vector \(\langle3,0,4\rangle\), so write \(3x+4z=d\). 3. The origin gives \(d=0\), so \(G:3x+4z=0\). 4. The family of distinct parallel planes is \(3x+4z=d\), where \(d\in\mathbb{R}\) and \(d\ne12\). The excluded value describes \(E\) itself.

a) \(E\) and \(F\) are identical. b) \(G:3x+4z=0\) c) \(3x+4z=d\), where \(d\in\mathbb{R}\) and \(d\ne12\)

- Compare the normal vectors as scalar multiples. - Substitute the given point into the family equation. - Normalize the resulting plane equation.

1. The normal vectors are \(\mathbf{n}_E=\langle5,-2,1\rangle\) and \(\mathbf{n}_{F_t}=\langle10,-4,2\rangle\). 2. Since \(\mathbf{n}_{F_t}=2\mathbf{n}_E\), every \(F_t\) is parallel to or identical with \(E\). 3. Substitute \(P=(1, 1, 1)\) into \(F_t\): \(10-4+2=t\), so \(t=8\). 4. Dividing \(F_8:10x-4y+2z=8\) by \(2\) gives \(5x-2y+z=4\), exactly the equation of \(E\). 5. Therefore, \(F_8\) and \(E\) are identical.

a) \(\mathbf{n}_{F_t}=2\mathbf{n}_E\), so every \(F_t\) is parallel to or identical with \(E\). b) \(t=8\) c) \(F_8\) and \(E\) are identical.

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

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

a) Yes: \(4(1)-2(-1)+2=8\), so \(A\in E\). b) \(k=2\) c) \(4(3)-2(5)+1=3<8\); thus \(C\) is on the side opposite \(\mathbf{n}\).

- Form a direction vector from the two given points. - Points in the xy-plane have z-coordinate \(0\). - Use the distance formula in three dimensions. - Simplify the radical exactly.

1. A direction vector is \(\overrightarrow{AB}=\langle 3,3,-6\rangle\), so \(g:\mathbf{r}(t)=\langle 2,-3,4\rangle+t\langle 3,3,-6\rangle\). 2. A point in the xy-plane has \(z=0\). Thus, \(4-6t=0\), so \(t=\frac{2}{3}\). 3. Substitution gives \(S=\left(2+3\left(\frac{2}{3}\right),-3+3\left(\frac{2}{3}\right),0\right)=(4,-1,0)\). 4. The displacement from \(S\) to \(Q\) is \(\overrightarrow{SQ}=\langle -4,2,0\rangle\). 5. The distance is \(\sqrt{(-4)^2+2^2+0^2}=\sqrt{20}=2\sqrt{5}\).

a) \(S(4, -1, 0)\) b) \(2\sqrt{5}\)

- 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 \((1, 2, 0)\).

- Which coordinate represents altitude above the ground? - How is the parameter \(t\) related to elapsed time? - How do you find the magnitude of a vector in three dimensions? - For the distance from the starting point, should you use the position vector or the displacement vector?

1. The altitude is the \(z\)-coordinate. Set \(50+5t=300\). Then \(5t=250\), so \(t=50\). The balloon reaches the altitude after \(50\,\text{s}\). 2. At \(t=50\), the displacement from the starting point is \(50\langle 4,2,5\rangle=\langle 200,100,250\rangle\). 3. Its magnitude is \(\sqrt{200^2+100^2+250^2}=\sqrt{112500}=150\sqrt{5}\approx335.4\). Therefore, the balloon is about \(335.4\,\text{m}\) from its starting point.

a) \(50\,\text{s}\) b) \(150\sqrt{5}\,\text{m}\approx335.4\,\text{m}\)

- What must the altitude coordinate equal when the plane reaches the ground? - How can you find the position at a given time? - Which distance formula gives the distance from a point to the origin?

1. The plane reaches the ground when its \(z\)-coordinate is \(0\). Solve \(600-20t=0\), which gives \(t=30\). Therefore, the plane reaches the ground after \(30\,\text{s}\). 2. At \(t=30\), the position is \(\mathbf{x}(30)=\langle -2000+30\cdot80,1500+30(-60),600+30(-20)\rangle=\langle 400,-300,0\rangle\). 3. The distance from the origin is \(\sqrt{400^2+(-300)^2}=\sqrt{250000}=500\).

a) \(30\,\text{s}\) b) \(500\,\text{m}\)

- 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.

- 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.

- Set the two three-dimensional distance expressions equal. - The perpendicular-bisector plane passes through the midpoint of the segment. - A vector joining the two fixed points is normal to the plane. - To find an axis intercept, set the other two coordinates equal to \(0\).

1. Let \(X(x, y, z)\) be a point on \(E\). Equidistance gives \((x-6)^2+(y+2)^2+(z-4)^2=(x-2)^2+(y-4)^2+z^2\). 2. Expand and simplify: \(-8x+12y-8z+36=0\), which is equivalent to \(2x-3y+2z=9\). 3. The midpoint of \(RS\) is \(M(4, 1, 2)\), and a normal vector is \(\overrightarrow{RS}=\langle -4,6,-4\rangle\), or the parallel vector \(\langle -2,3,-2\rangle\). Thus a point-normal equation is \(\langle -2,3,-2\rangle\cdot\langle x-4,y-1,z-2\rangle=0\). 4. For the \(x\)-intercept, set \(y=z=0\): \(2x=9\), so \((4.5, 0, 0)\). 5. For the \(y\)-intercept, set \(x=z=0\): \(-3y=9\), so \((0, -3, 0)\). 6. For the \(z\)-intercept, set \(x=y=0\): \(2z=9\), so \((0, 0, 4.5)\).

a) \(\langle -2,3,-2\rangle\cdot\langle x-4,y-1,z-2\rangle=0\) b) \(2x-3y+2z=9\); intercepts \((4.5, 0, 0)\), \((0, -3, 0)\), and \((0, 0, 4.5)\)

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

1. A direction vector for \(g\) is \(\mathbf{v}=\overrightarrow{AB}=\langle2,2,-1\rangle\). 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, \(a=-4\), \(b=0\).

- 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.

- Which points are connected by the sunlight ray? - A shadow must lie in the sunlight's forward direction, not on the backward extension of its line. - What two ground positions determine the length of the shadow?

1. The top of the pole is \(Q(2,4,9)\). 2. The sunlight ray through \(Q\) is \(\mathbf{r}(t)=\langle 2,4,9\rangle+t\langle 1,0,-2\rangle\), for \(t\ge0\). 3. Substitute this ray into the plane equation: \((2+t)+2(4)+2(9-2t)=20\). 4. This gives \(28-3t=20\), so \(t=\frac{8}{3}>0\); the ray reaches the plane in the direction of travel. 5. The shadow of the pole's top is \(S=\left(\frac{14}{3},4,\frac{11}{3}\right)\). 6. Thus \(\overrightarrow{PS}=\left\langle\frac{8}{3},0,-\frac{4}{3}\right\rangle\), and the shadow length is \(\sqrt{\frac{64}{9}+\frac{16}{9}}=\frac{4\sqrt{5}}{3}\,\text{m}\).

\(\frac{4\sqrt{5}}{3}\,\text{m}\)

- 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.

- Use the symmetry and diagonals of the square to find the missing vertex. - All vertices of the base have the same \(z\)-coordinate. - Compare two side lengths of the triangle. - Find a plane normal by requiring zero dot products with two vectors in the plane. - Use \(V=\frac{1}{3}Bh\), where \(h\) is perpendicular to the base plane.

1. The midpoint of diagonal \(\overline{AC}\) is \((0,0,2)\). Reflecting \(B\) across this midpoint gives \(D=(0,-6,2)\). The base plane is \(E:z=2\), parallel to the \(xy\)-plane. 2. \(AS=\sqrt{(-6)^2+0^2+8^2}=10\), and \(BS=\sqrt{0^2+(-6)^2+8^2}=10\). Therefore, triangle \(ABS\) is isosceles. 3. Two vectors in \(F\) are \(\overrightarrow{AB}=\langle -6,6,0\rangle\) and \(\overrightarrow{AS}=\langle -6,0,8\rangle\). Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. Then \(-6a+6b=0\) and \(-6a+8c=0\). One solution is \(\mathbf{n}=\langle 4,4,3\rangle\). Using \(S\), the plane equation is \(F:4x+4y+3z=30\). 4. The square base has side length \(AB=\sqrt{72}\), so its area is \(72\). The perpendicular height is \(10-2=8\). Thus, \(V=\frac{1}{3}\cdot72\cdot8=192\) cubic units.

a) \(D=(0,-6,2)\), and \(E:z=2\), which is parallel to the \(xy\)-plane. b) \(AS=BS=10\), so triangle \(ABS\) is isosceles. c) \(F:4x+4y+3z=30\) d) \(V=192\) cubic units

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

1. A normal vector to \(E\) is \(\mathbf{n}=\langle4,-1,8\rangle\). 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 \(\left(\frac{d}{3},\frac{d}{3},\frac{d}{3}\right)\).

- 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 \(S\notin\overline{PQ}\).

- Use one known point as the position vector and two connecting vectors as spanning directions. - Find a normal vector that is perpendicular to both spanning directions. - Use the normal-vector components as the coefficients in the Cartesian equation.

1. Use \(A\) as a point and form the spanning vectors \(\overrightarrow{AB}=\langle2,0,-2\rangle\) and \(\overrightarrow{AC}=\langle0,3,-2\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-2\rangle+s\langle0,3,-2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a-2c=0\) and \(3b-2c=0\). One solution is \(\mathbf{n}=\langle3,2,3\rangle\). 3. Using point \(A\), the Cartesian equation is \(E:3x+2y+3z=16\).

a) \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-2\rangle+s\langle0,3,-2\rangle\) b) \(E:3x+2y+3z=16\)

- Form two independent connecting vectors from one sail vertex. - Find a normal vector perpendicular to both spanning directions. - Convert the point-normal equation to Cartesian form before testing the sensor point.

1. Two spanning vectors are \(\overrightarrow{PQ}=\langle4,0,-4\rangle\) and \(\overrightarrow{PR}=\langle0,4,-4\rangle\). Thus, \(E:\mathbf{x}=\langle2,2,6\rangle+r\langle4,0,-4\rangle+s\langle0,4,-4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(4a-4c=0\) and \(4b-4c=0\), so a simplified normal vector is \(\langle1,1,1\rangle\). A point-normal equation is \((\mathbf{x}-\langle2,2,6\rangle)\cdot\langle1,1,1\rangle=0\). 3. The corresponding Cartesian equation is \(x+y+z=10\). Since \(3+3+4=10\), the sensor point lies in the plane.

a) \(E:\mathbf{x}=\langle2,2,6\rangle+r\langle4,0,-4\rangle+s\langle0,4,-4\rangle\) b) \((\mathbf{x}-\langle2,2,6\rangle)\cdot\langle1,1,1\rangle=0\) c) Yes, \(S=(3,3,4)\) lies in \(E\).

- First compare the fixed coordinate on the line with the corresponding coordinate of \(P\). - Use the line direction and one vector from the line to \(P\) as plane-spanning directions. - Find a normal perpendicular to both spanning directions, then use one known point to determine the Cartesian constant.

1. Every point on \(g\) has \(y=2\), while \(P\) has \(y=4\). Therefore, \(P\notin g\). 2. Use the line direction \(\langle2,0,-1\rangle\) and the vector from the line's given point to \(P\), \(\langle4,2,-2\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-1\rangle+s\langle4,2,-2\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a-c=0\) and \(4a+2b-2c=0\). One solution is \(\mathbf{n}=\langle1,0,2\rangle\). Using \(P\), the Cartesian equation is \(E:x+2z=7\).

a) \(P\notin g\). b) \(E:\mathbf{x}=\langle1,2,3\rangle+r\langle2,0,-1\rangle+s\langle4,2,-2\rangle\) c) \(E:x+2z=7\)

- Form two independent direction vectors in the plane. - Find a normal vector perpendicular to both directions. - Write the plane equation and substitute the coordinates of \(S\).

1. Two directions in the plane are \(\overrightarrow{PQ}=\langle-1,2,2\rangle\) and \(\overrightarrow{PR}=\langle-2,-1,4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be a normal vector. The dot-product conditions are \(-a+2b+2c=0\) and \(-2a-b+4c=0\). 3. Solving gives one normal vector \(\mathbf{n}=\langle2,0,1\rangle\). 4. Using point \(P\), the plane equation is \(2x+z=4\). 5. Substitute \(S(k,k,2)\): \(2k+2=4\), so \(k=1\).

\(k=1\)

- Find a vector orthogonal to both 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.

- Find a normal vector by setting its dot product with each plane direction equal to zero. - A parallel plane may use the same normal direction. - Substitute the given point on \(H\) to determine the constant term.

1. Let \(\mathbf{n}=\langle a,b,c\rangle\). Since \(\mathbf{n}\) is orthogonal to both direction vectors, \(2a+c=0\) and \(a-b+3c=0\). 2. Choose \(a=1\). Then \(c=-2\), and the second equation gives \(1-b-6=0\), so \(b=-5\). Thus, \(\mathbf{n}=\langle1,-5,-2\rangle\). 3. A plane parallel to \(E\) has a parallel normal vector, so \(H\) has the form \(x-5y-2z=d\). 4. Substitute \(Q(3,0,2)\): \(d=3-5(0)-2(2)=-1\). 5. Therefore, \(H:x-5y-2z=-1\).

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

- Read a normal vector from the plane equation. - Use 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 \(E_1\).

- Read the normal direction from the coordinate coefficients. - Relate each axis intercept to substitution of a point where the other coordinates are zero. - For the full \(y\)-axis to lie in the plane, the origin must also lie in it.

1. A normal vector is \(\mathbf{n}=\langle a,0,c\rangle\). Since \(\mathbf{n}\cdot\langle0,1,0\rangle=0\), every plane is parallel to the \(y\)-axis. Because \(d\ne0\), none of these planes contains the origin, so the \(y\)-axis is not contained in them. 2. The intercepts give \(4a=d\) and \(6c=d\). Choose \(d=12\); then \(a=3\) and \(c=2\). One equation is \(3x+2z=12\). 3. Keeping \(a=3\) and \(c=2\), choose \(d=0\). Then every point \((0,y,0)\) satisfies \(3(0)+2(0)=0\), so the plane contains the \(y\)-axis.

1) The planes are parallel to the \(y\)-axis: \(\mathbf{n}=\langle a,0,c\rangle\) and \(\mathbf{n}\cdot\langle0,1,0\rangle=0\). Since \(d\ne0\), they do not contain that axis. 2) One choice is \(a=3\), \(c=2\), and \(d=12\), giving \(3x+2z=12\). 3) \(d=0\)

- Read the normal vector from the coordinate coefficients. - Choose two different vectors whose dot product with that normal is zero. - Find one point satisfying the plane equation before writing the two-parameter form.

1. A normal vector is \(\mathbf{n}=\langle2,5,-3\rangle\). Every direction vector in the plane must satisfy \(\mathbf{n}\cdot\mathbf{w}=0\). 2. Choose \(\mathbf{u}=\langle5,-2,0\rangle\), since \(2(5)+5(-2)-3(0)=0\). 3. Choose \(\mathbf{v}=\langle3,0,2\rangle\), since \(2(3)+5(0)-3(2)=0\). The vectors are linearly independent because neither is a scalar multiple of the other. 4. Setting \(y=z=0\) gives \(x=5\), so \(A(5,0,0)\) is on the plane. 5. One vector equation is \(\mathbf{r}(s,t)=\langle5,0,0\rangle+s\langle5,-2,0\rangle+t\langle3,0,2\rangle\).

a) \(\mathbf{n}=\langle2,5,-3\rangle\); every direction vector satisfies \(\mathbf{n}\cdot\mathbf{w}=0\). b) One choice is \(\mathbf{u}=\langle5,-2,0\rangle\) and \(\mathbf{v}=\langle3,0,2\rangle\). c) \(\mathbf{r}(s,t)=\langle5,0,0\rangle+s\langle5,-2,0\rangle+t\langle3,0,2\rangle\)

- 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 endpoint-to-endpoint displacement gives a normal direction for the perpendicular-bisector plane. - Find the midpoint of the segment before writing the plane equation. - Combine the midpoint and normal in point-normal form, then expand.

1. Compute \(\overrightarrow{AB}=\langle4,-2,-4\rangle\). Use the parallel normal vector \(\mathbf{n}=\langle2,-1,-2\rangle\). 2. The midpoint is \(M\left(\frac{1+5}{2},\frac{2+0}{2},\frac{3+(-1)}{2}\right)=(3,1,1)\). 3. Point-normal form is \(\langle2,-1,-2\rangle\cdot\langle x-3,y-1,z-1\rangle=0\). 4. Expanding gives \(2x-y-2z=3\).

Point-normal form: \(\langle2,-1,-2\rangle\cdot\langle x-3,y-1,z-1\rangle=0\) Standard form: \(2x-y-2z=3\)

- 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.

- Read the normal vector from the coordinate coefficients. - Compare its current magnitude with the required magnitude before choosing scale factors. - For the axis relationship, test what happens when \(x=z=0\).

1. A normal vector is \(\mathbf{n}=\langle3,0,-4\rangle\), so the dot-product equation is \(\langle3,0,-4\rangle\cdot\mathbf{x}=0\). 2. Its magnitude is \(\|\mathbf{n}\|=\sqrt{3^2+0^2+(-4)^2}=5\). A parallel normal vector of magnitude \(10\) must be \(2\mathbf{n}\) or \(-2\mathbf{n}\). Thus, the vectors are \(\langle6,0,-8\rangle\) and \(\langle-6,0,8\rangle\). 3. The \(y\)-component of every normal vector is \(0\), so the \(y\)-axis direction lies in the plane. The equation also contains the origin, so the entire \(y\)-axis lies in the plane.

a) \(\langle3,0,-4\rangle\cdot\mathbf{x}=0\) b) \(\langle6,0,-8\rangle\) and \(\langle-6,0,8\rangle\) c) The plane contains the \(y\)-axis: its normal \(\langle3,0,-4\rangle\) is perpendicular to \(\langle0,1,0\rangle\), and the origin satisfies the plane equation.

- Form two independent direction vectors from the three points in the plane. - Find a vector perpendicular to both plane directions. - A line through the origin needs only that perpendicular direction in its parametric equation.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle2,2,0\rangle\) and \(\overrightarrow{AC}=\langle0,4,4\rangle\). 2. Let a normal vector be \(\mathbf{n}=\langle a,b,c\rangle\). Orthogonality requires \(\mathbf{n}\cdot\overrightarrow{AB}=2a+2b=0\) and \(\mathbf{n}\cdot\overrightarrow{AC}=4b+4c=0\). Thus, \(a=-b\) and \(c=-b\). Choosing \(b=-1\) gives \(\mathbf{n}=\langle1,-1,1\rangle\). 3. A line perpendicular to \(E\) can use this normal vector as its direction vector. Since the line passes through the origin, \(h:\mathbf{x}=t\langle1,-1,1\rangle\).

\(h:\mathbf{x}=t\langle1,-1,1\rangle\), where \(t\in\mathbb{R}\)

- 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 \(T(3,1,0)\).

- Treat two coordinates as free parameters after solving for the third. - Separate the fixed point from the two parameter-direction vectors. - Read the normal from the standard equation and verify both dot products.

1. Solve for \(x\): \(-x+3y+2z=6\) gives \(x=3y+2z-6\). 2. Let \(y=s\) and \(z=t\). Then \(x=-6+3s+2t\), so \(\mathbf{r}(s,t)=\langle-6,0,0\rangle+s\langle3,1,0\rangle+t\langle2,0,1\rangle\). 3. A normal vector is \(\mathbf{n}=\langle-1,3,2\rangle\). 4. \(\mathbf{n}\cdot\langle3,1,0\rangle=-3+3=0\). 5. \(\mathbf{n}\cdot\langle2,0,1\rangle=-2+2=0\). Both direction vectors are orthogonal to the normal vector.

a) \(x=3y+2z-6\) b) \(\mathbf{r}(s,t)=\langle-6,0,0\rangle+s\langle3,1,0\rangle+t\langle2,0,1\rangle\) c) \(\mathbf{n}=\langle-1,3,2\rangle\), and its dot product with each direction vector is \(0\).

- Find a normal perpendicular to both parametric direction vectors. - Use the initial point to determine the standard-equation constant. - Test each candidate point by substitution.

1. Let \(\mathbf{n}=\langle a,b,c\rangle\). Orthogonality to the direction vectors gives \(a+2b=0\) and \(b+2c=0\). 2. Choose \(c=1\). Then \(b=-2\) and \(a=4\), so \(\mathbf{n}=\langle4,-2,1\rangle\). 3. The plane has the form \(4x-2y+z=d\). Substitute \((2,1,-1)\): \(d=5\). Thus, \(E:4x-2y+z=5\). 4. For \(P\), \(4(3)-2(4)+1=5\), so \(P\in E\). 5. For \(Q\), \(4(1)-2(1)+1=3\ne5\), so \(Q\notin E\).

a) One normal is \(\mathbf{n}=\langle4,-2,1\rangle\), satisfying \(\mathbf{n}\cdot\langle1,2,0\rangle=0\) and \(\mathbf{n}\cdot\langle0,1,2\rangle=0\); the plane is \(E:4x-2y+z=5\). b) \(P\in E\), but \(Q\notin E\).

- Identify which coordinate is fixed for every point on the plane. - Use the plane's coordinate-axis normal to characterize in-plane directions. - Choose two independent coordinate directions and one convenient point on the plane.

1. All points in \(F\) have a constant \(y\)-coordinate of \(5\). Therefore, \(F\) is parallel to the \(xz\)-plane and is \(5\) units from it. 2. A normal vector to \(F\) is \(\mathbf{n}=\langle0,1,0\rangle\). For any \(\mathbf{v}=\langle v_x,0,v_z\rangle\), \(\mathbf{n}\cdot\mathbf{v}=0\), so \(\mathbf{v}\) is parallel to the plane. 3. Using \((0,5,0)\) and independent directions \(\langle1,0,0\rangle\), \(\langle0,0,1\rangle\), one equation is \(\mathbf{r}(s,t)=\langle0,5,0\rangle+s\langle1,0,0\rangle+t\langle0,0,1\rangle\).

a) \(F\) is parallel to the \(xz\)-plane and is \(5\) units from it. b) \(\langle0,1,0\rangle\cdot\langle v_x,0,v_z\rangle=0\), so every nonzero vector of that form is parallel to \(F\). c) \(\mathbf{r}(s,t)=\langle0,5,0\rangle+s\langle1,0,0\rangle+t\langle0,0,1\rangle\)

- Form two independent connecting vectors from the same point. - Find a normal vector perpendicular to both. - Use the normal components as Cartesian coefficients and substitute one known point for the constant.

1. Two spanning vectors are \(\overrightarrow{PQ}=\langle3,-2,2\rangle\) and \(\overrightarrow{PR}=\langle0,-3,4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(3a-2b+2c=0\) and \(-3b+4c=0\). One solution is \(\mathbf{n}=\langle2,12,9\rangle\). 3. Using \(P\), \(2\cdot2+12\cdot3+9\cdot0=40\). Therefore, \(E:2x+12y+9z=40\).

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

- Find a normal vector perpendicular to both parametric directions. - Use the initial point to determine the standard-equation constant. - For each coordinate-axis intercept, set the other two coordinates equal to zero.

1. Let \(\mathbf{n}=\langle a,b,c\rangle\). Orthogonality to the direction vectors gives \(a-2b+2c=0\) and \(2a+b=0\). 2. From \(b=-2a\), the first equation becomes \(a+4a+2c=0\), so \(2c=-5a\). Choose \(a=-2\), giving \(b=4\) and \(c=5\). Thus, \(\mathbf{n}=\langle-2,4,5\rangle\). 3. The plane has the form \(-2x+4y+5z=d\). Substitute \((2,2,1)\): \(d=9\). Therefore, \(E:-2x+4y+5z=9\). 4. The axis intercepts are \((-\frac92,0,0)\), \((0,\frac94,0)\), and \((0,0,\frac95)\).

a) One normal is \(\mathbf{n}=\langle-2,4,5\rangle\), satisfying both orthogonality conditions. The plane is \(E:-2x+4y+5z=9\). b) \((-\frac92,0,0)\), \((0,\frac94,0)\), and \((0,0,\frac95)\)

- Set corresponding line coordinates equal to find the intersection parameters. - Use the intersection point and both line directions for the plane's parametric form. - Find a normal perpendicular to both directions, then use the intersection point for Cartesian form.

1. Equating coordinates gives \(3+t=2+2s\), \(4+2t=2+s\), and \(1-t=2+s\). Solving gives \(t=-1\) and \(s=0\), so \(S=(2,2,2)\). 2. The intersection point and the two line directions give \(E:\mathbf{x}=\langle2,2,2\rangle+t\langle1,2,-1\rangle+s\langle2,1,1\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both direction vectors. Then \(a+2b-c=0\) and \(2a+b+c=0\). One solution is \(\mathbf{n}=\langle1,-1,-1\rangle\). Using \(S\), the Cartesian equation is \(E:x-y-z=-2\).

1. The lines meet at \(S=(2,2,2)\), with parameters \(t=-1\), \(s=0\). 2. \(E:\mathbf{x}=\langle2,2,2\rangle+t\langle1,2,-1\rangle+s\langle2,1,1\rangle\) 3. \(E:x-y-z=-2\)

- Compare the two line directions for scalar-multiple parallelism. - Use one line direction and a vector joining the two lines as plane-spanning directions. - Find a normal perpendicular to those two directions and use a known point for Cartesian form.

1. The first direction vector is \(-2\) times the second, so the lines are parallel. The point \((0,1,0)\) is not on \(g_1\): the \(x\)-coordinate would require \(r=-\frac12\), which gives \(y=3\), not \(1\). Thus, the lines are distinct. 2. Use direction \(\mathbf{u}=\langle-1,1,-2\rangle\) and the vector between the given points, \(\mathbf{v}=\langle-1,-1,-3\rangle\). One equation is \(E:\mathbf{x}=\langle0,1,0\rangle+s\langle-1,1,-2\rangle+k\langle-1,-1,-3\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(-a+b-2c=0\) and \(-a-b-3c=0\). One solution is \(\mathbf{n}=\langle-5,-1,2\rangle\). Using \((0,1,0)\) gives \(-5x-y+2z=-1\), or equivalently \(E:5x+y-2z=1\).

1. The directions are scalar multiples, but \((0,1,0)\) lies only on \(g_2\): setting \(x=0\) on \(g_1\) would make \(y=3\). Thus the lines are distinct and parallel. 2. \(E:\mathbf{x}=\langle0,1,0\rangle+s\langle-1,1,-2\rangle+k\langle-1,-1,-3\rangle\) 3. \(E:5x+y-2z=1\)

- 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 the line direction and a vector from the line to the additional point as plane-spanning directions. - Find a normal perpendicular to both directions. - Use that normal for Cartesian form, then test \(P\) by substitution.

1. Use the line direction \(\mathbf{u}=\langle1,0,1\rangle\) and the vector from the line's given point to \(Q\), \(\mathbf{v}=\langle1,-3,3\rangle\). Thus, \(E:\mathbf{x}=\langle1,2,0\rangle+r\langle1,0,1\rangle+s\langle1,-3,3\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(a+c=0\) and \(a-3b+3c=0\). One solution is \(\mathbf{n}=\langle3,-2,-3\rangle\). Using \((1,2,0)\), the Cartesian equation is \(E:3x-2y-3z=-1\). 3. For \(P\), \(3\cdot4-2\cdot(-1)-3\cdot5=-1\). Therefore, \(P\in E\).

1. \(E:\mathbf{x}=\langle1,2,0\rangle+r\langle1,0,1\rangle+s\langle1,-3,3\rangle\) 2. \(E:3x-2y-3z=-1\) 3. Yes, \(P=(4,-1,5)\) lies in \(E\).

- Use one point and two connecting vectors for parametric form. - Find a normal perpendicular to both spanning vectors. - Use the Cartesian equation for the reverse membership condition on \(D\).

1. Two spanning vectors are \(\overrightarrow{AB}=\langle-2,2,-1\rangle\) and \(\overrightarrow{AC}=\langle-4,0,4\rangle\). Thus, \(E:\mathbf{x}=\langle4,1,1\rangle+r\langle-2,2,-1\rangle+s\langle-4,0,4\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(-2a+2b-c=0\) and \(-4a+4c=0\). One solution is \(\mathbf{n}=\langle2,3,2\rangle\). Using \(A\), the Cartesian equation is \(E:2x+3y+2z=13\). 3. Substitute \(D\): \(2\cdot1+3\cdot2+2z=13\). Thus, \(8+2z=13\), so \(z=\frac52\).

a) \(E:\mathbf{x}=\langle4,1,1\rangle+r\langle-2,2,-1\rangle+s\langle-4,0,4\rangle\) b) \(E:2x+3y+2z=13\) c) \(z=\frac52\)

- Three points determine a unique plane when two connecting vectors are not scalar multiples. - Find a normal perpendicular to both connecting vectors. - For each axis intercept, set the other two coordinates equal to zero.

1. The vectors \(\overrightarrow{PQ}=\langle2,-1,2\rangle\) and \(\overrightarrow{PR}=\langle-1,2,-2\rangle\) are not scalar multiples, so the points are not collinear. 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both connecting vectors. Then \(2a-b+2c=0\) and \(-a+2b-2c=0\). One solution is \(\mathbf{n}=\langle-2,2,3\rangle\). Using \(P\), the plane equation is \(E:-2x+2y+3z=11\). 3. The axis intercepts are \((-\frac{11}{2},0,0)\), \((0,\frac{11}{2},0)\), and \((0,0,\frac{11}{3})\).

a) \(\overrightarrow{PQ}=\langle2,-1,2\rangle\) and \(\overrightarrow{PR}=\langle-1,2,-2\rangle\) are not scalar multiples, so the points are not collinear. b) One normal is \(\mathbf{n}=\langle-2,2,3\rangle\), perpendicular to both connecting vectors; \(E:-2x+2y+3z=11\). c) \((-\frac{11}{2},0,0)\), \((0,\frac{11}{2},0)\), and \((0,0,\frac{11}{3})\)

- Compare the line direction with the plane normal first. - If the dot product is nonzero, substitute the line coordinates into the plane equation. - Use the solved parameter to recover the intersection point.

1. A normal vector to \(E\) is \(\mathbf{n}=\langle2,2,-1\rangle\). 2. The line direction satisfies \(\langle1,-1,3\rangle\cdot\mathbf{n}=2-2-3=-3\ne0\), so the line intersects the plane exactly once. 3. Substitute \((t,4-t,-2+3t)\) into the plane equation: \(2t+2(4-t)-(-2+3t)=5\). 4. Simplify: \(10-3t=5\), so \(t=\frac53\). 5. The intersection point is \(S\left(\frac53,\frac73,3\right)\).

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

- Use a dot product between the line direction and the plane normal 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) \(S(1,-2,-5)\)

- Find a normal vector orthogonal to both plane direction vectors. - Use the dot product of the line direction and the normal vector to decide whether the line is parallel to the plane. - If the line is not parallel, substitute its coordinates into a standard equation of the plane.

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

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

- Form two direction vectors in the roof plane. - Find a vector orthogonal to both plane directions using dot products. - Parametrize the beam and substitute it into the plane equation.

1. Two directions in the roof plane are \(\overrightarrow{AB}=\langle-8,8,0\rangle\) and \(\overrightarrow{AC}=\langle-8,0,8\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\). The equations \(-8a+8b=0\) and \(-8a+8c=0\) give \(a=b=c\). Choose \(\mathbf{n}=\langle1,1,1\rangle\). 3. Using point \(C\), the plane equation is \(x+y+z=12\). 4. The beam direction is \(\overrightarrow{PQ}=\langle1,1,3\rangle\), so \(\mathbf{r}(t)=\langle1,1,0\rangle+t\langle1,1,3\rangle\). 5. Substitute into the plane equation: \((1+t)+(1+t)+3t=12\). 6. Thus \(2+5t=12\), so \(t=2\). The intersection point is \(D(3,3,6)\).

A plane normal is \(\mathbf{n}=\langle1,1,1\rangle\), since \(\mathbf{n}\cdot\overrightarrow{AB}=\mathbf{n}\cdot\overrightarrow{AC}=0\). The beam meets the plane at \(D(3,3,6)\).

- Form two direction vectors from the three plane points. - Find a vector orthogonal to both plane directions. - Use a dot product to test the line direction, 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.

- Compare the normal vectors as scalar multiples. - Identical equations require the same scalar factor on both sides. - Parallel normals with nonproportional constants give strictly parallel planes.

1. In part a, the normal vectors are \(\mathbf{n}_E=\langle3,2,-4\rangle\) and \(\mathbf{n}_F=\langle9,6,t\rangle\). 2. Parallel planes have normal vectors that are scalar multiples. The first two components show that \(\mathbf{n}_F=3\mathbf{n}_E\). 3. Therefore, \(t=3(-4)=-12\). 4. For the equations to describe the same plane, the right side must have the same scalar factor. Thus, \(s=3(6)=18\). 5. When \(t=-12\), the normals are parallel. When \(s=10\), the right side is not \(18\), so the equations do not describe the same plane. 6. Therefore, the planes are strictly parallel.

a) \(t=-12\) b) \(s=18\) c) The planes are strictly parallel.

- Compare the normal vectors. - Apply the same scale factor to the entire equation to test whether the planes are identical. - A parallel plane can use the same normal vector. - Substitute the given point to find the constant.

1. The normal vectors satisfy \(\langle-10,2,-4\rangle=-2\langle5,-1,2\rangle\), so the planes are parallel or identical. 2. Multiplying the equation of \(E_1\) by \(-2\) gives \(-10x+2y-4z=-20\), not \(5\). Therefore the planes are strictly parallel. 3. A plane parallel to \(E_1\) can use the same normal vector, so write \(5x-y+2z=d\). 4. Substituting \(P(2,-3,1)\) gives \(d=5(2)-(-3)+2(1)=15\). Thus \(E_3:5x-y+2z=15\). 5. Every plane with normal vector \(\langle5,-1,2\rangle\) has the form \(5x-y+2z=d\). To be distinct from \(E_1\), require \(d\ne10\).

a) \(E_1\) and \(E_2\) are strictly parallel. b) \(E_3:5x-y+2z=15\) c) \(5x-y+2z=d\), where \(d\in\mathbb{R}\setminus\{10\}\).

- Compare the normal vectors as scalar multiples. - Use the fixed y-coefficient to find the scalar factor. - Identical equations require the constant to have the same factor. - Nonparallel planes intersect in a line.

1. The normal vectors are \(\mathbf{n}_1=\langle6,-4,2\rangle\) and \(\mathbf{n}_2=\langle a,6,b\rangle\). 2. For the planes to be parallel or identical, the normals must be scalar multiples. Since the y-component satisfies \(6=q(-4)\), the scalar is \(q=-\frac32\). 3. Therefore, \(a=-\frac32(6)=-9\) and \(b=-\frac32(2)=-3\). 4. In part a, the planes are identical when the constant has the same scalar factor: \(c=-\frac32(12)=-18\). 5. In part b, the planes are strictly parallel when \(a=-9\), \(b=-3\), and \(c\ne-18\). 6. In part c, if \(a\ne-9\) or \(b\ne-3\), the normals are not scalar multiples. Two nonparallel planes in space intersect in a line, regardless of \(c\).

a) \(a=-9\), \(b=-3\), \(c=-18\) b) \(a=-9\), \(b=-3\), \(c\ne-18\) c) \(a\ne-9\) or \(b\ne-3\), with any real \(c\)

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

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

A shared normal vector is \(\mathbf{n}=\langle-1,1,1\rangle\). It is perpendicular to the direction vectors of both planes: \(\mathbf{n}\cdot\langle1,0,1\rangle=0\), \(\mathbf{n}\cdot\langle0,1,-1\rangle=0\), \(\mathbf{n}\cdot\langle1,1,0\rangle=0\), and \(\mathbf{n}\cdot\langle1,-1,2\rangle=0\). Both initial points satisfy \(-x+y+z=1\): \(-1+2+0=1\) and \(-2+3+0=1\). Thus both equations describe the same plane \(-x+y+z=1\).

- 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 \(k=1\).

\(k=1\)

- 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

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

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

The normal \(\mathbf{n}=\langle-3,-1,2\rangle\) of \(E_2\) is perpendicular to both directions of \(E_1\), since \(\mathbf{n}\cdot\langle1,-1,1\rangle=0\) and \(\mathbf{n}\cdot\langle2,0,3\rangle=0\). The point \((4,1,-2)\) of \(E_1\) also satisfies \(-3x-y+2z=-17\). Thus the parallel planes share a point and are identical.

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

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

Yes. With \(\mathbf{n}_F=\langle2,1,2.5\rangle\), \(\mathbf{n}_F\cdot\overrightarrow{AB}=0\) and \(\mathbf{n}_F\cdot\overrightarrow{AC}=0\), so \(E\) and \(F\) are parallel. Also, \(2(1)+0+2.5(2)=7\), so \(A\) lies in both planes. Therefore they are identical.

- Compare two connecting vectors for parallelism. - Use one point and two independent connecting vectors for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - A point on the x-axis has \(y=0\) and \(z=0\).

1. The vectors \(\overrightarrow{AB}=\langle 3,-1,-2\rangle\) and \(\overrightarrow{AC}=\langle -1,3,-1\rangle\) are not scalar multiples, so the points are not collinear. 2. One parametric equation is \(E:\mathbf{r}=\langle 2,1,3\rangle+r\langle 3,-1,-2\rangle+s\langle -1,3,-1\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(3a-b-2c=0\) and \(-a+3b-c=0\). One solution is \(\mathbf{n}=\langle 7,5,8\rangle\). Using \(A\), the Cartesian equation is \(E:7x+5y+8z=43\). 4. On the x-axis, \(y=z=0\), so \(7x=43\). The intersection point is \(\left(\frac{43}{7},0,0\right)\).

a) \(\overrightarrow{AB}=\langle3,-1,-2\rangle\) and \(\overrightarrow{AC}=\langle-1,3,-1\rangle\) are not scalar multiples (their first two component ratios differ), so the points are not collinear. b) \(E:\mathbf{r}=\langle 2,1,3\rangle+r\langle 3,-1,-2\rangle+s\langle -1,3,-1\rangle\) c) \(E:7x+5y+8z=43\) d) \(\left(\frac{43}{7},0,0\right)\)

- A point-on-line test must give one parameter value that works for every coordinate. - Use the line direction and a vector from the line to the outside point as spanning vectors. - Find a normal vector by requiring zero dot products with both spanning vectors. - Test \(R\) in the Cartesian equation.

1. Matching the x-coordinate of \(P\) would require \(t=-3\), while matching the y-coordinate would require \(t=2\). Therefore, \(P\notin h\). 2. Use the line direction \(\mathbf{u}=\langle 1,1,3\rangle\) and the vector from the line's given point to \(P\), \(\mathbf{v}=\langle -3,2,3\rangle\). Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. Then \(a+b+3c=0\) and \(-3a+2b+3c=0\). One solution is \(\mathbf{n}=\langle -3,-12,5\rangle\). 3. Using \((4, 0, -2)\), the plane equation is \(-3x-12y+5z=-22\), or equivalently \(F:3x+12y-5z=22\). 4. For \(R\), \(3\cdot2+12\cdot3-5\cdot4=22\). Therefore, \(R\in F\).

a) To match the x-coordinate of \(P\), \(t=-3\); to match its y-coordinate, \(t=2\). No single \(t\) works, so \(P\notin h\). b) \(F:3x+12y-5z=22\) c) Yes: \(3(2)+12(3)-5(4)=22\), so \(R\in F\).

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

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

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

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

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

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

- 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 \(t=0\).

- Use one point and two connecting vectors for parametric form. - Find a normal vector by requiring zero dot products with both spanning vectors. - Compare all three side lengths. - The centroid is the coordinate-wise average of the vertices.

1. Two spanning vectors are \(\overrightarrow{AB}=\langle 2,0,-2\rangle\) and \(\overrightarrow{AC}=\langle 0,2,-2\rangle\). Thus, \(E:\mathbf{x}=\langle 2,2,4\rangle+r\langle 2,0,-2\rangle+s\langle 0,2,-2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a-2c=0\) and \(2b-2c=0\). One solution is \(\mathbf{n}=\langle 1,1,1\rangle\). Using \(A\), the Cartesian equation is \(E:x+y+z=8\). 3. The three side lengths are \(AB=AC=BC=2\sqrt2\), so triangle \(ABC\) is equilateral. 4. The centroid is the coordinate-wise average: \(G=\left(\frac83,\frac83,\frac83\right)\).

a) \(E:\mathbf{x}=\langle 2,2,4\rangle+r\langle 2,0,-2\rangle+s\langle 0,2,-2\rangle\); \(E:x+y+z=8\) b) The triangle is equilateral with side length \(2\sqrt2\), and \(G=\left(\frac83,\frac83,\frac83\right)\).

- How far is each side of the square from its center? - Write a parametric equation using endpoints \(S\) and \(A\). - For a point on the segment, what range must the parameter satisfy?

1. The base is centered at the origin and has side length \(8\), so each base coordinate is \(4\) units from the corresponding axis. In counterclockwise order, \(A=(4,-4,0)\), \(B=(4,4,0)\), \(C=(-4,4,0)\), and \(D=(-4,-4,0)\). 2. A parametric equation of the line through \(S\) and \(A\) is \(\mathbf{x}=\langle 0,0,12\rangle+t\langle 4,-4,-12\rangle\). The edge corresponds to \(0\le t\le1\). 3. Substituting \(P\) gives \(2=4t\), \(-2=-4t\), and \(6=12-12t\). Each equation gives \(t=\frac{1}{2}\). 4. Because \(\frac{1}{2}\) is between \(0\) and \(1\), point \(P\) lies on edge \(\overline{SA}\).

a) \(A=(4,-4,0)\), \(B=(4,4,0)\), \(C=(-4,4,0)\), and \(D=(-4,-4,0)\) b) Yes. Point \(P(2,-2,6)\) lies on \(\overline{SA}\) when \(t=\frac{1}{2}\).

- 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) \(N\left(\frac32, \frac52, 6\right)\)

- 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, \(P=(2,3,1)\).

- Identify two nonparallel directions contained in the plane before looking for a normal. - Translate perpendicularity to both directions into two zero-dot-product equations. - Use the resulting normal with point \(P\) to write the plane equation.

1. The common line direction \(\mathbf{u}\) lies in the plane. Because the lines are distinct, \(\overrightarrow{PQ}\) supplies another direction in the plane that is not parallel to \(\mathbf{u}\). A plane normal must therefore be perpendicular to both. 2. \(\overrightarrow{PQ}=\langle-1,3,1\rangle\). For \(\mathbf{n}=\langle a,b,c\rangle\), the conditions are \(2a+b=0\) and \(-a+3b+c=0\). 3. One solution is \(\mathbf{n}=\langle1,-2,7\rangle\). 4. Using \(P\), \((\mathbf{x}-\langle1,-1,2\rangle)\cdot\langle1,-2,7\rangle=0\), which expands to \(x-2y+7z=17\).

a) The plane normal must be orthogonal to both independent directions contained in the plane. b) One normal vector is \(\mathbf{n}=\langle1,-2,7\rangle\). c) \(x-2y+7z=17\)

- Equal distances from the origin allow each axis intercept to be \(d\) or \(-d\). - Use intercept form to determine the possible signs of the coefficients. - Substitute \((2, 2, 2)\) and require \(d>0\). - Sign patterns that give a negative left-hand side at \(P\) cannot satisfy the positive-distance condition.

1. Each intercept is one of \((\pm d, 0, 0)\), \((0, \pm d, 0)\), and \((0, 0, \pm d)\). After normalizing the intercept form, an equation has the form \(\pm x\pm y\pm z=d\). 2. Substitute \(P(2, 2, 2)\): \(\pm2\pm2\pm2=d\). Since \(d>0\), only sign patterns with a positive sum can satisfy the condition. 3. With three positive signs, \(d=6\), giving \(x+y+z=6\). 4. With two positive signs and one negative sign, \(d=2\). The three placements of the negative sign give \(x+y-z=2\), \(x-y+z=2\), and \(-x+y+z=2\). 5. Every remaining sign pattern gives a negative value when \((2,2,2)\) is substituted, so it cannot equal the required positive value \(d\). Therefore, there are no additional planes.

\(E_1: x+y+z=6\) \(E_2: x+y-z=2\) \(E_3: x-y+z=2\) \(E_4: -x+y+z=2\)

- A line lies in a plane when one point is in the plane and its direction is orthogonal to the plane's normal. - The second plane needs a different normal, but that normal must still be perpendicular to the same line direction. - Distinct nonparallel plane normals ensure that the two planes intersect in a line rather than coincide.

1. \(P\) satisfies \(E_1\) because \(2(2)-0=4\). Also, \(\mathbf{n}_1\cdot\mathbf{d}=\langle 2,-1,0\rangle\cdot\langle 1,2,-1\rangle=2-2=0\). Thus the direction of \(g\) is parallel to \(E_1\), and because \(P\in E_1\), the whole line lies in \(E_1\). 2. One choice is \(\mathbf{n}_2=\langle 1,0,1\rangle\), since \(\langle 1,0,1\rangle\cdot\langle 1,2,-1\rangle=0\). It is not a scalar multiple of \(\mathbf{n}_1\). 3. Using \(P\), \((\mathbf{x}-\langle 2,0,1\rangle)\cdot\langle 1,0,1\rangle=0\), so one possible equation is \(E_2:x+z=3\). 4. The planes are distinct because their normals are not parallel, so they intersect in one line. Both contain \(g\), so that intersection line is \(g\).

a) \(P\in E_1\) and \(\mathbf{n}_1\cdot\mathbf{d}=0\), so \(g\subset E_1\). b) One choice is \(\mathbf{n}_2=\langle 1,0,1\rangle\). c) One possible plane is \(E_2:x+z=3\). The distinct planes both contain \(g\), so their intersection is \(g\).

- 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) \(\alpha\approx19.5^\circ\)

- 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) \(\phi\approx78.9^\circ\)

- Form two direction vectors from the three points. - Find a normal vector orthogonal to both directions. - Use the point-to-plane distance formula. - Substitute the line coordinates into the standard equation.

1. Two directions in the plane are \(\overrightarrow{AB}=\langle 3,8,-5\rangle\) and \(\overrightarrow{AC}=\langle 9,-2,-2\rangle\). 2. A parametric equation is \(\mathbf{r}=\langle 1,1,5\rangle+\lambda\langle 3,8,-5\rangle+\mu\langle 9,-2,-2\rangle\). 3. Let \(\mathbf{n}=\langle a,b,c\rangle\). Solving \(\mathbf{n}\cdot\overrightarrow{AB}=0\) and \(\mathbf{n}\cdot\overrightarrow{AC}=0\) gives \(\mathbf{n}=\langle 2,3,6\rangle\). 4. Using point \(A\), the standard equation is \(2x+3y+6z=35\). 5. The distance from the origin is \(\frac{|0-35|}{\sqrt{2^2+3^2+6^2}}=\frac{35}{7}=5\). 6. Substitute the line into the plane equation: \(2(1+\sigma)+3(2-2\sigma)+6(3+4\sigma)=35\). 7. This gives \(26+20\sigma=35\), so \(\sigma=\frac{9}{20}\). 8. The intersection point is \(S\left(\frac{29}{20},\frac{11}{10},\frac{24}{5}\right)\).

a) \(\mathbf{r}=\langle 1,1,5\rangle+\lambda\langle 3,8,-5\rangle+\mu\langle 9,-2,-2\rangle\); \(2x+3y+6z=35\) b) \(5\) c) \(S\left(\frac{29}{20},\frac{11}{10},\frac{24}{5}\right)\)

- 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 \(\langle-1,0,1\rangle\).

- Use two side vectors of the base to describe its plane. - Find a plane normal by requiring zero dot products with both side vectors. - Compare the side-vector lengths and dot product. - Complete the square using vector addition. - In a right pyramid, the apex lies on the line through the base center perpendicular to the base plane.

1. The vectors \(\overrightarrow{AB}=\langle 2,2,-1\rangle\) and \(\overrightarrow{BC}=\langle -1,2,2\rangle\) span the base plane. One parametric equation is \(E:\mathbf{x}=\langle 4,1,1\rangle+r\langle 2,2,-1\rangle+s\langle -1,2,2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both spanning vectors. Then \(2a+2b-c=0\) and \(-a+2b+2c=0\). One solution is \(\mathbf{n}=\langle 2,-1,2\rangle\). Using \(A\), the Cartesian equation is \(E:2x-y+2z=9\). 3. \(AB=3\), \(BC=3\), and \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\). Thus, triangle \(ABC\) is isosceles and right at \(B\). The fourth vertex is \(D=A+\overrightarrow{BC}=(3,3,3)\). 4. The center of the square is the midpoint of \(\overline{AC}\): \(M=\left(\frac92,3,\frac32\right)\). A unit normal vector is \(\frac13\langle 2,-1,2\rangle\). Therefore, \(S=M\pm3\langle 2,-1,2\rangle\), giving \(S_1=\left(\frac{21}{2},0,\frac{15}{2}\right)\) or \(S_2=\left(-\frac32,6,-\frac92\right)\).

a) \(E:\mathbf{x}=\langle4,1,1\rangle+r\langle2,2,-1\rangle+s\langle-1,2,2\rangle\); \(E:2x-y+2z=9\) b) \(AB=BC=3\), \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\), and \(D=(3,3,3)\) c) \(S_1=\left(\frac{21}{2},0,\frac{15}{2}\right)\) or \(S_2=\left(-\frac32,6,-\frac92\right)\)

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

- 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 \(\sqrt3\).