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

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

\(5\) units

- How can you compare perpendicular distances from two points to the same plane? - Does the sign of the plane expression change the distance? - Work exactly before comparing the two values.

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

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

- What does the normal vector tell you about the shortest distance to a plane? - How can a displacement vector be split into components parallel and perpendicular to a plane? - Check that the result cannot be negative and uses distance units.

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

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

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

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

\(3\) units

- How can the stated distance condition be expressed using the coordinates of the point and plane? - Which property of the plane is relevant to measuring perpendicular distance? - Could points on opposite sides of a plane have the same distance from it?

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

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

- Why should the distance between parallel lines be the same along their lengths? - Which connecting segment could give the shortest separation? - Check whether your result is independent of the chosen point.

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

\(\sqrt{5}\) units

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

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

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

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

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

\(5\) units

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

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

a) \(\frac{22}{9}\) units b) \(r=\frac{8}{9}\) unit

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

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

\(\langle3,3\rangle\)

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

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

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

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

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

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

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

1. A normal vector to the plane is \(\mathbf{n}=\langle1,1,1\rangle\). The perpendicular line through \(A\) is \(\mathbf{r}(t)=\langle1,2,3\rangle+t\langle1,1,1\rangle\). 2. At the point where this line meets the plane, \((1+t)+(2+t)+(3+t)=12\). 3. Thus, \(6+3t=12\), so \(t=2\). The foot of the perpendicular is \(S=(3, 4, 5)\). 4. Point \(S\) is the midpoint of \(A\) and its reflection. Therefore, \(A'=2S-A=(6, 8, 10)-(1, 2, 3)=(5, 6, 7)\).

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

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

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

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

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

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

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

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

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

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

- How can the plane's normal help identify the shortest route from the point to the plane? - What must be true about a closest point that lies on the plane? - Check whether your connecting segment is perpendicular to the plane.

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

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

- What geometric relationship connects a point, its reflection, and the mirror plane? - A reflected point must be equally far from the plane on the opposite side. - How could you verify your proposed reflected point using a midpoint?

1. A normal vector to the plane is \(\mathbf{n}=\langle2,-2,1\rangle\). The perpendicular line through \(B\) is \(\mathbf{x}=\langle-1,4,3\rangle+r\langle2,-2,1\rangle\). 2. Substitute into the plane equation: \(2(-1+2r)-2(4-2r)+(3+r)=2\). This simplifies to \(9r-7=2\), so \(r=1\). 3. Therefore, \(L=(1,2,4)\). 4. Since \(L\) is the midpoint of \(\overline{BB'}\), \(B'=2L-B=2(1,2,4)-(-1,4,3)=(3,0,5)\).

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

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

1. A general point on \(h\) is \(F(t)=(1+t,1,1-t)\), so \(\overrightarrow{AF}=\langle t-1,-2,-t-3\rangle\). 2. Require \(\overrightarrow{AF}\cdot\langle1,0,-1\rangle=0\): \((t-1)-(-t-3)=2t+2=0\). Thus, \(t=-1\). 3. The perpendicular foot is \(F=(0,1,2)\). 4. Since \(F\) is the midpoint of \(\overline{AA'}\), \(A'=2F-A=2(0,1,2)-(2,3,4)=(-2,-1,0)\).

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

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

1. A line direction is \(\overrightarrow{AB}=\langle1,-3,0\rangle\); a plane normal is \(\mathbf{n}=\langle6,2,-3\rangle\). 2. Their dot product is \(\mathbf{n}\cdot\overrightarrow{AB}=6(1)+2(-3)-3(0)=0\), so the line is parallel to or contained in the plane. 3. At \(A\), \(6(1)+2(5)-3(2)=10\neq12\). Thus, \(g\) is parallel to and distinct from \(E\). 4. Every point on \(g\) has the same distance from \(E\), so \(d(g,E)=\frac{|6(1)+2(5)-3(2)-12|}{\sqrt{6^2+2^2+(-3)^2}}=\frac{2}{7}\).

\(\mathbf{n}\cdot\overrightarrow{AB}=\langle6,2,-3\rangle\cdot\langle1,-3,0\rangle=0\). Since \(A\) gives \(10\neq12\) in the plane equation, \(g\) is strictly parallel to \(E\). The distance is \(\frac{2}{7}\) units.

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

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

\(\mathbf{n}\cdot\langle2,2,1\rangle=2-4+2=0\), so the line direction is parallel to the plane. Its point \((1,1,4)\) gives \(7\neq5\) in the plane equation, hence the line is strictly parallel. The distance is \(\frac{2}{3}\) units.

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

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

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

- How could you recognize the point on the line closest to \(Q\)? - What must the connecting segment have in common with the line's direction? - Check that the point you use for the distance actually lies on the line.

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

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

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

1. A line equation is \(\mathbf{x}=\langle2,2,1\rangle+r\langle4,0,3\rangle\), so \(L=(2+4r,2,1+3r)\). 2. Perpendicularity requires \(\overrightarrow{CL}\cdot\langle4,0,3\rangle=4(4r-2)+3(3r-4)=25r-20=0\), so \(r=\frac{4}{5}\). 3. Thus, \(L=\left(\frac{26}{5},2,\frac{17}{5}\right)\). 4. The altitude has length \(CL=\sqrt{\left(\frac{6}{5}\right)^2+(-3)^2+\left(-\frac{8}{5}\right)^2}=\sqrt{13}\). 5. Since \(AB=\sqrt{4^2+0^2+3^2}=5\), the area is \(\frac{1}{2}\cdot5\cdot\sqrt{13}=\frac{5\sqrt{13}}{2}\).

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

- What do the coordinates of the three base vertices tell you about the base plane? - Would the apex-to-base height change if you chose another point on the same plane? - Which distance must be perpendicular to the base to count as the height?

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

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

- Which feature of the given plane stays fixed when the plane moves parallel to itself? - Could planes on opposite sides of the original plane have the same separation? - Check that each plane you find has the requested distance, not merely a similar equation.

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

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

- Where could the closest point on the beam lie relative to the sensor? - How does the angle of a connecting segment affect its length? - Check that the closest point lies on the stated beam, not just an infinite extension.

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

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

- How can you describe all possible points on the given line? - What geometric property singles out the foot of the shortest segment? - How can you check whether your proposed segment is truly perpendicular?

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

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

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

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

a) \(2\) units b) \(B\) lies in \(E\), and \(C=(2,4,2)\). c) \(d(C,E)=\frac{|6(2)+3(4)+2(2)-14|}{7}=\frac{14}{7}=2\) units, equal to \(d(A,E)=\frac{|-14|}{7}=2\) units.

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

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

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

- Skew lines must be noncoplanar. - Reflection places an image the same distance from the mirror plane on the opposite side. - Which points remain fixed under reflection across a plane? - Consider lines contained in the plane and lines normal to the plane.

1. The lines cannot be skew. If \(g\) intersects \(E\), the intersection point is fixed by the reflection, so \(g\) and \(g'\) intersect there. If \(g\) is parallel to \(E\), its image is a parallel line. In either case, the two lines are coplanar. 2. When \(g\) is parallel to \(E\), its image lies the same distance from \(E\) on the opposite side. Therefore, \(g\) and \(g'\) are parallel, and their distance is \(2\cdot4.5\,\text{cm}=9\,\text{cm}\). 3. The image line is identical to the original line when \(g\) lies in \(E\), so every point is fixed, or when \(g\) is perpendicular to \(E\), so reflection maps the line onto itself.

a) No. The lines are always coplanar; they either intersect or are parallel. b) The lines are parallel and are \(9\,\text{cm}\) apart. c) \(g=g'\) when \(g\) lies in \(E\) or is perpendicular to \(E\).

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

1. The midpoint of \(\overline{AB}\) is \(M\left(\frac{4+0}{2},\frac{-2+4}{2},\frac{1+5}{2}\right)=(2,1,3)\). 2. The vector \(\overrightarrow{AB}=\langle-4,6,4\rangle\) is normal to the reflection plane. Use the parallel normal vector \(\mathbf{n}=\langle-2,3,2\rangle\). 3. Since \(E\) passes through \(M\), \(-2x+3y+2z=-2(2)+3(1)+2(3)=5\). Thus \(E:-2x+3y+2z=5\). 4. An axis for a half-turn mapping \(A\) to \(B\) must pass through \(M\) and be perpendicular to \(\overrightarrow{AB}\). Choose \(\mathbf{v}=\langle3,2,0\rangle\), because \(\langle-2,3,2\rangle\cdot\langle3,2,0\rangle=0\). 5. One possible axis is \(\mathbf{r}(t)=\langle2,1,3\rangle+t\langle3,2,0\rangle\). 6. Test \(C\): \(-2(1)+3(1)+2(3)=7\ne5\), so \(C\) is not in \(E\).

a) \(E:-2x+3y+2z=5\) b) One possible line is \(\mathbf{r}(t)=\langle2,1,3\rangle+t\langle3,2,0\rangle\), \(t\in\mathbb{R}\). c) No, \(C\notin E\).

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

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

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

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

1. Since \(\mathbf{v}\cdot\mathbf{v}=1^2+2^2+2^2=9\) and \(\mathbf{p}\cdot\mathbf{v}=3+6+6=15\), the projection is \(\operatorname{proj}_{\mathbf{v}}\mathbf{p}=\frac{15}{9}\mathbf{v}=\frac{5}{3}\langle1,2,2\rangle=\left\langle\frac{5}{3},\frac{10}{3},\frac{10}{3}\right\rangle\). 2. Therefore, \(F=\left(\frac{5}{3},\frac{10}{3},\frac{10}{3}\right)\). 3. The perpendicular vector is \(\overrightarrow{PF}=\left\langle-\frac{4}{3},\frac{1}{3},\frac{1}{3}\right\rangle\), so the distance is \(\sqrt{\frac{16}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{2}\).

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

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

1. For the \(x\)-axis, a common perpendicular direction to \(\langle0,1,1\rangle\) and \(\langle1,0,0\rangle\) is \(\langle0,1,-1\rangle\). Therefore, \(d_x=\frac{\lvert\langle3,0,4\rangle\cdot\langle0,1,-1\rangle\rvert}{\sqrt2}=2\sqrt2\). 2. For the \(y\)-axis, a common perpendicular direction to \(\langle0,1,1\rangle\) and \(\langle0,1,0\rangle\) is \(\langle1,0,0\rangle\). Thus, \(d_y=\lvert\langle3,0,4\rangle\cdot\langle1,0,0\rangle\rvert=3\). 3. For the \(z\)-axis, a common perpendicular direction to \(\langle0,1,1\rangle\) and \(\langle0,0,1\rangle\) is also \(\langle1,0,0\rangle\). Thus, \(d_z=3\).

x-axis: \(2\sqrt2\) units y-axis: \(3\) units z-axis: \(3\) units

- What feature distinguishes the closest point from other points on the line? - Can you describe a possible closest point using the line's parameter? - How can you test that your proposed segment is shortest?

1. A general point on \(h\) is \(L(t)=(1+t,1+2t,1+2t)\), so \(\overrightarrow{AL}=\langle t,2t-5,2t-4\rangle\). 2. Require \(\overrightarrow{AL}\cdot\langle1,2,2\rangle=0\): \(t+2(2t-5)+2(2t-4)=9t-18=0\), so \(t=2\). 3. Therefore, \(L=(3,5,5)\). 4. The distance is \(AL=\sqrt{2^2+(-1)^2+0^2}=\sqrt5\).

The perpendicular foot is \(L(3,5,5)\), and the distance is \(\sqrt5\) units.

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

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

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

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

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

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

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

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

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

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

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

\(5\) units

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

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

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

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

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

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

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

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

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

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

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

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

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

1. The line's direction vector is \(\mathbf{v}=\langle3,-2,-2\rangle\), and a normal vector to the plane is \(\mathbf{n}=\langle2,2,1\rangle\). Their dot product is \(3\cdot2+(-2)\cdot2+(-2)\cdot1=0\). 2. The point \((1,4,5)\) does not lie in the plane because \(2(1)+2(4)+5=15\ne6\). Therefore, \(g\) is parallel to and distinct from \(E\). 3. The distance is \(\frac{|2(1)+2(4)+5-6|}{3}=3\). 4. Substituting \(t=1\) into \(g\) gives \((1,4,5)+\langle3,-2,-2\rangle=(4,2,3)=M\), so \(M\in g\). 5. Because the sphere is tangent to the plane, its radius is the distance from \(M\) to \(E\), which is also \(3\). Therefore, its equation is \((x-4)^2+(y-2)^2+(z-3)^2=9\).

a) \(\langle3,-2,-2\rangle\cdot\langle2,2,1\rangle=0\), so the line direction is perpendicular to the plane normal. Also, \(2(1)+2(4)+5=15\ne6\), so \(g\) is strictly parallel to \(E\). b) \(3\) units c) \(M\in g\) when \(t=1\), and the sphere is \((x-4)^2+(y-2)^2+(z-3)^2=9\).

- Rewrite the coordinate equations as one vector equation. - Find a vector orthogonal to both direction vectors using dot products. - The shortest segment from a point to a plane is parallel to the plane's normal vector. - Intersect the perpendicular line through \(Q\) with the plane.

1. The equations give \(\mathbf{r}=\langle 2,5,-1\rangle+r\langle 2,-1,0\rangle+s\langle 0,1,-2\rangle\). The two direction vectors are not scalar multiples, so they determine a plane. 2. Let \(\mathbf{n}=\langle a,b,c\rangle\). Orthogonality gives \(2a-b=0\) and \(b-2c=0\). Choose \(a=1\), so \(b=2\) and \(c=1\). Thus, \(\mathbf{n}=\langle 1,2,1\rangle\). 3. Using \((2,5,-1)\), the plane equation is \(x+2y+z=11\). 4. The perpendicular line through \(Q\) is \(g(t)=\langle 4,1,8\rangle+t\langle 1,2,1\rangle\). 5. Substitute into the plane: \((4+t)+2(1+2t)+(8+t)=11\). This gives \(14+6t=11\), so \(t=-\frac{1}{2}\). 6. Therefore, \(P_0=\left(\frac{7}{2},0,\frac{15}{2}\right)\). 7. The distance is \(\|\overrightarrow{QP_0}\|=\left\|-\frac{1}{2}\langle 1,2,1\rangle\right\|=\frac{\sqrt{6}}{2}\) units.

a) \(M:x+2y+z=11\) b) \(P_0\left(\frac{7}{2},0,\frac{15}{2}\right)\) c) \(\frac{\sqrt{6}}{2}\) units

- A point on the symmetry axis has form \((5,5,z)\). - Build one lateral-face equation from two independent vectors in that face. - Use the full three-dimensional point-to-plane distance formula before exploiting symmetry. - Which distances should symmetry make equal, and which must you still calculate?

1. In face \(BCS\), use \(\overrightarrow{BC}=\langle-10,0,0\rangle\) and \(\overrightarrow{BS}=\langle-5,-5,12\rangle\). A normal vector is \(\langle0,12,5\rangle\), so \(BCS:12y+5z=120\). 2. Since \(P=(5,5,z)\) lies inside the pyramid, its distance to the base \(z=0\) is \(z\). 3. Its distance to face \(BCS\) is \(\frac{|12(5)+5z-120|}{13}=\frac{60-5z}{13}\). 4. Set \(z=\frac{60-5z}{13}\). Then \(18z=60\), so \(z=\frac{10}{3}\). 5. Thus \(P=\left(5,5,\frac{10}{3}\right)\). Rotating the regular pyramid by \(90^\circ\) about its symmetry axis maps each lateral face to the next while keeping \(P\) fixed. Rotation preserves distances, so \(P\) is equally far from all four lateral faces.

a) \(BCS:12y+5z=120\) b) \(d(P,\text{base})=z\), and \(d(P,BCS)=\frac{60-5z}{13}\) c) \(P\left(5,5,\frac{10}{3}\right)\). Each \(90^\circ\) rotation about the pyramid's symmetry axis fixes \(P\), maps one lateral face to another, and preserves distance. Therefore, all four lateral-face distances are equal.

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

1. A unit vector in the direction of \(h\) is \(\mathbf{u}=\left\langle\frac{3}{5},\frac{4}{5},0\right\rangle\). 2. Project \(\overrightarrow{OQ}\) onto the line: \(\overrightarrow{OQ}\cdot\mathbf{u}=7\left(\frac{3}{5}\right)+1\left(\frac{4}{5}\right)+5(0)=5\). 3. Thus, \(\overrightarrow{OF}=5\mathbf{u}=\langle3,4,0\rangle\), so \(F=(3,4,0)\). 4. Since \(F\) is the midpoint of \(\overline{QQ'}\), \(\overrightarrow{OQ'}=2\overrightarrow{OF}-\overrightarrow{OQ}=2\langle3,4,0\rangle-\langle7,1,5\rangle=\langle-1,7,-5\rangle\).

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

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

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

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

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

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

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

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

1. A general point on \(g\) is \(L(t)=(t,1+2t,1)\), so \(\overrightarrow{AL}=\langle t-3,2t-1,-5\rangle\). 2. Perpendicularity gives \(\overrightarrow{AL}\cdot\langle1,2,0\rangle=(t-3)+2(2t-1)=5t-5=0\). Thus, \(t=1\) and \(L=(1,3,1)\). 3. The distance is \(AL=\sqrt{(-2)^2+1^2+(-5)^2}=\sqrt{30}\). 4. Since \(L\) is the midpoint of \(\overline{AA''}\), \(A''=2L-A=2(1,3,1)-(3,2,6)=(-1,4,-4)\). 5. The tangent sphere has radius \(\sqrt{30}\), so its equation is \((x-3)^2+(y-2)^2+(z-6)^2=30\).

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

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

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

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

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

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

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

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

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

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

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

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

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

- Use two directions in lateral face \(ABS\) to construct its normal. - Apply the three-dimensional point-to-plane distance formula to \(L=(0,0,h)\). - Use that the LED lies inside the pyramid when simplifying the absolute value. - Symmetry handles the other three lateral faces only after one 3D distance is established.

1. In face \(ABS\), \(\overrightarrow{AB}=\langle0,-10,0\rangle\) and \(\overrightarrow{AS}=\langle-5,-5,12\rangle\). A normal vector is \(\langle12,0,5\rangle\), whose magnitude is \(13\). 2. Using point \(A\), the face equation is \(12x+5z=60\). 3. For \(L=(0,0,h)\) inside the pyramid, \(d(L,ABS)=\frac{|5h-60|}{13}=\frac{60-5h}{13}\). 4. Set \(\frac{60-5h}{13}=2\). Then \(60-5h=26\), so \(h=\frac{34}{5}\). 5. Therefore, \(L=\left(0,0,\frac{34}{5}\right)\). Rotating the pyramid by \(90^\circ\) about its symmetry axis keeps \(L\) fixed while taking each lateral face to the next. Rotation preserves perpendicular distance, so the distances to all four lateral faces are equal.

a) \(ABS:12x+5z=60\); one normal is \(\langle12,0,5\rangle\) with magnitude \(13\) b) \(d(L,ABS)=\frac{60-5h}{13}\) c) \(L\left(0,0,\frac{34}{5}\right)\). Quarter-turns about the symmetry axis fix \(L\), map each lateral face to another, and preserve distance. Therefore, the distances to all four faces are equal.

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

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

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

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

1. A general point on \(h\) is \(F(s)=(4,-1+4s,2+3s)\), so \(\overrightarrow{QF}=\langle0,4s-9,3s-13\rangle\). 2. Orthogonality gives \(\overrightarrow{QF}\cdot\langle0,4,3\rangle=4(4s-9)+3(3s-13)=25s-75=0\), so \(s=3\). 3. Therefore, \(F=(4,11,11)\) and \(QF=\sqrt{3^2+(-4)^2}=5\). 4. The cone has radius \(5\) and height \(BF=\sqrt{12^2+9^2}=15\). 5. Thus, \(V=\frac{1}{3}\pi\cdot5^2\cdot15=125\pi\).

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

- Divide each plane equation by the magnitude of its normal vector. - Use consistent signs before adding and subtracting. - Plane normals are perpendicular when their dot product is \(0\). - After measuring the two distances, what do you notice about them?

1. Both plane normals have magnitude \(3\). Choose normalized equations \(E_1:\frac{2}{3}x+\frac{2}{3}y-\frac{1}{3}z-2=0\) and \(E_2:-\frac{1}{3}x+\frac{2}{3}y-\frac{2}{3}z-1=0\). 2. Adding gives \(W_1:x+4y-3z-9=0\); subtracting gives \(W_2:3x+z-3=0\). 3. The normals \(\langle1,4,-3\rangle\) and \(\langle3,0,1\rangle\) have dot product \(3-3=0\), proving perpendicularity. 4. Substituting \(P\) into \(W_2\) gives \(3-3=0\), so \(P\in W_2\). 5. The distances are \(d(P,E_1)=\frac{|-3-6|}{3}=3\) and \(d(P,E_2)=\frac{|2(3)+3|}{3}=3\). They are equal, as expected on an angle-bisector plane.

a) \(E_1:\frac{2}{3}x+\frac{2}{3}y-\frac{1}{3}z-2=0\); \(E_2:-\frac{1}{3}x+\frac{2}{3}y-\frac{2}{3}z-1=0\) b) \(W_1:x+4y-3z-9=0\); \(W_2:3x+z-3=0\) c) \(\langle1,4,-3\rangle\cdot\langle3,0,1\rangle=3-3=0\), so \(W_1\perp W_2\). d) Substitution gives \(3(0)+3-3=0\), so \(P\in W_2\). Also, \(d(P,E_1)=\frac{|-3-6|}{3}=3\) and \(d(P,E_2)=\frac{|6+3|}{3}=3\). Thus, \(P\) is equally distant from the original planes.

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

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

a) \(\alpha\approx85.75^\circ\) b) \(W_1:10x+y+7z-27=0\); \(W_2:2x-13y-z-9=0\) c) \(g:(x,y,z)=(0,-1,4)+t\langle15,4,-22\rangle\). The starting point gives \(0-1+28-27=0\) in \(W_1\), and the direction satisfies \(\langle15,4,-22\rangle\cdot\langle10,1,7\rangle=150+4-154=0\). Therefore, the whole line lies in \(W_1\).

- Find a normal vector to the plane from two independent directions in it. - What equality of side lengths would establish that the triangle is isosceles? - Which area and perpendicular height correspond to the required base of the tetrahedron? - How does restricting a point to the \(z\)-axis simplify the final volume condition?

1. \(\overrightarrow{LM}=\langle-4,4,0\rangle\) and \(\overrightarrow{LN}=\langle-1,3,4\rangle\). A normal vector is \(\mathbf{n}=\langle2,2,-1\rangle\), so \(H:2x+2y-z=8\). 2. \(LN=\sqrt{1+9+16}=\sqrt{26}\) and \(MN=\sqrt{9+1+16}=\sqrt{26}\), so \(\triangle LMN\) is isosceles. 3. \(\overrightarrow{LM}\times\overrightarrow{LN}=\langle16,16,-8\rangle\), whose magnitude is \(24\). Thus the base area is \(12\). 4. The distance from \(O\) to \(H\) is \(\frac{|0-8|}{\sqrt{2^2+2^2+(-1)^2}}=\frac{8}{3}\). 5. Therefore, \(V=\frac13(12)\left(\frac83\right)=\frac{32}{3}\). 6. For tetrahedron \(OLMP\), triangle \(OLM\) has area \(8\) in the \(xy\)-plane. With \(P=(0,0,z)\), \(16=\frac13(8)|z|\), so \(|z|=6\). 7. Hence \(P=(0,0,6)\) or \(P=(0,0,-6)\).

a) \(H:2x+2y-z=8\) b) \(LN=MN=\sqrt{26}\) c) \(d(O,H)=\frac83\) and \(V=\frac{32}{3}\) cubic units d) \(P=(0,0,6)\) or \(P=(0,0,-6)\)

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

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle-2,2,0\rangle\) and \(\overrightarrow{AC}=\langle0,2,-2\rangle\). 2. Let \(\mathbf{n}=\langle a,b,c\rangle\) be perpendicular to both. The conditions \(-2a+2b=0\) and \(2b-2c=0\) give \(a=b=c\), so choose \(\mathbf{n}=\langle1,1,1\rangle\). 3. Using point \(A\), the plane equation is \(x+y+z=7\). 4. The line through \(P\) perpendicular to the plane is \((x,y,z)=(5,6,8)+t\langle1,1,1\rangle\). 5. At its intersection with the plane, \((5+t)+(6+t)+(8+t)=7\), so \(t=-4\). The perpendicular foot is \(F(1,2,4)\). 6. Since \(F\) is the midpoint of \(\overline{PP'}\), \(P'=2F-P=(-3,-2,0)\).

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

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

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

a) The directions are not parallel. The \(x\)- and \(y\)-equations give \(r=\frac{7}{3}\), \(s=-\frac{1}{3}\), but the \(z\)-equation requires \(2r-s=6\), while \(2r-s=5\). Thus the lines are skew. b) \(\frac{2}{3}\) unit c) \(F_g=\left(\frac{41}{9},\frac{46}{9},\frac{55}{9}\right)\), \(F_h=\left(\frac{37}{9},\frac{44}{9},\frac{59}{9}\right)\) d) \(r=\sqrt{2}\) units

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

1. A general point on \(h\) is \(F(k)=(1+k,1+2k,1+2k)\), so \(\overrightarrow{AF}=\langle k-1,2k-4,2k\rangle\). 2. Require \(\overrightarrow{AF}\cdot\langle1,2,2\rangle=0\): \((k-1)+2(2k-4)+2(2k)=9k-9=0\), so \(k=1\) and \(F=(2,3,3)\). 3. The distance is \(AF=\sqrt{0^2+(-2)^2+2^2}=2\sqrt2\). 4. The tangent sphere has radius squared \(8\), so its equation is \((x-2)^2+(y-5)^2+(z-1)^2=8\). 5. For a point on \(h\) to be \(3\) units from \(A\), require \((k-1)^2+(2k-4)^2+(2k)^2=9\). This simplifies to \(9k^2-18k+8=0\), giving \(k=\frac23\) or \(k=\frac43\). 6. The points are \((\frac53,\frac73,\frac73)\) and \((\frac73,\frac{11}{3},\frac{11}{3})\).

a) \(F(2,3,3)\) b) \(2\sqrt2\) units c) \((x-2)^2+(y-5)^2+(z-1)^2=8\) d) \((\frac53,\frac73,\frac73)\) and \((\frac73,\frac{11}{3},\frac{11}{3})\)

- Compare the two three-dimensional side vectors from \(A\). - Parameterize line \(AB\) and impose a zero dot product with the residual from \(D\). - Use the parallelogram relation for the fourth vertex. - The perpendicular distance from \(D\) to \(AB\) is a height for the rhombus.

1. \(\overrightarrow{AB}=\langle1,2,2\rangle\) and \(\overrightarrow{AD}=\langle2,2,1\rangle\), so \(AB=AD=3\). 2. A point on line \(AB\) is \(L=A+t\overrightarrow{AB}\). Perpendicularity requires \((D-L)\cdot\overrightarrow{AB}=0\). 3. This gives \(t=\frac89\), so \(L=\left(\frac{17}{9},\frac{34}{9},\frac{25}{9}\right)\). 4. The distance is \(DL=\frac{\sqrt{17}}{3}\). 5. For the parallelogram/rhombus, \(C=B+D-A=(4,6,4)\). 6. The diagonals bisect each other, so \(M=\frac{A+C}{2}=\left(\frac52,4,\frac52\right)\). 7. Using base \(AB=3\) and height \(DL=\frac{\sqrt{17}}3\), the area is \(\sqrt{17}\).

a) \(AB=AD=3\), so \(\triangle ABD\) is isosceles. b) \(L\left(\frac{17}{9},\frac{34}{9},\frac{25}{9}\right)\); distance \(\frac{\sqrt{17}}3\) c) \(C(4,6,4)\); \(M\left(\frac52,4,\frac52\right)\); area \(\sqrt{17}\) square units

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

1. A general point on \(g\) is \(G(k)=(1+2k,1+2k,k)\), so \(\overrightarrow{PG}=\langle2k-4,2k-3,k-4\rangle\). 2. Require \(\overrightarrow{PG}\cdot\langle2,2,1\rangle=0\): \(2(2k-4)+2(2k-3)+(k-4)=9k-18=0\), so \(k=2\) and \(L=(5,5,2)\). 3. The distance is \(PL=\sqrt{0^2+1^2+(-2)^2}=\sqrt5\). 4. For part c), require \((2k-4)^2+(2k-3)^2+(k-4)^2=41\). This simplifies to \(9k(k-4)=0\), so \(k=0\) or \(k=4\). 5. The two points are \(Q_1=(1,1,0)\) and \(Q_2=(9,9,4)\).

a) \(L(5,5,2)\) b) \(\sqrt5\) units c) \(Q_1(1,1,0)\) and \(Q_2(9,9,4)\)

- Represent an arbitrary point on each line and think about their separation. - What must be true about the shortest possible connection if you move either endpoint slightly along its line? - After determining the relevant conditions, use them to find both points.

1. Write \(P=(1+r,r,1)\) and \(Q=(0,1+s,4+s)\). Then \(\overrightarrow{PQ}=\langle-1-r,1+s-r,3+s\rangle\). 2. For the shortest connection, \(\overrightarrow{PQ}\) must be perpendicular to both line directions. 3. From \(\overrightarrow{PQ}\cdot\langle1,1,0\rangle=0\), obtain \(-2r+s=0\). 4. From \(\overrightarrow{PQ}\cdot\langle0,1,1\rangle=0\), obtain \(-r+2s=-4\). 5. Solving gives \(r=-\frac43\) and \(s=-\frac83\). Therefore, \(P=(-\frac13,-\frac43,1)\) and \(Q=(0,-\frac53,\frac43)\). 6. The connecting vector is \(\langle\frac13,-\frac13,\frac13\rangle\), so the minimum distance is \(\frac{\sqrt3}{3}\).

\(P=(-\frac13,-\frac43,1)\), \(Q=(0,-\frac53,\frac43)\), and \(d=\frac{\sqrt3}{3}\) unit

- Consider how the shortest segment could be oriented relative to both lines. - Form a vector joining a given point on each line. - How could the component along a common perpendicular yield their separation?

1. A vector perpendicular to both direction vectors \(\langle1,1,0\rangle\) and \(\langle0,1,1\rangle\) is \(\mathbf{n}=\langle1,-1,1\rangle\), with magnitude \(\sqrt3\). 2. The vector between the given points is \(\mathbf{w}=\langle1,-2,1\rangle\). 3. The shortest distance is the magnitude of the scalar projection of \(\mathbf{w}\) onto \(\mathbf{n}\): \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac4{\sqrt3}=\frac{4\sqrt3}{3}\).

\(\frac{4\sqrt3}{3}\) units

- Look for a direction that measures the separation between the laser paths. - A vector between points on the paths may have components along their directions. Which part contributes to the minimum distance? - For the closest points, consider what happens if you move one endpoint along its line.

1. A vector perpendicular to both direction vectors is \(\mathbf{n}=\langle1,-1,1\rangle\), with magnitude \(\sqrt3\). 2. The vector between the given points is \(\langle2,0,2\rangle\). Therefore, the distance is \(d=\frac{\lvert\langle2,0,2\rangle\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac4{\sqrt3}=\frac{4\sqrt3}{3}\,\text{m}\). 3. Let \(P=(r,r,0)\) and \(Q=(2,s,2+s)\). Then \(\overrightarrow{PQ}=\langle2-r,s-r,2+s\rangle\). 4. The shortest connecting vector is perpendicular to both line directions. Solving \(\overrightarrow{PQ}\cdot\langle1,1,0\rangle=0\) and \(\overrightarrow{PQ}\cdot\langle0,1,1\rangle=0\) gives \(r=\frac23\) and \(s=-\frac23\). 5. Thus, \(P=(\frac23,\frac23,0)\) and \(Q=(2,-\frac23,\frac43)\).

a) \(\frac{4\sqrt3}{3}\,\text{m}\) b) \(P=(\frac23,\frac23,0)\), \(Q=(2,-\frac23,\frac43)\)

- Compare adjacent side lengths and use a dot product to check for a right angle. - For each distance, write a general point on the relevant line. - The shortest connecting vector must be orthogonal to the line's direction vector.

1. \(\overrightarrow{AB}=\langle6,0,0\rangle\) and \(\overrightarrow{AD}=\langle0,6,0\rangle\). They have equal magnitude \(6\) and dot product \(0\). Also, opposite side vectors are equal, so \(ABCD\) is a square with side length \(6\,\text{m}\). 2. A general point on \(AB\) is \(F(t)=(2+6t,2,0)\). The condition \(\overrightarrow{SF}\cdot\overrightarrow{AB}=0\) gives \(t=\frac12\), so \(F=(5,2,0)\). Thus, the distance is \(SF=\sqrt{0^2+(-3)^2+(-10)^2}=\sqrt{109}\,\text{m}\). 3. A general point on \(CS\) is \(G(r)=(8-3r,8-3r,10r)\). Its direction vector is \(\langle-3,-3,10\rangle\). 4. Require \(\overrightarrow{AG}\cdot\langle-3,-3,10\rangle=0\). This gives \(r=\frac{18}{59}\), so \(G=(\frac{418}{59},\frac{418}{59},\frac{180}{59})\). 5. Therefore, \(AG=\frac{60}{\sqrt{59}}\,\text{m}\).

a) Adjacent side vectors have length \(6\,\text{m}\) and dot product \(0\), and opposite sides are parallel and equal. b) \(\sqrt{109}\,\text{m}\) c) \(\frac{60}{\sqrt{59}}\,\text{m}\)

- Determine whether the three vectors from \(O\) can lie in one plane. - Use a dot product to find the perpendicular foot on \(QR\). - For the face area, use the triangle base-height relationship.

1. The vectors \(\overrightarrow{OP}=\langle0,0,4\rangle\), \(\overrightarrow{OQ}=\langle6,0,0\rangle\), and \(\overrightarrow{OR}=\langle0,8,0\rangle\) point along three different coordinate axes. They do not lie in one plane through \(O\), so the four points are noncoplanar and form a tetrahedron. 2. A general point on \(QR\) is \(F(t)=(6-6t,8t,0)\). For the perpendicular from \(O\), require \(\overrightarrow{OF}\cdot\langle-6,8,0\rangle=0\). This gives \(t=\frac{9}{25}\), so \(F=(\frac{96}{25},\frac{72}{25},0)\). 3. Thus, the distance from \(O\) to \(QR\) is \(OF=\frac{24}{5}\). 4. The same parameter gives the perpendicular foot from \(P\) because \(P\) differs from \(O\) only in the \(z\)-coordinate, while line \(QR\) lies in the \(xy\)-plane. The altitude from \(P\) to \(QR\) is \(PF=\frac{4\sqrt{61}}{5}\). 5. Since \(QR=10\), the area is \(\frac12(10)(\frac{4\sqrt{61}}{5})=4\sqrt{61}\) square units.

a) The three vectors from \(O\) to \(P\), \(Q\), and \(R\) lie along different coordinate axes, so the four points are noncoplanar. b) \(\frac{24}{5}\) units c) \(4\sqrt{61}\) square units

- Use \(B-A\) for the direction of \(g\). - Test perpendicularity with a dot product. - Check whether the two parametric lines can share all three coordinates. - Consider how a common perpendicular and the side vectors can measure distance and area.

1. \(\overrightarrow{AB}=\langle1,2,-2\rangle\), so \(g: (x,y,z)=(2,1,3)+r\langle1,2,-2\rangle\). 2. The direction vectors have dot product \(\langle1,2,-2\rangle\cdot\langle2,2,1\rangle=4\), so they are not perpendicular. 3. The directions are not scalar multiples. Matching \(x\)-coordinates gives \(2+r=2s\), and matching \(y\)-coordinates gives \(1+2r=4+2s\), so \(r=5\) and \(s=\frac72\). But then the \(z\)-coordinates are \(3-2(5)=-7\) and \(1+\frac72=\frac92\), which differ. Thus the lines do not meet and are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle6,-5,-2\rangle\), with magnitude \(\sqrt{65}\). The vector from the point on \(g\) to the point on \(h\) is \(\mathbf{w}=\langle-2,3,-2\rangle\). Therefore, \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{23}{\sqrt{65}}=\frac{23\sqrt{65}}{65}\). 5. For triangle \(ABC\), \(\overrightarrow{AB}=\langle1,2,-2\rangle\) and \(\overrightarrow{AC}=\langle0,5,-1\rangle\). Its area is \(\frac12\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=\frac{3\sqrt{10}}{2}\).

a) \(g: (x,y,z)=(2,1,3)+r\langle1,2,-2\rangle\) b) No; the direction vectors' dot product is \(4\). c) The lines are skew: matching \(x,y\) forces \(r=5\), \(s=\frac72\), but their \(z\)-coordinates then disagree. Their distance is \(\frac{23\sqrt{65}}{65}\) units. d) \(\frac{3\sqrt{10}}{2}\) square units

- Use the two given points to find the direction of \(h\). - Test perpendicularity with a dot product. - Equate coordinates to check for an intersection. - Project the vector between the lines onto a common perpendicular direction.

1. \(\overrightarrow{P_1P_2}=\langle2,1,-2\rangle\), so \(h: (x,y,z)=(1,2,2)+s\langle2,1,-2\rangle\). 2. The direction vectors have dot product \(\langle2,-2,1\rangle\cdot\langle2,1,-2\rangle=0\), so they are perpendicular. 3. The directions are not scalar multiples. Matching the \(x\)-coordinates gives \(2k=1+2s\), and matching the \(y\)-coordinates gives \(-2k=2+s\). Hence \(k=-\frac12\), \(s=-1\). The \(z\)-coordinates would then be \(5+k=\frac92\) and \(2-2s=4\), a contradiction. The lines are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle1,2,2\rangle\), with magnitude \(3\). The vector between the given points is \(\mathbf{w}=\langle1,2,-3\rangle\). 5. The distance is \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac13\).

a) \(h: (x,y,z)=(1,2,2)+s\langle2,1,-2\rangle\) b) \(\langle2,-2,1\rangle\cdot\langle2,1,-2\rangle=0\), so the directions are perpendicular. c) They are skew: the first two coordinates require \(k=-\frac12\), \(s=-1\), but the \(z\)-coordinates are then \(\frac92\) and \(4\). d) \(\frac13\) unit

- Use two adjacent vectors in the parallelogram base. - The area can be found from their lengths and dot product. - Find a vector orthogonal to both base directions. - Use the point-to-plane distance as the pyramid's height, then apply \(V=\frac{1}{3}Gh\).

1. Adjacent base vectors are \(\mathbf{u}=\overrightarrow{AB}=\langle 4,0,2\rangle\) and \(\mathbf{v}=\overrightarrow{AD}=\langle 0,4,4\rangle\). 2. Their dot products are \(\mathbf{u}\cdot\mathbf{u}=20\), \(\mathbf{v}\cdot\mathbf{v}=32\), and \(\mathbf{u}\cdot\mathbf{v}=8\). The parallelogram area is \(G=\sqrt{(\mathbf{u}\cdot\mathbf{u})(\mathbf{v}\cdot\mathbf{v})-(\mathbf{u}\cdot\mathbf{v})^2}=\sqrt{20(32)-8^2}=24\). 3. A normal vector to the base is \(\mathbf{n}=\langle 1,2,-2\rangle\), because \(\mathbf{n}\cdot\mathbf{u}=0\) and \(\mathbf{n}\cdot\mathbf{v}=0\). The base plane is \(x+2y-2z-1=0\). 4. The pyramid's height is the distance from \(S\) to the base plane: \(h=\frac{\lvert 6+2(9)-2(10)-1\rvert}{\sqrt{1^2+2^2+(-2)^2}}=1\). 5. Therefore, \(V=\frac{1}{3}Gh=\frac{1}{3}(24)(1)=8\) cubic units.

\(8\) cubic units

- Use two vectors along the base triangle. - The Gram determinant gives the area of the parallelogram formed by two vectors; divide by \(2\) for the triangle. - The base points are the coordinate-axis intercepts of its plane. - Use the point-to-plane distance as the tetrahedron's height.

1. Let \(\mathbf{u}=\overrightarrow{PQ}=\langle -2,4,0\rangle\) and \(\mathbf{v}=\overrightarrow{PR}=\langle -2,0,6\rangle\). 2. Their dot products are \(\mathbf{u}\cdot\mathbf{u}=20\), \(\mathbf{v}\cdot\mathbf{v}=40\), and \(\mathbf{u}\cdot\mathbf{v}=4\). Therefore, the triangle's area is \(G=\frac{1}{2}\sqrt{(\mathbf{u}\cdot\mathbf{u})(\mathbf{v}\cdot\mathbf{v})-(\mathbf{u}\cdot\mathbf{v})^2}=\frac{1}{2}\sqrt{20(40)-4^2}=14\). 3. The base plane has intercept form \(\frac{x}{2}+\frac{y}{4}+\frac{z}{6}=1\), so its coordinate equation is \(6x+3y+2z=12\). 4. A normal vector is \(\langle 6,3,2\rangle\), with magnitude \(7\). The height is the distance from \(S\) to the base plane: \(h=\frac{\lvert 6(5)+3(1)+2(0)-12\rvert}{7}=3\). 5. Thus, \(V=\frac{1}{3}Gh=\frac{1}{3}(14)(3)=14\) cubic units.

Base area: \(14\) square units Height: \(3\) units Volume: \(14\) cubic units

- Find a vector orthogonal to both direction vectors of the plane. - Test the line's direction vector with that normal vector. - Convert the plane to coordinate form. - Use a point on the line in the point-to-plane distance formula.

1. Let \(\mathbf{u}=\langle 0,0,1\rangle\) and \(\mathbf{v}=\langle 4,-3,0\rangle\) be the plane's direction vectors. A vector \(\mathbf{n}=\langle a,b,c\rangle\) perpendicular to both must satisfy \(c=0\) and \(4a-3b=0\). Choose \(\mathbf{n}=\langle 3,4,0\rangle\). 2. The line's direction vector is \(\mathbf{d}=\langle 4,-3,7\rangle\). Since \(\mathbf{d}\cdot\mathbf{n}=4(3)+(-3)(4)+7(0)=0\), the line is parallel to the plane or lies in it. 3. Using point \((4,0,0)\), the plane equation is \(3x+4y=12\). The point \((1,1,5)\) on \(g\) gives \(3(1)+4(1)=7\ne12\), so the line is parallel to and distinct from the plane. 4. The distance is \(d=\frac{\lvert 3(1)+4(1)-12\rvert}{\sqrt{3^2+4^2}}=1\).

The line is parallel to the plane, and the distance is \(1\) unit.

- First describe all points at distance \(2\) from plane \(F\). - A fixed distance from a plane gives two planes parallel to it. - Intersect each of those planes with \(E\). - Parameterize each intersection line.

1. A normal vector to \(F\) is \(\langle 2,1,-2\rangle\), with magnitude \(3\). 2. A point \((x,y,z)\) is \(2\) units from \(F\) when \(\frac{\lvert 2x+y-2z-6\rvert}{3}=2\). 3. Therefore, the point lies in one of the two planes \(2x+y-2z=12\) or \(2x+y-2z=0\). 4. Intersect \(x+y+z=3\) with \(2x+y-2z=12\). Subtracting the equations gives \(x-3z=9\). Let \(z=t\). Then \(x=9+3t\) and \(y=-6-4t\), so one line is \((x,y,z)=(9,-6,0)+t\langle 3,-4,1\rangle\). 5. Intersect \(x+y+z=3\) with \(2x+y-2z=0\). Subtracting gives \(x-3z=-3\). Let \(z=s\). Then \(x=-3+3s\) and \(y=6-4s\), so the other line is \((x,y,z)=(-3,6,0)+s\langle 3,-4,1\rangle\).

\((x,y,z)=(9,-6,0)+t\langle 3,-4,1\rangle\) or \((x,y,z)=(-3,6,0)+s\langle 3,-4,1\rangle\), where \(s,t\in\mathbb{R}\)

- Use the distance formula to identify two planes parallel to \(F\). - Substitute the parametric coordinates of \(E\) into each parallel-plane equation. - Each resulting linear relationship between \(\lambda\) and \(\mu\) describes a line.

1. A normal vector to \(F\) is \(\langle 2,-1,2\rangle\), with magnitude \(3\). 2. Points at distance \(4\) from \(F\) satisfy \(\frac{\lvert 2x-y+2z-9\rvert}{3}=4\). Thus, they lie in one of the parallel planes \(2x-y+2z=21\) or \(2x-y+2z=-3\). 3. A point in \(E\) has coordinates \(x=2+\lambda\), \(y=1+\mu\), and \(z=\lambda-\mu\). 4. For \(2x-y+2z=21\), substitution gives \(4\lambda-3\mu=18\). One solution is \((\lambda,\mu)=(6,2)\), giving the point \((8,3,4)\). Increasing \(\lambda\) by \(3\) and \(\mu\) by \(4\) preserves the equation, producing direction vector \(3\langle 1,0,1\rangle+4\langle 0,1,-1\rangle=\langle 3,4,-1\rangle\). 5. For \(2x-y+2z=-3\), substitution gives \(4\lambda-3\mu=-6\). One solution is \((\lambda,\mu)=(0,2)\), giving \((2,3,-2)\). The direction vector is again \(\langle 3,4,-1\rangle\).

\((x,y,z)=(8,3,4)+t\langle 3,4,-1\rangle\) or \((x,y,z)=(2,3,-2)+s\langle 3,4,-1\rangle\), where \(s,t\in\mathbb{R}\)

- Face \(OAB\) lies in the xy-plane. - Find a normal vector to \(ABC\) using dot-product orthogonality. - Apply the point-to-plane distance formula. - Use two different faces as the base in \(V=\frac{1}{3}Gh\).

1. Triangle \(OAB\) is right with legs of length \(6\), so its area is \(\frac{1}{2}(6)(6)=18\). 2. For plane \(ABC\), \(\overrightarrow{AB}=\langle -6,6,0\rangle\) and \(\overrightarrow{AC}=\langle 0,6,6\rangle\). A vector \(\langle a,b,c\rangle\) orthogonal to both satisfies \(-6a+6b=0\) and \(6b+6c=0\). Choose \(\mathbf{n}=\langle 1,1,-1\rangle\). 3. Using point \(A\), the plane equation is \(x+y-z=6\). 4. The distance from the origin is \(\frac{\lvert -6\rvert}{\sqrt{1^2+1^2+(-1)^2}}=2\sqrt{3}\). 5. Using base \(OAB\), the height from \(C\) is \(6\), so \(V=\frac{1}{3}(18)(6)=36\). 6. For a second method, the area of triangle \(ABC\) is \(\frac{1}{2}\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=18\sqrt{3}\). Using the height from \(O\), \(V=\frac{1}{3}(18\sqrt{3})(2\sqrt{3})=36\).

a) \(18\) square units b) \(E_{ABC}: x+y-z=6\) c) \(2\sqrt{3}\) units d) \(36\) cubic units by both methods

- Notice that \(A\), \(B\), and \(C\) have the same z-coordinate. - Use \(V=\frac{1}{3}Gh\) with face \(ABC\) as the base first. - For face \(ABD\), separate the part of \(\overrightarrow{AD}\) perpendicular to \(\overrightarrow{AB}\), then reuse the same volume with a different base.

1. \(\overrightarrow{AB}=\langle 4,0,0\rangle\) and \(\overrightarrow{AC}=\langle 0,6,0\rangle\) are perpendicular, so the area of \(ABC\) is \(\frac{1}{2}(4)(6)=12\) square units. 2. Plane \(ABC\) is \(z=2\). Therefore, the distance from \(D(3,3,6)\) to the plane is \(6-2=4\) units. 3. The volume is \(V=\frac{1}{3}(12)(4)=16\) cubic units. 4. For face \(ABD\), use \(\overrightarrow{AB}=\langle 4,0,0\rangle\) as the base. The component of \(\overrightarrow{AD}=\langle 2,2,4\rangle\) perpendicular to \(\overrightarrow{AB}\) is \(\langle 0,2,4\rangle\), with length \(2\sqrt{5}\). Thus the area of \(ABD\) is \(\frac{1}{2}(4)(2\sqrt{5})=4\sqrt{5}\) square units. 5. Let \(h\) be the distance from \(C\) to plane \(ABD\). From \(16=\frac{1}{3}(4\sqrt{5})h\), \(h=\frac{12}{\sqrt{5}}=\frac{12\sqrt{5}}{5}\) units.

a) Area of \(ABC\): \(12\) square units; distance from \(D\) to plane \(ABC\): \(4\) units b) \(V=16\) cubic units c) Area of \(ABD\): \(4\sqrt{5}\) square units d) Distance from \(C\) to plane \(ABD\): \(\frac{12\sqrt{5}}{5}\) units

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

1. A normal vector to \(E\) is \(\mathbf{n}=\langle 0,-1,1\rangle\), so a coordinate equation is \(-y+z=1\). 2. Substituting \(g\) into the plane equation gives \(-\lambda+2=1\), so \(\lambda=1\). Thus \(S=(2, 1, 2)\). 3. The line direction is \(\mathbf{v}=\langle 1,1,0\rangle\). Hence \(\sin\alpha=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\|\mathbf{n}\|}=\frac{1}{2}\), so \(\alpha=30^\circ\). 4. Project the point \(A=(1, 0, 2)\) onto \(E\) along the normal. The perpendicular line is \(\mathbf{x}=\langle 1,0,2\rangle+t\langle 0,-1,1\rangle\). Substitution into \(-y+z=1\) gives \(t=-\frac{1}{2}\), so the projection is \(A'=\left(1, \frac{1}{2}, \frac{3}{2}\right)\). 5. The projected line passes through \(S\) and \(A'\). A direction vector is \(\langle 2,1,1\rangle\), so one equation is \(h:\mathbf{x}=\langle 2,1,2\rangle+\sigma\langle 2,1,1\rangle\).

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

- Find a vector orthogonal to both direction vectors in the plane. - Use the point-to-plane distance formula. - Find the perpendicular foot from \(Q\) to the plane. - The foot is the midpoint of \(Q\) and its reflection.

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle -2,4,0\rangle\) and \(\overrightarrow{AC}=\langle 0,4,-2\rangle\). 2. A normal vector \(\mathbf{n}=\langle a,b,c\rangle\) satisfies \(-2a+4b=0\) and \(4b-2c=0\). Choose \(\mathbf{n}=\langle 2,1,2\rangle\). 3. Using point \(A\), the plane equation is \(2x+y+2z=6\). 4. The distance from \(Q\) is \(d(Q,E)=\frac{|2(5)+5+2(9)-6|}{3}=9\). 5. The perpendicular foot is \(F=Q-\frac{2(5)+5+2(9)-6}{\|\mathbf{n}\|^2}\mathbf{n}=Q-3\mathbf{n}=(-1,2,3)\). 6. Since \(F\) is the midpoint of \(\overline{QQ'}\), \(Q'=2F-Q=(-7,-1,-3)\).

a) \(E:2x+y+2z=6\) b) \(9\) units c) \(Q'=(-7,-1,-3)\)

- Substitute the line into the plane to find the fixed intersection point. - Reflect one additional point along a line parallel to the plane normal. - Use the fixed point and reflected point to construct the image line. - Apply the point-to-plane distance formula to a general point of the line.

1. Substitute the line into the plane: \(2\lambda-(2+\lambda)+2=6\). Thus \(\lambda=6\), so \(D=(6, 8, 1)\). 2. Reflect the initial point \(A=(0, 2, 1)\). A perpendicular line through \(A\) has direction \(\mathbf{n}=\langle 2,-1,2\rangle\): \(\mathbf{r}(t)=\langle 0,2,1\rangle+t\mathbf{n}\). 3. Its intersection with \(H\) occurs when \(9t=6\), so \(t=\frac23\). The reflected point is twice as far along the perpendicular line, at \(t=\frac43\): \(A'=\left(\frac83,\frac23,\frac{11}{3}\right)\). 4. The reflected line passes through fixed point \(D\) and \(A'\). A direction vector is \(D-A'=\frac13\langle 10,22,-8\rangle\), so one equation is \(f':\mathbf{r}(\mu)=\langle 6,8,1\rangle+\mu\langle 5,11,-4\rangle\). 5. A general point on \(f\) is \((\lambda,2+\lambda,1)\). Its distance from \(H\) is \(\frac{|2\lambda-(2+\lambda)+2-6|}{\sqrt{2^2+(-1)^2+2^2}}=\frac{|\lambda-6|}{3}\). 6. Set this equal to \(3\): \(|\lambda-6|=9\), so \(\lambda=15\) or \(\lambda=-3\). 7. The points are \(P_1=(15, 17, 1)\) and \(P_2=(-3, -1, 1)\).

a) \(D=(6, 8, 1)\) b) \(f':\mathbf{r}(\mu)=\langle 6,8,1\rangle+\mu\langle 5,11,-4\rangle\) c) \(P_1=(15, 17, 1)\) and \(P_2=(-3, -1, 1)\)

- A perpendicular-bisector plane passes through the segment midpoint and has a normal parallel to the segment. - A plane perpendicular to a line can use the line direction as its normal. - Use a dot product to prove the right angle. - Under a half-turn about a line, the perpendicular foot is the midpoint of a point and its image.

1. The midpoint of \(\overline{AB}\) is \(M=(4, 2, -1)\). A normal to the perpendicular-bisector plane is parallel to \(\overrightarrow{AB}=\langle -2,4,-4\rangle\), so use \(\mathbf{n}_E=\langle 1,-2,2\rangle\). 2. Using \(M\), the equation is \(x-2y+2z=-2\). 3. A general point of \(g\) is \((2\lambda,1+2\lambda,\lambda)\). Substitution gives \(2\lambda-2(1+2\lambda)+2\lambda=-2\), so every point of \(g\) lies in \(E\). 4. A plane perpendicular to \(g\) can use the line direction \(\langle 2,2,1\rangle\) as its normal. Using \(A\), \(F:2x+2y+z=11\). Point \(B\) also satisfies this equation. 5. Substitute \(g\) into \(F\): \(4\lambda+2+4\lambda+\lambda=11\), so \(\lambda=1\) and \(S=(2, 3, 1)\). 6. The vectors \(\overrightarrow{SA}=\langle 3,-3,0\rangle\) and \(\overrightarrow{SB}=\langle 1,1,-4\rangle\) have dot product \(3-3=0\), so \(\angle ASB=90^\circ\). 7. Point \(S\) is the foot of the perpendicular from \(A\) to the rotation axis \(g\). Under a half-turn about \(g\), \(S\) is the midpoint of \(A\) and \(A'\). 8. Therefore \(A'=2S-A=\langle 4,6,2\rangle-\langle 5,0,1\rangle=(-1, 6, 1)\).

a) \(E:x-2y+2z=-2\) b) \(g\subset E\); \(F:2x+2y+z=11\) c) \(S=(2, 3, 1)\), and \(\angle ASB=90^\circ\) d) \(A'=(-1, 6, 1)\)