Free math worksheets

- Each coordinate plane is characterized by one coordinate being zero. - Identify which coordinate of \(P\) is zero. - Match that zero coordinate to the corresponding coordinate plane.

A point lies in the \(yz\)-plane when its \(x\)-coordinate is \(0\). Since \(P\) has \(x=0\), it lies in the \(yz\)-plane.

The \(yz\)-plane

- Compare the three coordinates of the two points. - Notice which coordinates stay unchanged. - When only one coordinate changes, the distance is the absolute change in that coordinate.

The points have the same \(x\)- and \(y\)-coordinates, so their distance is the change in the \(z\)-coordinate: \(\lvert 9-5\rvert=4\).

\(4\) units

- Moving from \(O\) to the opposite vertex uses all three edge lengths. - The opposite vertex's coordinates are the signed changes in \(x\), \(y\), and \(z\). - Then apply the three-dimensional distance formula.

1. The displayed side length is \(4\), so moving from \(O\) to the opposite vertex changes each coordinate by \(4\). Thus \(P=(4,4,4)\). 2. Use the three-dimensional distance formula: \(OP=\sqrt{4^2+4^2+4^2}=\sqrt{48}=4\sqrt3.\)

\(P=(4,4,4)\) and \(OP=4\sqrt3\) units.

- A centered base places each coordinate at half of the corresponding base dimension. - The apex is directly above the origin. - Use the three-dimensional distance formula between the two points.

1. The displayed square base has side length \(6\), so each coordinate from the center to a side is \(3\). Thus \(C=(3,3,0)\). 2. The displayed height is \(4\), so \(A=(0,0,4)\). 3. Therefore, \(AC=\sqrt{(3-0)^2+(3-0)^2+(0-4)^2}=\sqrt{9+9+16}=\sqrt{34}.\)

\(C=(3,3,0)\), \(A=(0,0,4)\), and \(AC=\sqrt{34}\) units.

- Use the origin as the second point. - Square the changes in all three coordinates and add them.

1. The origin is \(O=(0,0,0)\). 2. Apply the three-dimensional distance formula: \(OP=\sqrt{(-2)^2+3^2+6^2}=\sqrt{49}=7.\)

\(7\) units.

- How are a point's coordinates related to its position vector? - To find a vector from one point to another, which coordinates do you subtract? - What formula gives the magnitude of a vector in three dimensions?

1. The position vector uses the coordinates of \(A\): \(\overrightarrow{OA}=\langle 5, -2, 1\rangle\). 2. Subtract the coordinates of \(A\) from those of \(B\): \(\overrightarrow{AB}=\langle 7-5, 1-(-2), 7-1\rangle=\langle 2, 3, 6\rangle\). 3. Find the magnitude: \(\lVert\overrightarrow{AB}\rVert=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).

\(\overrightarrow{OA}=\langle 5, -2, 1\rangle\), \(\overrightarrow{AB}=\langle 2, 3, 6\rangle\), and \(\lVert\overrightarrow{AB}\rVert=7\)

- 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. - Substitute \((0, 0, 0)\) into the general equation.

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

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

- Where do the diagonals of a rectangle intersect? - What equation describes the yz-plane? - How does a point's x-coordinate determine its distance from the yz-plane?

1. The diagonals of a rectangle bisect each other, so their intersection is the midpoint of \(A\) and \(C\): \(M=\left(\frac{10+2}{2},\frac{2+8}{2},\frac{3+9}{2}\right)=(6,5,6)\). 2. The yz-plane has equation \(x=0\). 3. The perpendicular distance from \(M=(6,5,6)\) to the yz-plane is the absolute value of its x-coordinate, \(\lvert 6\rvert=6\). 4. Therefore, the support rod must be \(6\) units long.

\(6\) units

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

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

\(5\) units

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

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

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

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

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

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

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

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

\(3\) units

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

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

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

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

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

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

- 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 \(\sqrt{2^2+0^2+1^2}=\sqrt{5}\approx 2.24\,\text{m}\).

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

- First find the distance from the center to the given point. - A sphere consists of all points a fixed distance from its center. - Square the radius in the coordinate equation.

1. The radius is the distance from \(M\) to \(P\): \(r=\sqrt{(7-3)^2+(2-(-1))^2+(4-4)^2}=5\). 2. In vector form, the sphere is \(\lVert\mathbf{x}-\langle 3,-1,4\rangle\rVert=5\). 3. In coordinate form, the sphere is \((x-3)^2+(y+1)^2+(z-4)^2=25\).

Vector form: \(\lVert\mathbf{x}-\langle 3,-1,4\rangle\rVert=5\) Coordinate form: \((x-3)^2+(y+1)^2+(z-4)^2=25\)

- Read the center and radius from the equation. - Compare each point's squared distance from the center with \(r^2\). - You do not need to take square roots.

1. The center is \(M(1,-4,2)\), and \(r^2=49\). 2. For \(A\), the squared distance to \(M\) is \(0^2+6^2+(-3)^2=45\). Since \(45<49\), \(A\) is inside. 3. For \(B\), the squared distance is \(6^2+(-2)^2+3^2=49\). Therefore, \(B\) is on the sphere. 4. For \(C\), the squared distance is \((-4)^2+4^2+6^2=68\). Since \(68>49\), \(C\) is outside.

\(A\) is inside; \(B\) is on the sphere; \(C\) is outside.

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

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

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

- The radius to a tangent point is perpendicular to the tangent plane. - There is one possible center on each side of the plane. - Use the standard sphere equation.

1. The center must lie on the line through \(B\) perpendicular to the xy-plane. 2. Because the radius is \(3\), the center is \(3\) units above or below the plane. Thus, the possible centers are \(M_1(2,1,3)\) and \(M_2(2,1,-3)\). 3. Using standard form, the two sphere equations are \((x-2)^2+(y-1)^2+(z-3)^2=9\) and \((x-2)^2+(y-1)^2+(z+3)^2=9\).

\((x-2)^2+(y-1)^2+(z-3)^2=9\) and \((x-2)^2+(y-1)^2+(z+3)^2=9\)

- The center is the midpoint of the diameter. - The radius is half the diameter length. - Substitute \(P\) into the sphere equation.

1. The center is the midpoint of \(\overline{AB}\): \(M=\left(\frac{1+7}{2},\frac{1+9}{2},\frac{1+1}{2}\right)=(4,5,1)\). 2. The squared radius is \(MA^2=(4-1)^2+(5-1)^2+(1-1)^2=25\), so \(r=5\). 3. Vector form is \(\lVert\mathbf{x}-\langle 4,5,1\rangle\rVert^2=25\). 4. Coordinate form is \((x-4)^2+(y-5)^2+(z-1)^2=25\). 5. For \(P\), \((8-4)^2+(2-5)^2+(1-1)^2=25\), so \(P\) lies on the sphere.

a) Vector form: \(\lVert\mathbf{x}-\langle 4,5,1\rangle\rVert^2=25\); coordinate form: \((x-4)^2+(y-5)^2+(z-1)^2=25\) b) Yes, \(P\) lies on the sphere.

- 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.

- How do you find the distance between two points in three dimensions? - Apply the distance formula to each pair of vertices. - You can compare exact square roots by comparing their radicands.

1. Find \(AB\): \(AB=\sqrt{(4-1)^2+(2-2)^2+(7-3)^2}=\sqrt{3^2+0^2+4^2}=5\). 2. Find \(AC\): \(AC=\sqrt{(1-1)^2+(8-2)^2+(3-3)^2}=\sqrt{0^2+6^2+0^2}=6\). 3. Find \(BC\): \(BC=\sqrt{(1-4)^2+(8-2)^2+(3-7)^2}=\sqrt{9+36+16}=\sqrt{61}\approx 7.81\). 4. Therefore, \(AB<AC<BC\).

The side lengths are \(AB=5\), \(AC=6\), and \(BC=\sqrt{61}\approx 7.81\). From least to greatest: \(AB<AC<BC\).

- 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 \((0, 4, 0)\).

- Which coordinate changes sign when a point is reflected across the \(yz\)-plane? - What happens to all three coordinates under a reflection through the origin? - Find a midpoint by averaging corresponding coordinates.

1. Reflecting across the \(yz\)-plane changes the sign of the \(x\)-coordinate and leaves the other coordinates unchanged. Thus, \(A'=(-5, -4, 2)\). 2. Reflecting through the origin changes the sign of every coordinate. Thus, \(B'=(1, -6, 3)\). 3. Average corresponding coordinates: \(M=\left(\frac{-5+1}{2}, \frac{-4+(-6)}{2}, \frac{2+3}{2}\right)=(-2, -5, 2.5)\).

1. \(A'=(-5, -4, 2)\) 2. \(B'=(1, -6, 3)\) 3. \(M=(-2, -5, 2.5)\)

- For each coordinate plane, identify the coordinate perpendicular to that plane. - Record the coordinates after each reflection. - Compare the signs of the original and final coordinates.

1. Reflection across the \(xy\)-plane changes the sign of \(z\), so \(P_1=(3, 8, 4)\). 2. Reflection across the \(xz\)-plane changes the sign of \(y\), so \(P_2=(3, -8, 4)\). 3. Reflection across the \(yz\)-plane changes the sign of \(x\), so \(P_3=(-3, -8, 4)\). 4. Every coordinate of \(P_3\) is the opposite of the corresponding coordinate of \(P\). Therefore, one reflection through the origin maps \(P\) to \(P_3\).

1. \(P_1=(3, 8, 4)\) 2. \(P_2=(3, -8, 4)\) 3. \(P_3=(-3, -8, 4)\) 4. A reflection through the origin.

- For a reflection across a coordinate plane, only the coordinate perpendicular to the plane changes sign. - A reflection through the origin changes the sign of every coordinate. - Under a \(180^\circ\) rotation about an axis, the coordinate on that axis stays fixed. - Compare the coordinates one position at a time.

1. Across the \(yz\)-plane, the \(y\)- and \(z\)-coordinates stay the same while the \(x\)-coordinate changes sign. Therefore, \(A\) and \(B\) are reflections of each other, and \(C\) and \(D\) are reflections of each other. 2. A reflection through the origin changes the sign of every coordinate. Since \(-D=(5, -2, 4)\), point \(A\) is the image of \(D\). 3. A \(180^\circ\) rotation about the \(y\)-axis keeps the \(y\)-coordinate and changes the signs of \(x\) and \(z\). Therefore, \(E=(-5, -2, -4)\).

a) \(A\) and \(B\), and \(C\) and \(D\). b) \(A\); each coordinate of \(A\) is the opposite of the corresponding coordinate of \(D\). c) \(E=(-5, -2, -4)\)

- Subtract coordinates to form each side vector. - Use the three-dimensional distance formula. - Compare the resulting side lengths.

1. The connecting vectors are \(\overrightarrow{PQ}=\begin{pmatrix}3\\0\\4\end{pmatrix}\), \(\overrightarrow{QR}=\begin{pmatrix}-3\\4\\-1\end{pmatrix}\), and \(\overrightarrow{PR}=\begin{pmatrix}0\\4\\3\end{pmatrix}\). 2. Their lengths are \(PQ=5\), \(QR=\sqrt{26}\), and \(PR=5\). 3. Since \(PQ=PR\), the triangle is isosceles. It is not equilateral because \(\sqrt{26}\ne5\).

a) \(PQ=5\), \(QR=\sqrt{26}\), \(PR=5\) b) The triangle is isosceles but not equilateral.

- Compare corresponding coordinates. - Find the midpoint of the segment joining the points. - Which coordinate plane contains that midpoint and lies between the points? - Recall the coordinate rule for reflection across the \(xy\)-plane.

1. The \(x\)- and \(y\)-coordinates of \(A\) and \(B\) are equal, while their \(z\)-coordinates are opposites. 2. Their midpoint is \(M=(7, 2, 0)\), which lies in the \(xy\)-plane. 3. The segment \(\overline{AB}\) is perpendicular to the \(xy\)-plane, and the plane bisects the segment at \(M\). Therefore, \(A\) and \(B\) are reflections of each other across the \(xy\)-plane. 4. The same reasoning applies to \(P(x, y, z)\) and \(Q(x, y, -z)\): changing only the sign of \(z\) is reflection across the \(xy\)-plane.

a) \(A\) and \(B\) are reflections of each other across the \(xy\)-plane. b) Any points \(P(x, y, z)\) and \(Q(x, y, -z)\) are reflections of each other across the \(xy\)-plane.

- Identify the coordinate that remains unchanged. - Project the points onto the \(xy\)-plane and compare them with the origin. - What rotation changes the signs of two coordinates while leaving the third unchanged? - Use the coordinate rule for a rotation about the \(z\)-axis.

1. The points have the same \(z\)-coordinate, and their \(x\)- and \(y\)-coordinates are opposites. 2. Both points lie in the plane \(z=6\). Their midpoint is \((0, 0, 6)\), a point on the \(z\)-axis. 3. A \(180^\circ\) rotation about the \(z\)-axis changes \((x, y, z)\) to \((-x, -y, z)\). Therefore, this rotation maps \(C\) to \(D\) and, in general, maps \(M\) to \(N\).

a) \(C\) and \(D\) are related by a \(180^\circ\) rotation about the \(z\)-axis. b) The transformation is \((x, y, z)\mapsto(-x, -y, z)\), a \(180^\circ\) rotation about the \(z\)-axis.

- A fixed distance from one point defines a sphere. - Substitute each test point into the equation. - In part c), replace the fixed coordinates before solving for y.

1. The distance condition is \((x-2)^2+(y+3)^2+(z-5)^2=49\). 2. For \(P(9,-3,5)\), the left side is \(7^2=49\), so \(P\) satisfies the condition. 3. For \(Q(6,0,5)\), the left side is \(4^2+3^2=25\), so \(Q\) does not satisfy it. 4. In part c, set \(x=2\) and \(z=5\). Then \((y+3)^2=49\), so \(y=4\) or \(y=-10\). 5. The required points are \((2,4,5)\) and \((2,-10,5)\).

a) \((x-2)^2+(y+3)^2+(z-5)^2=49\) b) \(P\) satisfies the condition; \(Q\) does not. c) \((2,4,5)\) and \((2,-10,5)\)

- An equilateral triangle has three equal side lengths. - Start with the known length \(OA\). - Compare the squared distances \(OB^2\) and \(AB^2\) with \(OA^2\). - Include both signs when solving for \(z\).

1. \(OA=6\). 2. Also, \(OB^2=3^2+3^2+z^2=18+z^2\). 3. For \(OB=6\), \(18+z^2=36\), so \(z^2=18\). 4. Thus, \(z=\pm3\sqrt{2}\). 5. Since \(AB^2=(-3)^2+3^2+z^2=18+z^2\), the same values also give \(AB=6\). Therefore, all three sides are equal.

\(z=3\sqrt{2}\) or \(z=-3\sqrt{2}\)

- What do you notice about the \(x\)-coordinates? - Which two coordinates must be \(0\) for a point to lie on a coordinate axis? - What plane is described by \(x=0\)? - Determine which coordinate changes along the segment.

1. Average corresponding coordinates: \(M=\left(\frac{0+0}{2}, \frac{5+(-5)}{2}, \frac{2+2}{2}\right)=(0, 0, 2)\). 2. Since \(x=0\) for both endpoints, \(P\) and \(Q\) lie in the \(yz\)-plane. Since both \(x\) and \(y\) are \(0\) at \(M\), the midpoint lies on the \(z\)-axis. 3. Only the \(y\)-coordinate changes along the segment, while \(x=0\) and \(z=2\) remain constant. Thus, the segment lies in the \(yz\)-plane and is parallel to the \(y\)-axis.

1. \(M=(0, 0, 2)\) 2. \(P\) and \(Q\) lie in the \(yz\)-plane, and \(M\) lies on the \(z\)-axis. 3. \(\overline{PQ}\) lies in the \(yz\)-plane and is parallel to the \(y\)-axis.

- Compare the \(z\)-coordinates of the endpoints. - What does a zero \(z\)-component in a direction vector mean? - What axis contains points whose \(x\)- and \(y\)-coordinates are both \(0\)?

1. Average corresponding coordinates: \(M=\left(\frac{3+(-3)}{2}, \frac{4+(-4)}{2}, \frac{-1+(-1)}{2}\right)=(0, 0, -1)\). 2. The endpoints have the same \(z\)-coordinate. Equivalently, \(\overrightarrow{AB}=\begin{pmatrix}-6\\-8\\0\end{pmatrix}\), whose \(z\)-component is \(0\). Therefore, the segment lies in the plane \(z=-1\), which is parallel to the \(xy\)-plane. 3. Since the \(x\)- and \(y\)-coordinates of \(M\) are \(0\), \(M\) lies on the \(z\)-axis.

1. \(M=(0, 0, -1)\) 2. The segment lies in \(z=-1\), a plane parallel to the \(xy\)-plane. 3. \(M\) lies on the \(z\)-axis.

- 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 \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

- Which coordinate becomes \(0\) when projecting onto each coordinate plane? - Which coordinate measures perpendicular distance from a coordinate plane? - A distance is always nonnegative.

1. To project onto a coordinate plane, set the coordinate perpendicular to that plane equal to \(0\). 2. Therefore, \(P_{xy}=(-4, 7, 0)\), \(P_{xz}=(-4, 0, 12)\), and \(P_{yz}=(0, 7, 12)\). 3. The distance to a coordinate plane is the absolute value of the perpendicular coordinate. 4. The distances are \(12\) to the \(xy\)-plane, \(7\) to the \(xz\)-plane, and \(4\) to the \(yz\)-plane.

a) \(P_{xy}=(-4, 7, 0)\), \(P_{xz}=(-4, 0, 12)\), \(P_{yz}=(0, 7, 12)\) b) \(12\) units to the \(xy\)-plane, \(7\) units to the \(xz\)-plane, and \(4\) units to the \(yz\)-plane.

- 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 \(z=1\)

- A vector in the \(xy\)-plane has no vertical component. - A vector normal to the \(xy\)-plane points along the \(z\)-axis. - Add vectors componentwise, then use the magnitude formula.

1. A vector in the \(xy\)-plane has third component \(0\). Adding a zero component does not change the magnitude, so its three-dimensional magnitude remains \(8\). 2. A vector perpendicular to the \(xy\)-plane is parallel to the \(z\)-axis. One possible vector is \(\mathbf{v}=\langle 0,0,6\rangle\). 3. \(\mathbf{s}=\langle 3,4,12\rangle\), so \(\|\mathbf{s}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

1. Third component \(0\); magnitude \(8\) 2. \(\langle 0,0,6\rangle\) 3. \(13\)

- Compare a point on the first base with its corresponding point on the second base. - Apply the same translation vector to every vertex. - The length of a lateral edge equals the magnitude of the translation vector.

1. Subtract the coordinates of \(A\) from those of \(A'\): \(\mathbf{v}=\overrightarrow{AA'}=\langle 0,0,8\rangle\). 2. Translate \(B\) and \(C\) by \(\mathbf{v}\): \(B'=(6,1,9)\) and \(C'=(2,4,9)\). 3. A lateral edge has length \(\|\mathbf{v}\|=\sqrt{0^2+0^2+8^2}=8\).

a) \(\mathbf{v}=\langle 0,0,8\rangle\) b) \(B'(6,1,9)\) and \(C'(2,4,9)\) c) \(8\) units

- The center of a point reflection is the midpoint of a point and its image. - Rearrange the midpoint formula to solve for the unknown endpoint. - You can also think of continuing the same displacement beyond the midpoint.

1. Since \(S\) is the midpoint of \(\overline{AA'}\), \(A'=2S-A\). 2. Thus, \(A'=2(1,3,-5)-(4,-1,2)=(-2,7,-12)\). 3. Since \(A\) is the midpoint of \(\overline{SB}\), \(A=\frac{S+B}{2}\), so \(B=2A-S\). 4. Therefore, \(B=2(4,-1,2)-(1,3,-5)=(7,-5,9)\).

a) \(A'(-2,7,-12)\) b) \(B(7,-5,9)\)

- The reflection center is the midpoint of each point and its image. - Use \(X'=2Z-X\) for each vertex. - Apply the same rule independently to all three coordinates.

1. For a point reflection across \(Z\), each image satisfies \(X'=2Z-X\). 2. \(A'=2(2,2,1)-(1,3,0)=(3,1,2)\). 3. \(B'=2(2,2,1)-(4,1,2)=(0,3,0)\). 4. \(C'=2(2,2,1)-(2,5,3)=(2,-1,-1)\).

\(A'(3,1,2)\), \(B'(0,3,0)\), and \(C'(2,-1,-1)\)

- Use the midpoint formula in three dimensions. - Rearrange the formula to isolate the unknown endpoint. - The displacement from \(P\) to \(M\) equals the displacement from \(M\) to \(Q\).

1. The midpoint relation is \(M=\frac{P+Q}{2}\). 2. Solve for \(Q\): \(Q=2M-P\). 3. Therefore, \(Q=2(5, 1, 2)-(2, -3, 5)=(8, 5, -1)\).

\(Q(8, 5, -1)\)

- Recall the relationship between the diagonals of a parallelogram. - Apply the midpoint formula to the known diagonal. - Decide whether one complete diagonal is enough to locate the intersection.

1. The diagonals of a parallelogram bisect each other, so \(S\) is the midpoint of \(\overline{AC}\). 2. \(S=\left(\frac{1+7}{2},\frac{4+0}{2},\frac{-2+6}{2}\right)=(4, 2, 2)\). 3. One diagonal is enough because the common intersection is the midpoint of each diagonal. Since \(A\) and \(C\) determine \(\overline{AC}\), they also determine \(S\).

a) \(S(4, 2, 2)\) b) The diagonals of a parallelogram bisect each other, so the midpoint of \(\overline{AC}\) determines \(S\) without \(B\) or \(D\).

- A \(180^\circ\) rotation about the \(y\)-axis keeps the \(y\)-coordinate fixed. - A reflection through the origin changes all three signs. - Compare one coordinate position at a time.

1. A \(180^\circ\) rotation about the \(y\)-axis keeps \(y\) unchanged and changes the signs of \(x\) and \(z\): \(P'=(-2,3,-4)\), \(Q'=(1,0,-5)\), and \(R'=(0,-2,-1)\). 2. A reflection through the origin changes the sign of every coordinate: \(P''=(-2,-3,-4)\), \(Q''=(1,0,-5)\), and \(R''=(0,2,-1)\). 3. In both transformations, the \(x\)- and \(z\)-coordinates change sign. The \(y\)-coordinate stays the same in part a) but changes sign in part b).

a) \(P'=(-2,3,-4)\), \(Q'=(1,0,-5)\), \(R'=(0,-2,-1)\) b) \(P''=(-2,-3,-4)\), \(Q''=(1,0,-5)\), \(R''=(0,2,-1)\) c) \(x\) and \(z\) change sign in both; \(y\) stays fixed in a) and changes sign in b).

- Identify the coordinate perpendicular to each reflecting plane. - A reflection across a coordinate plane changes only the sign of the perpendicular component. - Compare the original and final vectors component by component.

1. Reflection across the \(xy\)-plane changes the sign of the \(z\)-component: \(\mathbf{r}_1=\langle 4,-3,1\rangle\). 2. Reflection across the \(yz\)-plane changes the sign of the \(x\)-component: \(\mathbf{r}_2=\langle -4,-3,1\rangle\). 3. Relative to \(\mathbf{r}\), the \(x\)- and \(z\)-components changed sign, while the \(y\)-component did not. 4. This combined transformation is equivalent to a \(180^\circ\) rotation about the \(y\)-axis.

a) \(\mathbf{r}_1=\langle 4,-3,1\rangle\) b) \(\mathbf{r}_2=\langle -4,-3,1\rangle\) c) The \(x\)- and \(z\)-components change sign. The result is equivalent to a \(180^\circ\) rotation about the \(y\)-axis.

- A coordinate-plane reflection changes the component perpendicular to that plane. - Under a \(180^\circ\) rotation about the \(y\)-axis, which component remains unchanged? - Apply the transformations in order.

1. Reflection across the \(xy\)-plane changes the sign of the \(z\)-component: \(\mathbf{r}_1=\langle 4,-1,-3\rangle\). 2. A \(180^\circ\) rotation about the \(y\)-axis changes the signs of the \(x\)- and \(z\)-components while leaving the \(y\)-component unchanged. 3. Therefore, \(\mathbf{r}_2=\langle -4,-1,3\rangle\).

a) \(\mathbf{r}_1=\langle 4,-1,-3\rangle\) b) \(\mathbf{r}_2=\langle -4,-1,3\rangle\)

- 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 \((13,-17,-14)\)

- Use the distance formula in three dimensions for each segment. - First find the difference between corresponding coordinates. - Compare the two distances after calculating them.

1. Find \(AB\): \(AB=\sqrt{(5-2)^2+(1-(-3))^2+(5-5)^2}=\sqrt{3^2+4^2+0^2}=5\). 2. Find \(BC\): \(BC=\sqrt{(9-5)^2+(4-1)^2+(-7-5)^2}=\sqrt{4^2+3^2+(-12)^2}=13\). 3. Since \(13>5\), \(\overline{BC}\) is longer than \(\overline{AB}\).

\(AB=5\) units and \(BC=13\) units, so \(\overline{BC}\) is longer.

- What form do the coordinates of a point on the z-axis have? - Substitute that form into the distance formula. - Isolate the squared expression before taking square roots. - Taking a square root may produce two cases.

1. A point on the z-axis has the form \(P(0, 0, z)\). 2. Apply the distance formula: \(\sqrt{(3-0)^2+(-4-0)^2+(12-z)^2}=13\). 3. Square both sides and simplify: \(9+16+(12-z)^2=169\), so \((12-z)^2=144\). 4. Therefore, \(12-z=12\) or \(12-z=-12\), which gives \(z=0\) or \(z=24\). 5. The possible points are \(P_1(0, 0, 0)\) and \(P_2(0, 0, 24)\).

\(P_1(0, 0, 0)\) and \(P_2(0, 0, 24)\)

- How can you find the displacement vector between two points in space? - The distance between the drones is the magnitude of that vector. - A point's distance from the origin is a special case of the distance formula. - To compare positive square roots, compare their radicands.

1. Subtract the coordinates to find the displacement from \(P_1\) to \(P_2\): \(\overrightarrow{P_1P_2}=\langle 14-10, 17-20, 5-5\rangle=\langle 4, -3, 0\rangle\). 2. Its magnitude is \(\sqrt{4^2+(-3)^2+0^2}=5\), so the drones are \(5\,\text{m}\) apart. 3. The distance from \(P_1\) to the origin is \(\sqrt{10^2+20^2+5^2}=\sqrt{525}=5\sqrt{21}\approx 22.91\,\text{m}\). 4. The distance from \(P_2\) to the origin is \(\sqrt{14^2+17^2+5^2}=\sqrt{510}\approx 22.58\,\text{m}\). 5. Since \(525>510\), drone \(P_1\) is farther from the origin.

a) \(5\,\text{m}\) b) \(P_1\) is \(5\sqrt{21}\approx 22.91\,\text{m}\) from the origin, and \(P_2\) is \(\sqrt{510}\approx 22.58\,\text{m}\) from the origin. Drone \(P_1\) is farther away.

- Write the coordinate differences in terms of \(t\). - Use the three-dimensional distance formula. - How can you eliminate the square root from the equation? - An equation of the form \(u^2=1\) has two solutions.

1. The coordinate differences from \(A\) to \(B_t\) are \(\langle 2-2, t-(-1), 1-5\rangle=\langle 0, t+1, -4\rangle\). 2. Set the distance equal to the given value: \(\sqrt{(t+1)^2+16}=\sqrt{17}\). 3. Square both sides: \((t+1)^2+16=17\), so \((t+1)^2=1\). 4. Thus, \(t+1=1\) or \(t+1=-1\), giving \(t=0\) or \(t=-2\).

\(t=0\) or \(t=-2\)

- Use the three-dimensional distance formula for each pair of adjacent vertices. - The perimeter is the sum of all three side lengths. - Keep radical lengths exact until the final step.

1. Find \(AB\): \(AB=\sqrt{(3-0)^2+(4-0)^2+(12-0)^2}=\sqrt{169}=13\). 2. Find \(BC\): \(BC=\sqrt{(-4-3)^2+(3-4)^2+(0-12)^2}=\sqrt{49+1+144}=\sqrt{194}\). 3. Find \(CA\): \(CA=\sqrt{(0-(-4))^2+(0-3)^2+(0-0)^2}=5\). 4. Add the side lengths: \(13+\sqrt{194}+5=18+\sqrt{194}\approx 31.93\).

\(18+\sqrt{194}\approx 31.93\) units

- Apply the three-dimensional distance formula to each side. - What equality among side lengths makes a triangle isosceles? - What additional equality would make it equilateral? - Compute coordinate differences before squaring.

1. Find \(AB\): \(AB=\sqrt{(1-5)^2+(5-2)^2+(1-1)^2}=\sqrt{16+9}=5\). 2. Find \(BC\): \(BC=\sqrt{(1-1)^2+(2-5)^2+(5-1)^2}=\sqrt{9+16}=5\). 3. Find \(AC\): \(AC=\sqrt{(1-5)^2+(2-2)^2+(5-1)^2}=\sqrt{32}=4\sqrt{2}\approx 5.66\). 4. Since \(AB=BC\), the triangle is isosceles. It is not equilateral because \(AC\ne 5\).

a) \(AB=5\), \(BC=5\), and \(AC=4\sqrt{2}\approx 5.66\) b) The triangle is isosceles but not equilateral because exactly two side lengths are equal.

- Calculate \(PQ\), \(QR\), and \(PR\). - What do three different side lengths imply about the triangle? - Be careful with signs when finding coordinate differences.

1. Find \(PQ\): \(PQ=\sqrt{(3-1)^2+(4-0)^2+(6-2)^2}=\sqrt{36}=6\). 2. Find \(QR\): \(QR=\sqrt{(2-3)^2+(3-4)^2+(6-6)^2}=\sqrt{2}\). 3. Find \(PR\): \(PR=\sqrt{(2-1)^2+(3-0)^2+(6-2)^2}=\sqrt{26}\). 4. Because \(6\), \(\sqrt{2}\), and \(\sqrt{26}\) are all different, the triangle is not isosceles.

The triangle is not isosceles because its side lengths are \(6\), \(\sqrt{2}\), and \(\sqrt{26}\), which are all different.

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

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

- 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 \(2x-3y+2z=3\).

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

- 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 \(r\in\mathbb{R}\)

- 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 \(S=(8,4,5)\).

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

- 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.

- 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. a) Multiplying the equation of \(F\) by \(-2\) gives \(3x+4z=12\), which is exactly the equation of \(E\). Therefore the planes are identical. 2. b) A plane parallel to \(E\) can use the normal vector \(\langle 3,0,4\rangle\), so write \(3x+4z=d\). 3. The origin gives \(d=0\), so \(G:3x+4z=0\). 4. c) The family of all parallel planes is \(3x+4z=d\), where \(d\in\mathbb{R}\).

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

- 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.

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

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

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

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

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

- Trace a path from \(A\) to the opposite vertex along three edges. - Add each side length to the corresponding coordinate. - Match each edge direction with its coordinate axis.

1. Moving from \(A\) to the opposite vertex adds \(6\) to the \(x\)-coordinate, \(4\) to the \(y\)-coordinate, and \(3\) to the \(z\)-coordinate. 2. Therefore, \(x_G=2+6=8\), \(y_G=1+4=5\), and \(z_G=0+3=3\). 3. Thus, \(G=(8, 5, 3)\).

\(G=(8, 5, 3)\)

- 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, \(A'=2S-A=(6, 8, 10)-(1, 2, 3)=(5, 6, 7)\).

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

- Every point on a sphere is one radius from its center. - Use the three-dimensional distance formula. - A squared equation can produce two values.

1. Because \(P\) lies on the sphere, its distance from \(M\) is \(9\). 2. Set up the distance equation: \((x-2)^2+(5-1)^2+(12-4)^2=9^2\). 3. Simplify: \((x-2)^2+16+64=81\), so \((x-2)^2=1\). 4. Therefore, \(x-2=1\) or \(x-2=-1\), giving \(x=3\) or \(x=1\).

\(x=1\) or \(x=3\)

- 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) \(2\sqrt{5}\)

- 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) \(500\,\text{m}\)

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

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

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

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

1. A normal vector to the plane is \(\mathbf{n}=\langle2,-2,1\rangle\). 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) \(B'(3,0,5)\)

- Find the coordinates of the top of the pole. - Write the sunlight line through the top. - Find where that line intersects the plane. - The shadow length is the distance from the pole's base to that intersection point.

1. The top of the pole is \(Q(2,4,9)\). 2. The sunlight line through \(Q\) is \(\mathbf{r}(t)=\langle 2,4,9\rangle+t\langle 1,2,-2\rangle\). 3. Substitute this line into the plane equation: \((2+t)+2(4+2t)+2(9-2t)=20\). 4. Simplifying gives \(28+t=20\), so \(t=-8\). 5. The shadow point of the pole's top is \(S=\langle 2,4,9\rangle-8\langle 1,2,-2\rangle=\langle -6,-12,25\rangle\). 6. The shadow vector is \(\overrightarrow{PS}=\langle -8,-16,20\rangle\). Its length is \(\sqrt{(-8)^2+(-16)^2+20^2}=\sqrt{720}=12\sqrt{5}\,\text{m}\).

\(12\sqrt{5}\,\text{m}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- Complete the square in the x- and y-terms. - Compare the point's squared distance from the center with \(r^2\). - A larger squared distance means the point is outside.

1. Complete the square: \(x^2+6x=(x+3)^2-9\) and \(y^2-4y=(y-2)^2-4\). 2. The equation becomes \((x+3)^2+(y-2)^2+z^2=25\). 3. Therefore, the center is \(M(-3,2,0)\) and the radius is \(5\). 4. For \(A\), the squared distance from the center is \((1+3)^2+(0-2)^2+(4-0)^2=36\). 5. Since \(36>25\), point \(A\) lies outside the sphere.

a) Center \(M(-3,2,0)\); radius \(5\) b) \(A\) lies outside the sphere.

- Complete the square in all three variables. - Compare the center-to-point distance with the radius. - For an exterior point, subtract the radius from the center-to-point distance.

1. Complete the square to obtain \((x+3)^2+(y-2)^2+(z+1)^2=25\). 2. The center is \(M(-3,2,-1)\), and the radius is \(5\). 3. The distance from \(P\) to \(M\) is \(\sqrt{(9+3)^2+(7-2)^2+(-1+1)^2}=\sqrt{169}=13\). 4. Because \(13>5\), \(P\) is outside the sphere. 5. The shortest distance to the surface is \(13-5=8\).

a) Center \(M(-3,2,-1)\); radius \(5\) b) \(13\) units c) \(P\) is outside; its distance to the surface is \(8\) units.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- A tangent coordinate plane is one radius from the center. - Each of the x- and y-coordinates can have either sign. - The z-coordinate is fixed by the given plane.

1. Tangency to the xz-plane \(y=0\) requires the center's y-coordinate to satisfy \(\lvert y_M\rvert=5\). 2. Tangency to the yz-plane \(x=0\) requires \(\lvert x_M\rvert=5\). 3. The center lies in \(z=10\), so \(z_M=10\). 4. The four possible centers are \((5,5,10)\), \((5,-5,10)\), \((-5,5,10)\), and \((-5,-5,10)\). 5. Substitute each center and \(r^2=25\) into the standard sphere equation.

\((x-5)^2+(y-5)^2+(z-10)^2=25\) \((x-5)^2+(y+5)^2+(z-10)^2=25\) \((x+5)^2+(y-5)^2+(z-10)^2=25\) \((x+5)^2+(y+5)^2+(z-10)^2=25\)

- Compare the squared distance from \(P\) to the center with \(r^2\). - Solve the equality first to find the boundary values. - Use the corresponding quadratic inequality for the inside and outside cases.

1. The squared distance from \(P\) to \(M\) is \(d^2=(4-1)^2+(1-1)^2+(c-1)^2=9+(c-1)^2\). 2. On the sphere, \(9+(c-1)^2=25\), so \((c-1)^2=16\). Thus, \(c=-3\) or \(c=5\). 3. Inside the sphere, \((c-1)^2<16\), which gives \(-3<c<5\). 4. Outside the sphere, \((c-1)^2>16\), which gives \(c<-3\) or \(c>5\).

Inside: \(-3<c<5\) On: \(c=-3\) or \(c=5\) Outside: \(c<-3\) or \(c>5\)

- Write the squared distance as a quadratic expression in \(c\). - Solve the equality to find the boundary values. - Use the sign of the upward-opening quadratic between and outside its roots.

1. The squared distance is \(d^2=c^2+(c-1)^2+1=2c^2-2c+2\). 2. On the sphere, \(2c^2-2c+2=6\). This simplifies to \(c^2-c-2=0\), so \(c=-1\) or \(c=2\). 3. Because \(2c^2-2c-4\) opens upward, \(d^2<6\) between the roots: \(-1<c<2\). 4. Outside the sphere, \(c<-1\) or \(c>2\).

Inside: \(-1<c<2\) On: \(c=-1\) or \(c=2\) Outside: \(c<-1\) or \(c>2\)

- Write the sphere equation from its center and radius. - Substitute the parametric coordinates of the line into the sphere equation. - The number of real parameter values determines the number of intersections. - Substitute each parameter value back into the line equation.

1. The sphere equation is \((x-2)^2+y^2+(z-7)^2=9\). 2. Substitute \(x=1+t\), \(y=2-2t\), and \(z=5+2t\): \((t-1)^2+(2-2t)^2+(2t-2)^2=9\). 3. Simplifying gives \(9(t-1)^2=9\), so \((t-1)^2=1\). Therefore, \(t=0\) or \(t=2\). 4. For \(t=0\), the point is \((1, 2, 5)\). For \(t=2\), the point is \((3, -2, 9)\). 5. Because there are two intersection points, \(g\) is a secant line of the sphere.

Line \(g\) is a secant. The intersection points are \((1, 2, 5)\) and \((3, -2, 9)\).

- Substitute the line's coordinate expressions into the sphere equation. - Expand the squared binomials carefully. - Use the number of real solutions of the resulting quadratic equation. - One real parameter value means the line is tangent.

1. Substitute \(x=6+2\lambda\), \(y=-\lambda\), and \(z=5\) into the sphere equation: \((2\lambda+3)^2+(1-\lambda)^2+3^2=14\). 2. Expanding and simplifying gives \(5\lambda^2+10\lambda+5=0\). 3. Divide by \(5\): \((\lambda+1)^2=0\). Thus, there is exactly one parameter value, \(\lambda=-1\), so the line is tangent to the sphere. 4. Substitution into \(h\) gives the point of tangency \((4, 1, 5)\).

Line \(h\) is tangent to the sphere at \((4, 1, 5)\).

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

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

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

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

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

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

- Tangency to a coordinate plane fixes the corresponding center coordinate at one radius from the plane. - Substitute the known point into the sphere equation. - Use the positive-coordinate condition to choose the valid solution.

1. Tangency to the yz-plane \(x=0\), together with positivity, gives \(m_1=9\). 2. Tangency to the xy-plane \(z=0\) gives \(m_3=9\). 3. Because \(P\) lies on the sphere, \((5-9)^2+(1-m_2)^2+(5-9)^2=81\). 4. Thus, \((1-m_2)^2=49\), so \(m_2=-6\) or \(m_2=8\). 5. The center must have positive coordinates, so \(m_2=8\). Therefore, \(M=(9,8,9)\).

\(M=(9,8,9)\)

- Tangency to all three coordinate planes makes all center coordinates equal to the radius. - Substitute the given point into the sphere equation. - Solve the resulting quadratic equation for the radius.

1. Because the sphere is tangent to all three coordinate planes in the first octant, its center has the form \(M(r,r,r)\), where \(r\) is the radius. 2. Since \(A\) lies on the sphere, \((1-r)^2+(2-r)^2+(1-r)^2=r^2\). 3. Expanding gives \(2r^2-8r+6=0\), or \(r^2-4r+3=0\). 4. Therefore, \(r=1\) or \(r=3\). 5. The possible centers are \(M_1=(1,1,1)\) and \(M_2=(3,3,3)\).

\(M_1=(1,1,1)\) and \(M_2=(3,3,3)\)

- Read the center and radius from the sphere equation. - Use the vector from the center to \(Q\) as the line direction. - Parameterize the entire line and impose the sphere equation; do not assume \(Q\) lies on the sphere.

1. The sphere has center \(M=(5,-2,1)\) and radius \(7\). 2. A direction vector from \(M\) toward \(Q\) is \(\langle4,6,12\rangle\), whose magnitude is \(14\). 3. Parameterize the line as \(\mathbf{r}(t)=\langle5,-2,1\rangle+t\langle4,6,12\rangle\). 4. Substituting in the sphere gives \(196t^2=49\), so \(t=\pm\frac12\). 5. The two intersection points are \((7,1,7)\) and \((3,-5,-5)\).

\((7,1,7)\) and \((3,-5,-5)\)

- The two centers and the tangency point are collinear. - For external tangency, the tangency point lies between the centers. - Move one new radius beyond the tangency point.

1. Sphere \(K\) has center \(M_K=(1,2,3)\) and radius \(5\). 2. The vector from \(M_K\) to the tangency point is \(\overrightarrow{M_KB}=\langle 0,4,3\rangle\), with magnitude \(5\). 3. For external tangency, the new center lies beyond \(B\) along the same ray and is \(10\) units from \(B\). 4. A length-\(10\) vector in that direction is \(2\langle 0,4,3\rangle=\langle 0,8,6\rangle\). 5. The new center is \(M_S=B+\langle 0,8,6\rangle=(1,14,12)\). 6. Therefore, the sphere is \((x-1)^2+(y-14)^2+(z-12)^2=100\).

\((x-1)^2+(y-14)^2+(z-12)^2=100\)

- Find the center and radius of \(K\). - The centers and tangency point are collinear. - Move one new radius from \(P\) in each direction along the center-to-point line.

1. Sphere \(K\) has center \(M_K=(1,-2,2)\) and radius \(12\). 2. The vector \(\overrightarrow{M_KP}=\langle 4,8,-8\rangle\) has magnitude \(12\). 3. A vector of length \(6\) in this direction is \(\frac{1}{2}\overrightarrow{M_KP}=\langle 2,4,-4\rangle\). 4. The two possible centers are \(M_1=P+\langle 2,4,-4\rangle=(7,10,-10)\) and \(M_2=P-\langle 2,4,-4\rangle=(3,2,-2)\). 5. Therefore, the sphere equations are \((x-7)^2+(y-10)^2+(z+10)^2=36\) and \((x-3)^2+(y-2)^2+(z+2)^2=36\).

\((x-7)^2+(y-10)^2+(z+10)^2=36\) and \((x-3)^2+(y-2)^2+(z+2)^2=36\)

- 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

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

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

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

- The top vertices are vertical translations of the base vertices. - Use symmetry between the base and top planes. - The circumcenter of a right triangle is the midpoint of its hypotenuse.

1. Moving each base vertex vertically by \(6\) gives \(D(4,0,6)\), \(E(0,4,6)\), and \(F(0,0,6)\). 2. The sphere center lies halfway between the base and top planes, so its z-coordinate is \(3\). 3. In the xy-plane, the center is the circumcenter of right triangle \(ABC\), which is the midpoint of hypotenuse \(AB\). Thus, the center is \(M(2,2,3)\). 4. The squared radius is the squared distance from \(M\) to \(C\): \(r^2=2^2+2^2+3^2=17\). 5. The sphere equation is \((x-2)^2+(y-2)^2+(z-3)^2=17\).

a) \(D=(4,0,6)\), \(E=(0,4,6)\), \(F=(0,0,6)\) b) \((x-2)^2+(y-2)^2+(z-3)^2=17\)

- Use the diameter to find the radius. - Relate the center's height to the sphere's highest point. - Set \(z=0\) to find the ground-level cross section. - Use the circle circumference formula.

1. The sphere's radius is \(10\,\text{m}\). 2. Let the center be \((0,0,h)\). The highest point has z-coordinate \(h+10=16\), so \(h=6\). 3. The sphere equation is \(x^2+y^2+(z-6)^2=100\). 4. At ground level, \(z=0\). Substitution gives \(x^2+y^2+36=100\), so the ground circle has radius \(8\,\text{m}\). 5. Its circumference is \(2\pi(8)=16\pi\,\text{m}\approx50.27\,\text{m}\).

a) \(x^2+y^2+(z-6)^2=100\) b) \(16\pi\,\text{m}\approx50.27\,\text{m}\)

- Complete the square in all three variables. - Match the right side with \(r^2=49\). - Read the center from the completed-square form.

1. Complete the square: \((x-4)^2-16+(y+5)^2-25+(z+2)^2-4+c=0\). 2. Therefore, \((x-4)^2+(y+5)^2+(z+2)^2=45-c\). 3. The center is \(M=(4,-5,-2)\). 4. For radius \(7\), set \(45-c=49\). Thus, \(c=-4\).

\(c=-4\); center \(M=(4,-5,-2)\)

- First divide by the common coefficient of the squared terms. - Complete the square in each variable. - A real sphere requires a nonnegative squared radius.

1. Divide by \(3\): \(x^2+y^2+z^2+4x-6y+8z+31=0\). 2. Complete the square: \((x+2)^2-4+(y-3)^2-9+(z+4)^2-16+31=0\). 3. This becomes \((x+2)^2+(y-3)^2+(z+4)^2=-2\). 4. A sum of real squares cannot equal a negative number. Therefore, the equation has no real points and does not describe a sphere in \(\mathbb{R}^3\).

It does not describe a real sphere because the completed-square form is \((x+2)^2+(y-3)^2+(z+4)^2=-2\), which has no real solutions.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

- 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 \(E\).

- Complete the square in all three variables. - A real sphere with positive radius requires \(r^2>0\). - For part c), set the squared radius equal to \(25\).

1. Complete the square to obtain \((x-2k)^2+(y+1)^2+(z-3)^2=4k^2-3\). 2. Therefore, \(M_k=(2k,-1,3)\) and \(r^2=4k^2-3\). 3. A sphere with positive radius requires \(4k^2-3>0\), so \(\lvert k\rvert>\frac{\sqrt{3}}{2}\). 4. For radius \(5\), set \(4k^2-3=25\). Then \(k^2=7\), so \(k=\pm\sqrt{7}\).

a) \(M_k=(2k,-1,3)\), \(r^2=4k^2-3\) b) \(\lvert k\rvert>\frac{\sqrt{3}}{2}\) c) \(k=\pm\sqrt{7}\)

- Complete the square in x and z. - Express the center coordinates using the parameter. - Tangency to the xy-plane means the absolute z-coordinate of the center equals the radius.

1. Complete the square: \((x-2a)^2+y^2+(z-a)^2=16\). 2. Thus, every member is a sphere of radius \(4\), with center \(M_a=(2a,0,a)\). 3. The centers lie on the line \((x,y,z)=t\langle 2,0,1\rangle\). 4. A sphere is tangent to the xy-plane when the center's distance from \(z=0\) equals the radius. Therefore, \(\lvert a\rvert=4\), so \(a=4\) or \(a=-4\).

a) Radius \(4\) b) \((x,y,z)=t\langle 2,0,1\rangle\) c) \(a=4\) or \(a=-4\)

- 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, \(C\notin E\).

- What form do the coordinates of a point on the y-axis have in three dimensions? - Substitute that form into the three-dimensional distance formula. - Remember that an equation of the form \(u^2=a\) can have two solutions.

1. A point on the y-axis has the form \(P(0, y, 0)\). 2. Use the distance formula: \(\sqrt{(4-0)^2+(3-y)^2+(-2-0)^2}=6\). 3. Square both sides and simplify: \(16+(3-y)^2+4=36\), so \((3-y)^2=16\). 4. Thus, \(3-y=4\) or \(3-y=-4\), giving \(y=-1\) or \(y=7\). 5. The points are \(P_1(0, -1, 0)\) and \(P_2(0, 7, 0)\).

\(P_1(0, -1, 0)\) and \(P_2(0, 7, 0)\)

- For a reflection across a coordinate plane, only the perpendicular coordinate changes sign. - Apply the two reflections in order. - Compare the original and final coordinates. - Which rotation keeps one coordinate fixed and reverses the other two?

1. Reflection across the \(xy\)-plane changes the sign of the third coordinate, so \(P'=(p_1,p_2,-p_3)\). 2. Reflection across the \(xz\)-plane changes the sign of the second coordinate. Therefore, \(P''=(p_1,-p_2,-p_3)\). 3. The first coordinate is unchanged while the other two coordinates change sign. This is exactly a \(180^\circ\) rotation about the \(x\)-axis. In general, \((x,y,z)\mapsto(x,-y,-z)\).

a) \(P'=(p_1,p_2,-p_3)\) b) \(P''=(p_1,-p_2,-p_3)\) c) A \(180^\circ\) rotation about the \(x\)-axis; \((x,y,z)\mapsto(x,-y,-z)\).

- Find the center of the square base. - The vertex of a right pyramid lies on the perpendicular line through the base center. - Use the distance from \(A\) to the parameterized vertex. - Remember both signs when solving for the height.

1. The center of the square base is the midpoint of \(A\) and \(C\): \(M=(4, 4, 2)\). 2. In a right pyramid, the vertex lies on the line perpendicular to the base through \(M\). Since the base lies in \(z=2\), write \(S=(4, 4, 2+h)\). 3. Then \(\overrightarrow{AS}=\begin{pmatrix}3\\3\\h\end{pmatrix}\). The edge-length condition gives \(3^2+3^2+h^2=43\). 4. Thus, \(h^2=25\), so \(h=5\) or \(h=-5\). 5. Therefore, the possible vertices are \(S_1=(4, 4, 7)\) and \(S_2=(4, 4, -3)\).

\(S_1(4, 4, 7)\) and \(S_2(4, 4, -3)\)

- Identify the plane containing the base. - Find the center of the square. - Parameterize the perpendicular line through the center. - Use the given lateral-edge length and include both possible signs.

1. The base lies in the plane \(x=5\), and its center is \(M=(5, 2, 2)\). 2. The vertex lies on the line through \(M\) perpendicular to the base, so write \(S=(5+h, 2, 2)\). 3. Then \(\overrightarrow{AS}=\begin{pmatrix}h\\2\\2\end{pmatrix}\). The edge-length condition is \(h^2+2^2+2^2=6^2\). 4. Thus, \(h^2=28\), so \(h=\pm2\sqrt{7}\). 5. Therefore, \(S_1=(5+2\sqrt{7}, 2, 2)\) and \(S_2=(5-2\sqrt{7}, 2, 2)\).

\(S_1(5+2\sqrt{7}, 2, 2)\) and \(S_2(5-2\sqrt{7}, 2, 2)\)

- Form the vector from \(A\) to \(B\). - Set its squared magnitude equal to \(36\). - Solve both cases produced by the square.

1. \(\overrightarrow{AB}=\begin{pmatrix}2\\1-k\\-4\end{pmatrix}\). 2. The length condition gives \(2^2+(1-k)^2+(-4)^2=6^2\). 3. Thus, \((1-k)^2=16\). 4. Therefore, \(1-k=4\) or \(1-k=-4\), giving \(k=-3\) or \(k=5\).

\(k=-3\) or \(k=5\)

- Form the two edge vectors from \(A\). - Use magnitudes and a dot product. - Complete the base parallelogram to find \(C\). - A face parallel to the \(yz\)-plane has a normal parallel to the \(x\)-axis.

1. \(\overrightarrow{AB}=\begin{pmatrix}0\\4\\3\end{pmatrix}\) and \(\overrightarrow{AD}=\begin{pmatrix}-5\\0\\0\end{pmatrix}\). Both have length \(5\). 2. Their dot product is \(0\), so the edges are perpendicular. 3. Since \(\overrightarrow{BC}=\overrightarrow{AD}\), \(C=B+\overrightarrow{AD}=(-1, 5, 5)\). 4. Edge \(AD\) is parallel to the \(x\)-axis because only its \(x\)-coordinate changes. The faces perpendicular to that edge, \(ABFE\) and \(DCGH\), are therefore parallel to the \(yz\)-plane.

a) \(AB=AD=5\) b) \(\overrightarrow{AB}\cdot\overrightarrow{AD}=0\) c) \(C(-1, 5, 5)\) d) \(ABFE\) and \(DCGH\)

- Read the center and radius from standard form. - Points on the x-axis have y- and z-coordinates equal to \(0\). - Substitute the known coordinates of \(S\).

1. The center is \(M=(-1,4,0)\), and the radius is \(5\). 2. Points on the x-axis have \(y=z=0\). Substitution gives \((x+1)^2+16=25\), so \((x+1)^2=9\). 3. Therefore, \(x=2\) or \(x=-4\), giving \((2,0,0)\) and \((-4,0,0)\). 4. Substitute \(S\): \(3^2+4^2+z^2=25\). Thus, \(z^2=0\), so \(z=0\).

a) Center \((-1,4,0)\); radius \(5\) b) \((2,0,0)\) and \((-4,0,0)\) c) \(z=0\)

- Compare squared distances to \(A\) and \(B\). - On the \(z\)-axis, two coordinates are zero. - Parameterize \(Q\) from the line equation. - Use a dot product for the right angle.

1. For \(P=(0,y,z)\), \(PA^2=3^2+y^2+z^2\) and \(PB^2=(-3)^2+y^2+z^2\). Thus, \(PA=PB\). 2. On the \(z\)-axis, \(P=(0,0,z)\). Since \(AB=6\), an equilateral triangle requires \(9+z^2=36\). Thus, \(z=\pm3\sqrt{3}\). 3. A point on \(g\) is \(Q=(k,2,0)\). Then \(\overrightarrow{QA}=\langle 3-k,-2,0\rangle\) and \(\overrightarrow{QB}=\langle -3-k,-2,0\rangle\). 4. Their dot product is \((3-k)(-3-k)+4=k^2-5\). Setting it equal to \(0\) gives \(k=\pm\sqrt{5}\).

a) \(PA=PB=\sqrt{9+y^2+z^2}\) b) \((0,0,3\sqrt{3})\) and \((0,0,-3\sqrt{3})\) c) \((\sqrt{5},2,0)\) and \((-\sqrt{5},2,0)\)

- Compare the squared lengths of the two sides meeting at \(C_k\). - Compare a leg with the base for the equilateral condition. - Use a dot product at \(C_k\) for the right-angle condition.

1. \(\overrightarrow{AC_k}=\begin{pmatrix}1+k\\2-k\\-1-k\end{pmatrix}\) and \(\overrightarrow{BC_k}=\begin{pmatrix}-1+k\\-2-k\\1-k\end{pmatrix}\). 2. Both squared lengths equal \(3k^2+6\), so \(AC_k=BC_k\). Therefore, the base is \(AB\). The value \(k=0\) is excluded because \(C_0\) is the midpoint of \(AB\). 3. Since \(AB^2=24\), an equilateral triangle requires \(3k^2+6=24\). Thus, \(k=\pm\sqrt{6}\). 4. For a right angle at \(C_k\), \(\overrightarrow{C_kA}\cdot\overrightarrow{C_kB}=3k^2-6\). Setting this equal to \(0\) gives \(k=\pm\sqrt{2}\).

a) \(AC_k=BC_k\), so the base is \(AB\). b) \(k=\pm\sqrt{6}\) c) \(k=\pm\sqrt{2}\)

- Find displacement vectors for the sides. - Check whether two sides from the same vertex are perpendicular. - Use the dot-product angle formula with both vectors starting at \(B\).

1. \(\overrightarrow{AB}=(3, 0, 4)\), \(\overrightarrow{AC}=(0, 5, 0)\), and \(\overrightarrow{BC}=(-3, 5, -4)\). 2. Therefore, \(AB=5\), \(AC=5\), and \(BC=\sqrt{50}=5\sqrt2\). 3. Since \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), the triangle is right at \(A\). Its area is \(\frac12\cdot5\cdot5=12.5\). 4. At \(B\), use \(\overrightarrow{BA}=(-3, 0, -4)\) and \(\overrightarrow{BC}=(-3, 5, -4)\). 5. \(\overrightarrow{BA}\cdot\overrightarrow{BC}=25\), and \(\cos B=\frac{25}{5\cdot5\sqrt2}=\frac{1}{\sqrt2}\). 6. Thus, \(B=45^\circ\).

a) \(AB=5\), \(AC=5\), \(BC=5\sqrt2\) b) \(12.5\) square units c) \(45^\circ\)

- Form the three side vectors from the coordinates. - All three coordinates matter in the magnitude calculations. - Use a dot product to test whether two sides meeting at one vertex are perpendicular. - Equal side lengths give equal opposite angles.

1. \(\overrightarrow{AB}=\langle1,2,2\rangle\), \(\overrightarrow{AC}=\langle2,1,-2\rangle\), and \(\overrightarrow{BC}=\langle1,-1,-4\rangle\). 2. Hence \(AB=3\), \(AC=3\), and \(BC=\sqrt{18}=3\sqrt2\). 3. Also \(\overrightarrow{AB}\cdot\overrightarrow{AC}=2+2-4=0\), so \(\angle A=90^\circ\). 4. Since \(AB=AC\), the remaining two angles are equal. They sum to \(90^\circ\), so \(\angle B=\angle C=45^\circ\). 5. Therefore, the triangle is an isosceles right triangle.

a) \(AB=3\), \(AC=3\), and \(BC=3\sqrt2\) b) \(\angle A=90^\circ\), \(\angle B=45^\circ\), \(\angle C=45^\circ\) c) Isosceles right triangle

- Compute all three side lengths first. - The largest angle lies opposite the longest side. - Use two vectors that begin at the vertex of that angle. - A negative cosine indicates an obtuse angle.

1. The side lengths are \(KL=4\), \(KM=\sqrt{2^2+1^2+1^2}=\sqrt{6}\), and \(LM=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6}\). 2. Since \(KM=LM\), the triangle is isosceles. Its longest side is \(KL\), so the largest angle is at \(M\). 3. Use \(\overrightarrow{MK}=\begin{pmatrix}-2\\-1\\-1\end{pmatrix}\) and \(\overrightarrow{ML}=\begin{pmatrix}2\\-1\\-1\end{pmatrix}\). 4. Then \(\cos\angle M =\frac{-4+1+1}{\sqrt{6}\sqrt{6}} =-\frac{1}{3}\). 5. Therefore, \(\angle M=\cos^{-1}\left(-\frac{1}{3}\right)\approx109.47^\circ\). 6. The triangle is isosceles and obtuse.

The side lengths are \(KL=4\) and \(KM=LM=\sqrt{6}\). The largest angle is \(\angle M\approx109.47^\circ\). The triangle is isosceles and obtuse.

- Compare the vectors for the two possible bases. - Parallel vectors are scalar multiples. - A parallelogram has congruent opposite sides. - Compare the lengths of the nonparallel sides.

1. \(\overrightarrow{AB}=\begin{pmatrix}4\\0\\0\end{pmatrix}\) and \(\overrightarrow{DC}=\begin{pmatrix}2\\0\\0\end{pmatrix}\). Since \(\overrightarrow{AB}=2\overrightarrow{DC}\), sides \(AB\) and \(DC\) are parallel. 2. Their lengths are \(4\) and \(2\), so they are not equal. Therefore, the quadrilateral is not a parallelogram. 3. The nonparallel side vectors are \(\overrightarrow{AD}=\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-1\\2\\2\end{pmatrix}\). 4. Both have length \(3\). Therefore, the trapezoid has congruent legs and is isosceles.

Since \(AB\parallel DC\), \(AB\ne DC\), and \(AD=BC=3\), the quadrilateral is an isosceles trapezoid but not a parallelogram.

- Compare the squared lengths of \(AC_k\) and \(BC_k\); all three coordinates contribute. - For a right angle at \(C_k\), form the two vectors that begin at \(C_k\) and set their dot product to zero. - For an equilateral triangle, compare a leg length with \(AB\).

1. \(AC_k^2=(k-1)^2+(k+1)^2+1=2k^2+3\), and \(BC_k^2=(k+1)^2+(k-1)^2+1=2k^2+3\). Therefore, \(AC_k=BC_k\). 2. At \(C_k\), \(\overrightarrow{C_kA}\cdot\overrightarrow{C_kB}=2k^2-3\). A right angle occurs when \(2k^2-3=0\), so \(k=\pm\frac{\sqrt6}{2}\). 3. \(AB^2=12\). For an equilateral triangle, set \(2k^2+3=12\). Thus \(k^2=\frac92\), so \(k=\pm\frac{3\sqrt2}{2}\). 4. The excluded value \(k=0\) would place \(C_k=(1,1,1)\), the midpoint of \(AB\), producing a degenerate triangle.

a) \(AC_k=BC_k=\sqrt{2k^2+3}\) b) \(k=\pm\frac{\sqrt6}{2}\) c) \(k=\pm\frac{3\sqrt2}{2}\)

- Subtract coordinates to form the side vectors. - Compare an opposite-side vector to \(\overrightarrow{PQ}\). - A parallelogram with equal adjacent sides is a rhombus. - Multiply the side length by \(4\).

1. \(\overrightarrow{PQ}=\begin{pmatrix}2\\1\\2\end{pmatrix}\) and \(\overrightarrow{PS}=\begin{pmatrix}1\\2\\2\end{pmatrix}\). Both vectors have length \(3\). 2. \(\overrightarrow{SR}=\begin{pmatrix}2\\1\\2\end{pmatrix}=\overrightarrow{PQ}\), so \(PQRS\) is a parallelogram. 3. Since the adjacent sides \(PQ\) and \(PS\) both have length \(3\), the parallelogram is a rhombus. 4. Its perimeter is \(4\cdot3=12\) units.

a) \(\overrightarrow{PQ}=\begin{pmatrix}2\\1\\2\end{pmatrix}\), \(\overrightarrow{PS}=\begin{pmatrix}1\\2\\2\end{pmatrix}\), and both lengths are \(3\). b) \(PQRS\) is a rhombus. c) \(12\) units

- Which coordinates are preserved by projection onto the \(yz\)-plane? - How is distance from the \(yz\)-plane related to the \(x\)-coordinate? - Which coordinate remains when projecting onto the \(z\)-axis?

1. Projection onto the \(yz\)-plane preserves the \(y\)- and \(z\)-coordinates, so \(Q=(x,5,-3)\). 2. The distance to the \(yz\)-plane is \(|x|\). Since \(|x|=8\), \(x=8\) or \(x=-8\). Thus, the two possible points are \((8,5,-3)\) and \((-8,5,-3)\). 3. For part b), use \(Q=(8,5,-3)\). Projection onto the \(z\)-axis sets \(x\) and \(y\) equal to \(0\), so \(Q''=(0,0,-3)\).

a) \(Q=(8,5,-3)\) or \(Q=(-8,5,-3)\) b) \(Q''=(0,0,-3)\)

- What value must \(z\) have for a point in the \(xy\)-plane? - Compute a displacement vector as terminal point minus initial point. - Average corresponding endpoint coordinates to find a midpoint. - Moving directly upward changes only the \(z\)-coordinate.

1. Every point in the \(xy\)-plane has \(z=0\), so \(A=(5, -1, 0)\) and \(B=(2, 3, 0)\). 2. In two dimensions, \(\overrightarrow{AB}=\begin{pmatrix}2-5\\3-(-1)\end{pmatrix}=\begin{pmatrix}-3\\4\end{pmatrix}\). In three dimensions, \(\overrightarrow{AB}=\begin{pmatrix}2-5\\3-(-1)\\0-0\end{pmatrix}=\begin{pmatrix}-3\\4\\0\end{pmatrix}\). The three-dimensional vector has the same first two components and an added zero component. 3. \(M=\left(\frac{5+2}{2}, \frac{-1+3}{2}, \frac{0+0}{2}\right)=(3.5, 1, 0)\). 4. Moving \(4\) units in the positive \(z\)-direction gives \(P=(3.5, 1, 4)\).

1. \(A=(5, -1, 0)\), \(B=(2, 3, 0)\) 2. \(\overrightarrow{AB}_{2D}=\begin{pmatrix}-3\\4\end{pmatrix}\), \(\overrightarrow{AB}_{3D}=\begin{pmatrix}-3\\4\\0\end{pmatrix}\) 3. \(M=(3.5, 1, 0)\) 4. \(P=(3.5, 1, 4)\)

- Every point of a prism's base is translated by the same vector. - The stated correspondence determines the translation vector. - What form must a vector have to be perpendicular to the \(xy\)-plane?

1. In part a), the translation vector is \(\overrightarrow{PS}=\langle 2,2,5\rangle\). Applying this vector gives \(Q'=(7,3,5)\) and \(R'=(3,6,5)\). 2. In part b), the translation vector is \(\overrightarrow{QS}=\langle -2,2,5\rangle\). Applying this vector gives \(P''=(-1,3,5)\) and \(R''=(-1,6,5)\). 3. The base lies in the \(xy\)-plane. A vector perpendicular to that plane has zero \(x\)- and \(y\)-components. Both translation vectors have nonzero horizontal components, so the lateral edges are not perpendicular to the base. Thus, each prism is oblique.

a) \(Q'(7,3,5)\) and \(R'(3,6,5)\) b) \(P''(-1,3,5)\) and \(R''(-1,6,5)\) c) In both cases, the translation vector has nonzero \(x\)- and \(y\)-components, so it is not perpendicular to the \(xy\)-plane. Therefore, the prism is oblique.

- Which opposite side vectors of a parallelogram are equal? - Use the midpoint formula for each diagonal. - Two segments bisect each other when they have the same midpoint.

1. For consecutive vertices of a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\), so \(D=A+C-B\). 2. Therefore, \(D=(1, 4, -2)+(2, -1, 5)-(5, 2, 2)=(-2, 1, 1)\). 3. The midpoint of \(\overline{AC}\) is \(\left(\frac{1+2}{2},\frac{4+(-1)}{2},\frac{-2+5}{2}\right)=(1.5, 1.5, 1.5)\). 4. The midpoint of \(\overline{BD}\) is \(\left(\frac{5+(-2)}{2},\frac{2+1}{2},\frac{2+1}{2}\right)=(1.5, 1.5, 1.5)\). 5. Since the diagonals have the same midpoint, they bisect each other.

a) \(D(-2, 1, 1)\) b) Both diagonals have midpoint \(M(1.5, 1.5, 1.5)\).

- The reflection center is the midpoint of a point and its image. - Use the same center to reflect \(Q\). - Two nonzero vectors are parallel when one is a scalar multiple of the other.

1. The center \(M\) is the midpoint of \(\overline{PP'}\): \(M=\left(\frac{3+1}{2},\frac{-2+6}{2},\frac{4+0}{2}\right)=(2,2,2)\). 2. Use \(Q'=2M-Q\): \(Q'=2(2,2,2)-(5,1,2)=(-1,3,2)\). 3. \(\overrightarrow{PQ}=Q-P=\langle 2,3,-2\rangle\). 4. \(\overrightarrow{P'Q'}=Q'-P'=\langle -2,-3,2\rangle=-\overrightarrow{PQ}\). 5. Since one vector is a scalar multiple of the other, the vectors are parallel.

1. \(M(2,2,2)\) 2. \(Q'(-1,3,2)\) 3. \(\overrightarrow{P'Q'}=-\overrightarrow{PQ}\), so the vectors are parallel.

- Which coordinate changes sign under reflection across the \(xy\)-plane? - In a reflection through a point, that point is the midpoint of a point and its image. - Use \(\mathbf{x}''=2\mathbf{s}-\mathbf{x}\) for the point reflection.

1. Reflection across the \(xy\)-plane changes the sign of the \(z\)-coordinate: \(A'=(5,-2,-1)\), \(B'=(0,4,3)\), and \(C'=(-3,1,-2)\). 2. For a reflection through \(S\), the point \(S\) is the midpoint of each point and its image, so \(\mathbf{x}''=2\mathbf{s}-\mathbf{x}\). 3. For \(A\), \(2(1,2,3)-(5,-2,1)=(-3,6,5)\). 4. For \(B\), \(2(1,2,3)-(0,4,-3)=(2,0,9)\). 5. For \(C\), \(2(1,2,3)-(-3,1,2)=(5,3,4)\).

a) \(A'=(5,-2,-1)\), \(B'=(0,4,3)\), \(C'=(-3,1,-2)\) b) \(A''=(-3,6,5)\), \(B''=(2,0,9)\), \(C''=(5,3,4)\)

- A point reflection across the origin changes every coordinate's sign. - For each coordinate plane, identify the perpendicular coordinate. - Apply the reflections one at a time.

1. A point reflection across the origin multiplies every component by \(-1\), so \(\mathbf{v}'=\langle-v_1,-v_2,-v_3\rangle\). 2. Reflection across the \(xy\)-plane gives \(\langle v_1,v_2,-v_3\rangle\). 3. Reflection across the \(xz\)-plane then gives \(\langle v_1,-v_2,-v_3\rangle\). 4. Reflection across the \(yz\)-plane finally gives \(\langle-v_1,-v_2,-v_3\rangle\), which equals \(\mathbf{v}'\).

a) \(\mathbf{v}'=\langle-v_1,-v_2,-v_3\rangle\) b) The successive vectors are \(\langle v_1,v_2,-v_3\rangle\), \(\langle v_1,-v_2,-v_3\rangle\), and \(\langle-v_1,-v_2,-v_3\rangle\), so the final result equals \(\mathbf{v}'\).

- Reflect one point on \(h\) through \(Z\). - A point reflection does not change the direction of a line. - For parallel lines, find a perpendicular segment joining them. - Here both lines are parallel to the z-axis.

1. Reflect the point \(A=(1, 1, 1)\) on \(h\) through \(Z\): \(A_r=2Z-A=2(4, 5, 1)-(1, 1, 1)=(7, 9, 1)\). 2. A point reflection preserves the line's direction, so \(h_r: \mathbf{x}=\langle 7, 9, 1\rangle+t\langle 0, 0, 1\rangle\). 3. The two lines are parallel to the z-axis. Points \((1,1,1)\) and \((7,9,1)\) have the same z-coordinate, so the segment joining them is perpendicular to both lines. 4. Their distance is \(\sqrt{(7-1)^2+(9-1)^2}=\sqrt{36+64}=10\).

a) \(h_r: \mathbf{x}=\langle 7, 9, 1\rangle+t\langle 0, 0, 1\rangle\) b) \(10\) units

- Calculate the distance between each pair of adjacent vertices. - What condition defines an isosceles triangle? - Compare exact side lengths rather than rounded values. - Subtract the decimal coordinates before squaring.

1. Find the side lengths: \(PQ=\sqrt{(4.2-1.2)^2+(4-0)^2+(3.5-3.5)^2}=\sqrt{3^2+4^2}=5\). 2. \(QR=\sqrt{(1.2-4.2)^2+(4-4)^2+(7.5-3.5)^2}=\sqrt{(-3)^2+4^2}=5\). 3. \(RP=\sqrt{(1.2-1.2)^2+(0-4)^2+(3.5-7.5)^2}=\sqrt{32}=4\sqrt{2}\). 4. The perimeter is \(5+5+4\sqrt{2}=10+4\sqrt{2}\approx 15.66\). 5. Because \(PQ=QR=5\), the triangle has two congruent sides and is isosceles.

a) \(10+4\sqrt{2}\approx 15.66\) units b) Yes. The triangle is isosceles because \(PQ=QR=5\).

- Identify which pairs of vertices form sides and which form diagonals. - Apply the three-dimensional distance formula to each required segment. - The perimeter is the sum of the four side lengths.

1. Use the distance formula for the sides: \(AB=\sqrt{(6-2)^2+(3-1)^2+(-2-(-3))^2}=\sqrt{21}\), \(BC=\sqrt{(5-6)^2+(7-3)^2+(0-(-2))^2}=\sqrt{21}\), \(CD=\sqrt{(1-5)^2+(5-7)^2+(-1-0)^2}=\sqrt{21}\), and \(DA=\sqrt{(2-1)^2+(1-5)^2+(-3-(-1))^2}=\sqrt{21}\). 2. Find the diagonals: \(AC=\sqrt{(5-2)^2+(7-1)^2+(0-(-3))^2}=\sqrt{54}=3\sqrt{6}\), and \(BD=\sqrt{(1-6)^2+(5-3)^2+(-1-(-2))^2}=\sqrt{30}\). 3. Add the four side lengths: \(P=4\sqrt{21}\approx 18.33\).

Sides: \(AB=BC=CD=DA=\sqrt{21}\approx 4.58\) Diagonals: \(AC=3\sqrt{6}\approx 7.35\), \(BD=\sqrt{30}\approx 5.48\) Perimeter: \(P=4\sqrt{21}\approx 18.33\)

- Use the three-dimensional distance formula for each segment. - Opposite vertices determine the diagonals. - Add only the four exterior side lengths to find the perimeter.

1. Find the side lengths: \(PQ=\sqrt{(7-3)^2+(3-0)^2+(4-2)^2}=\sqrt{29}\), \(QR=\sqrt{(4-7)^2+(7-3)^2+(10-4)^2}=\sqrt{61}\), \(RS=\sqrt{(0-4)^2+(4-7)^2+(8-10)^2}=\sqrt{29}\), and \(SP=\sqrt{(3-0)^2+(0-4)^2+(2-8)^2}=\sqrt{61}\). 2. Add the side lengths: \(P=2\sqrt{29}+2\sqrt{61}\approx 26.39\). 3. Find the diagonals: \(PR=\sqrt{(4-3)^2+(7-0)^2+(10-2)^2}=\sqrt{114}\), and \(QS=\sqrt{(0-7)^2+(4-3)^2+(8-4)^2}=\sqrt{66}\).

Sides: \(PQ=RS=\sqrt{29}\approx 5.39\), \(QR=SP=\sqrt{61}\approx 7.81\) Diagonals: \(PR=\sqrt{114}\approx 10.68\), \(QS=\sqrt{66}\approx 8.12\) Perimeter: \(P=2\sqrt{29}+2\sqrt{61}\approx 26.39\)

- A closest point on the \(z\)-axis has coordinates \((0,0,z)\). - A point on the \(x\)-axis has coordinates \((x,0,0)\). - Use the three-dimensional distance formula and solve the resulting quadratic equation.

1. The closest point on the \(z\)-axis has the same \(z\)-coordinate as \(A\), so it is \((0,0,12)\). The distance is \(\sqrt{3^2+4^2}=5\). 2. A point on the \(x\)-axis has form \((x,0,0)\). The distance condition is \((x-3)^2+4^2+12^2=13^2\). 3. Thus \((x-3)^2=9\), so \(x=0\) or \(x=6\). 4. The required points are \((0,0,0)\) and \((6,0,0)\).

a) \(5\) units; closest point \((0,0,12)\) b) \((0,0,0)\) and \((6,0,0)\)

- Picture one right triangle in a horizontal plane and a second right triangle containing the space diagonal. - What coordinate relationship holds for two points in the same plane parallel to the xy-plane? - What happens to \((q_3-p_3)^2\) when the third coordinates are equal?

1. Project \(Q\) vertically onto the plane through \(P\) that is parallel to the xy-plane. In that plane, the horizontal separation is the hypotenuse of a right triangle with leg lengths \(\lvert q_1-p_1\rvert\) and \(\lvert q_2-p_2\rvert\). Thus, \(d_{xy}^2=(q_1-p_1)^2+(q_2-p_2)^2\). 2. The horizontal separation \(d_{xy}\) and the vertical separation \(\lvert q_3-p_3\rvert\) form the legs of a second right triangle whose hypotenuse is the distance \(d\). Therefore, \(d^2=d_{xy}^2+(q_3-p_3)^2\). 3. Substitution gives \(d^2=(q_1-p_1)^2+(q_2-p_2)^2+(q_3-p_3)^2\), so \(d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2+(q_3-p_3)^2}\). 4. If both points lie in a plane parallel to the xy-plane, then \(p_3=q_3\). Hence, \(q_3-p_3=0\), and the formula becomes \(d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}\).

a) First use the Pythagorean Theorem to obtain \(d_{xy}^2=(q_1-p_1)^2+(q_2-p_2)^2\). Then use it again with vertical leg \(\lvert q_3-p_3\rvert\) to obtain \(d^2=d_{xy}^2+(q_3-p_3)^2\). b) In a plane parallel to the xy-plane, \(p_3=q_3\), so the third squared difference is \(0\). The formula reduces to \(d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}\).

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

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

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

- Use the distance formula in three dimensions. - For the angle at \(B\), form two vectors beginning at \(B\). - Use two adjacent sides and their included angle for the area.

1. The side vectors are \(\overrightarrow{AB}=\begin{pmatrix}3\\4\\0\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-3\\0\\4\end{pmatrix}\), and \(\overrightarrow{AC}=\begin{pmatrix}0\\4\\4\end{pmatrix}\). 2. Thus, \(AB=5\), \(BC=5\), and \(AC=\sqrt{32}=4\sqrt{2}\). 3. For the angle at \(B\), use \(\overrightarrow{BA}=\begin{pmatrix}-3\\-4\\0\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-3\\0\\4\end{pmatrix}\). 4. Their dot product is \(9\), so \(\cos\beta=\frac{9}{5\cdot5}=\frac{9}{25}\). 5. Therefore, \(\beta=\cos^{-1}\left(\frac{9}{25}\right)\approx68.90^\circ\). 6. Using the two sides adjacent to \(\beta\), \(K=\frac{1}{2}\cdot5\cdot5\sin\beta=2\sqrt{34}\approx11.66\).

a) \(AB=5\), \(BC=5\), and \(AC=4\sqrt{2}\) b) \(\beta\approx68.90^\circ\) c) The area is \(2\sqrt{34}\approx11.66\) square units.

- Build the two side vectors that start at \(A\). - Report their dot product before using the angle formula. - Use the included-angle area formula only after part b. - Compute the remaining side separately at the end.

1. \(\overrightarrow{AB}=\langle2,2,0\rangle\) and \(\overrightarrow{AC}=\langle2,0,2\rangle\), so \(AB=AC=2\sqrt2\). 2. Their dot product is \(4\). Hence \(\cos\angle BAC=\frac4{(2\sqrt2)(2\sqrt2)}=\frac12\), so \(\angle BAC=60^\circ\). 3. The area is \(\frac12(2\sqrt2)(2\sqrt2)\sin60^\circ=2\sqrt3\). 4. \(\overrightarrow{BC}=\langle0,-2,2\rangle\), so \(BC=2\sqrt2\). Thus all three sides are equal and the triangle is equilateral.

a) \(AB=AC=2\sqrt2\); yes, it is isosceles at \(A\). b) \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\), so \(\angle BAC=60^\circ\). c) Area \(=2\sqrt3\); \(BC=2\sqrt2\), so the triangle is equilateral.

- Use equal opposite-side vectors to find \(C\). - Work with squared lengths directly. - Each of the two distinct side lengths occurs twice.

1. \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\2\end{pmatrix}\), so \(C=B+\overrightarrow{AD}=(6, 5, 2)\). 2. \(\overrightarrow{AC}=\begin{pmatrix}5\\4\\1\end{pmatrix}\), so \(AC^2=25+16+1=42\). 3. \(\overrightarrow{BD}=\begin{pmatrix}-3\\2\\3\end{pmatrix}\), so \(BD^2=9+4+9=22\). The diagonal-square sum is \(42+22=64\). 4. \(\overrightarrow{AB}=\begin{pmatrix}4\\1\\-1\end{pmatrix}\), so \(AB^2=18\), and \(AD^2=14\). 5. Opposite sides are equal, so the side-square sum is \(2\cdot18+2\cdot14=64\). The two sums are equal, as predicted by the parallelogram law.

a) \(C(6, 5, 2)\) b) \(AC^2=42\), \(BD^2=22\) c) Both sums equal \(64\).

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

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

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

- Use the center-to-surface point distance for the radius. - Substitute \(Q\) into the sphere equation. - Compare the center's distance from the xy-plane with the radius.

1. The squared radius is \((5-3)^2+(-1+2)^2+(2-4)^2=9\), so \(r=3\). 2. The sphere equation is \((x-3)^2+(y+2)^2+(z-4)^2=9\). 3. For \(Q\), the left side is \((1-3)^2+(-4+2)^2+(5-4)^2=9\), so \(Q\) lies on the sphere. 4. The center is \(4\) units from the xy-plane, while the radius is only \(3\). Therefore, the sphere has no points in the xy-plane.

a) \((x-3)^2+(y+2)^2+(z-4)^2=9\) b) \(Q\) lies on the sphere. c) The center-to-plane distance is \(4>3\), so there is no intersection.

- Complete the square separately for the \(x\)-, \(y\)-, and \(z\)-terms. - Compare the result with \((x-a)^2+(y-b)^2+(z-c)^2=r^2\). - Use the distance formula in three dimensions.

1. Group the terms by variable and complete each square: \((x^2-4x+4)+(y^2+6y+9)+(z^2-2z+1)=2+4+9+1\). 2. This gives \((x-2)^2+(y+3)^2+(z-1)^2=16\), which is the standard equation of a sphere. 3. Therefore, the center is \(M=(2, -3, 1)\), and the radius is \(r=4\). 4. The requested distance equals \(MO=\sqrt{(2-0)^2+(-3-0)^2+(1-0)^2}=\sqrt{14}\approx3.74\).

a) \((x-2)^2+(y+3)^2+(z-1)^2=16\) b) \(M=(2, -3, 1)\) and \(r=4\) c) \(\sqrt{14}\approx3.74\)

- Use the three-dimensional distance formula for each point. - A sphere with center \((h, k, \ell)\) and radius \(r\) has equation \((x-h)^2+(y-k)^2+(z-\ell)^2=r^2\). - Substitute the coordinates of \(P\) into the sphere equation. - Use the inequality in the problem to choose between the two possible values.

1. Compute the squared distances from \(M\): \(MA^2=4^2+0^2+(-3)^2=25\), \(MB^2=0^2+5^2+0^2=25\), and \(MC^2=3^2+4^2+0^2=25\). 2. Thus, all three points are \(5\) units from \(M\), so the sphere equation is \((x-2)^2+(y+1)^2+(z-5)^2=25\). 3. Substitute \(P=(2, y, 8)\): \((y+1)^2+3^2=25\), so \((y+1)^2=16\). 4. Therefore, \(y=3\) or \(y=-5\). The condition \(y>-1\) gives \(y=3\).

a) Each distance is \(5\). b) \((x-2)^2+(y+1)^2+(z-5)^2=25\) c) \(y=3\)

- Compare each point's squared distance from the origin with \(100\). - Points in the xy-plane have \(z=0\). - Points on the z-axis have \(x=y=0\). - Recall the standard equation of a circle centered at the origin.

1. Compare \(x^2+y^2+z^2\) with \(100\). For \(P\), the value is \(36+0+64=100\), so \(P\) is on the sphere. For \(Q\), it is \(49+49+1=99\), so \(Q\) is inside. 2. For \(R\), the value is \(100+1+0=101\), so \(R\) is outside. For \(S\), it is \(50+25+25=100\), so \(S\) is on the sphere. 3. In the xy-plane, \(z=0\). The intersection therefore satisfies \(x^2+y^2=100\), a circle centered at the origin with radius \(10\). 4. On the z-axis, \(x=y=0\). Then \(z^2=100\), so \(z=\pm10\).

a) \(P\) and \(S\) are on the sphere; \(Q\) is inside; \(R\) is outside. b) A circle with equation \(x^2+y^2=100\) c) \((0, 0, 10)\) and \((0, 0, -10)\)

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

- Read the normal vector from the 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) \(\mathbf{r}(s,t)=\langle5,0,0\rangle+s\langle5,-2,0\rangle+t\langle3,0,2\rangle\)

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

- 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.

- 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 \(t\in\mathbb{R}\)

- 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) \(E:4x-2y+z=5\) 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 \(\left(-\frac92,0,0\right)\), \(\left(0,\frac94,0\right)\), and \(\left(0,0,\frac95\right)\).

a) \(E:-2x+4y+5z=9\) b) \(\left(-\frac92,0,0\right)\), \(\left(0,\frac94,0\right)\), and \(\left(0,0,\frac95\right)\)

- 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. \(S=(2,2,2)\) 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 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. \(E:5x+y-2z=1\)

- 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 \(\left(-\frac{11}{2},0,0\right)\), \(\left(0,\frac{11}{2},0\right)\), and \(\left(0,0,\frac{11}{3}\right)\).

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

- 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 \(S\left(\frac53,\frac73,3\right)\).

- 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 \(\mathbf{n}=\langle 1,1,1\rangle\). 3. Using point \(C\), the plane equation is \(x+y+z=12\). 4. The beam direction is \(\overrightarrow{PQ}=\langle 1,1,3\rangle\), so \(\mathbf{r}(t)=\langle 1,1,0\rangle+t\langle 1,1,3\rangle\). 5. Substitute into the plane equation: \((1+t)+(1+t)+3t=12\). 6. Thus \(2+5t=12\), so \(t=2\). The intersection point is \(D(3, 3, 6)\).

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

- Express the tetrahedron as coordinate inequalities. - Substitute the line coordinates into each inequality. - Intersect the resulting parameter intervals. - Evaluate the line at the interval endpoints.

1. The tetrahedron consists of points satisfying \(x\ge0\), \(y\ge0\), \(z\ge0\), and \(x+y+z\le6\). 2. On the line, \(x=1+t\), \(y=1+t\), and \(z=-1+2t\). 3. The conditions \(x\ge0\) and \(y\ge0\) require \(t\ge-1\). The condition \(z\ge0\) requires \(t\ge\frac12\). 4. The final inequality gives \((1+t)+(1+t)+(-1+2t)\le6\), so \(1+4t\le6\) and \(t\le\frac54\). 5. Combining the restrictions gives \(\frac12\le t\le\frac54\). 6. At \(t=\frac12\), the point is \(P_1\left(\frac32,\frac32,0\right)\). 7. At \(t=\frac54\), the point is \(P_2\left(\frac94,\frac94,\frac32\right)\).

The line enters and exits at \(P_1\left(\frac32,\frac32,0\right)\) and \(P_2\left(\frac94,\frac94,\frac32\right)\). The portion inside corresponds to \(t\in\left[\frac12,\frac54\right]\).

- 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. a) The normal vectors satisfy \(\langle -10,2,-4\rangle=-2\langle 5,-1,2\rangle\), 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. b) A plane parallel to \(E_1\) can use the same normal vector, so write \(5x-y+2z=d\). 4. Substitute \(P(2, -3, 1)\): \(d=5(2)-(-3)+2(1)=15\). Thus \(E_3:5x-y+2z=15\). 5. c) Every plane parallel to \(E_1\) has the form \(5x-y+2z=d\), where \(d\in\mathbb{R}\).

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

- 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 \(c\)

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

1. Let \(\mathbf{n}=\langle p,q,w\rangle\) 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 standard equation of \(E_1\) is \(-x+y+z=d\). Using \((1, 2, 0)\), \(d=1\). 5. The initial point of \(E_2\) also satisfies this equation because \(-2+3+0=1\). 6. Parallel planes that share a point are identical.

The two equations describe the same plane.

- 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

- 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 planes are identical.

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

1. Two directions in \(E\) are \(\overrightarrow{AB}=\langle 2,1,-2\rangle\) and \(\overrightarrow{AC}=\langle -1,2,0\rangle\). 2. A normal vector to \(F\) is \(\mathbf{n}_F=\langle 2,1,2.5\rangle\). 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. Planes \(E\) and \(F\) 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) The two connecting vectors are not scalar multiples. 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) \(P\notin h\). b) \(F:3x+12y-5z=22\) c) Yes, \(R=(2, 3, 4)\) lies 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 \(E:2x-y+5z=9\).

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

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

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

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

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

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

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

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

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

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

- 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 \(S_2=\left(-\frac32,6,-\frac92\right)\)

- 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 \(t=-\frac12\). 6. Therefore, \(P_0=\left(\frac72,0,\frac{15}{2}\right)\). 7. The distance is \(\|\overrightarrow{QP_0}\|=\left\|-\frac12\langle 1,2,1\rangle\right\|=\frac12\sqrt6=\sqrt{\frac32}\approx1.22\).

a) \(M:x+2y+z=11\) b) \(P_0\left(\frac72,0,\frac{15}{2}\right)\) c) \(\sqrt{\frac32}\approx1.22\) units

- 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. One consistent labeling is \(A(4,-4,0)\), \(B(4,4,0)\), \(C(-4,4,0)\), 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 \(t=\frac12\). 4. Because \(\frac12\) is between \(0\) and \(1\), point \(P\) lies on edge \(\overline{SA}\).

a) One consistent labeling is \(A(4,-4,0)\), \(B(4,4,0)\), \(C(-4,4,0)\), and \(D(-4,-4,0)\). b) Yes. Point \(P(2,-2,6)\) lies on \(\overline{SA}\) when \(t=\frac12\).

- A vertical translation changes only the \(z\)-coordinate. - Use the three-dimensional distance formula for \(\overline{BF}\). - Multiply the area of the triangular base by the prism height.

1. Add \(10\) to the \(z\)-coordinate of each base vertex: \(D=(0, 0, 10)\), \(E=(8, 0, 10)\), and \(F=(0, 6, 10)\). 2. \(\overrightarrow{BF}=F-B=\begin{pmatrix}-8\\6\\10\end{pmatrix}\). Therefore, \(\lVert\overrightarrow{BF}\rVert=\sqrt{(-8)^2+6^2+10^2}=\sqrt{200}=10\sqrt{2}\approx14.14\) units. 3. The base is a right triangle with area \(\frac{1}{2}\cdot 8\cdot 6=24\) square units. The volume is \(24\cdot 10=240\) cubic units.

a) \(D=(0, 0, 10)\), \(E=(8, 0, 10)\), \(F=(0, 6, 10)\) b) \(10\sqrt{2}\) units, or approximately \(14.14\) units c) \(240\) cubic 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. - Equate the base distance and lateral-face distance.

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 \(P=\left(5,5,\frac{10}{3}\right)\). Because the point lies on the symmetry axis of the regular pyramid, the four lateral-face distances are equal.

a) \(BCS:12y+5z=120\) b) \(d(P,\text{base})=z\), and \(d(P,BCS)=\frac{60-5z}{13}\) c) \(P\left(5,5,\frac{10}{3}\right)\)

- 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.

- 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, \(L=\left(0,0,\frac{34}{5}\right)\). By symmetry about the \(z\)-axis, the distance to each lateral face is the same.

a) \(ABS:12x+5z=60\); one normal is \(\langle12,0,5\rangle\) with magnitude \(13\) b) \(d(L,ABS)=\frac{60-5h}{13}\) c) \(L\left(0,0,\frac{34}{5}\right)\)

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

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

\(r=\frac34\)

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

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

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

- Find a normal vector to the plane from two independent directions in it. - Compare \(LN\) and \(MN\). - For part c, use the actual oblique face \(LMN\) as the base and compute the point-to-plane distance from \(O\). - For part d, every point on the \(z\)-axis has the form \((0,0,z)\).

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 \(\frac{|0-8|}{\sqrt{2^2+2^2+(-1)^2}=3}=\frac83\). 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 \(P=(0,0,-6)\)

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

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

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

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

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

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

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

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

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

- Substitute the four vertices into a general sphere equation. - Complete the square to identify the center and radius. - For one sphere to lie inside another, compare the center distance plus the smaller radius with the larger radius.

1. Let \(K_1\) have equation \(x^2+y^2+z^2+ax+by+cz+d=0\). 2. Substituting \(O\) gives \(d=0\). Substituting \(P\), \(Q\), and \(R\) gives \(a=b=c=-8\). 3. Thus, \(x^2+y^2+z^2-8x-8y-8z=0\). Completing the square gives \((x-4)^2+(y-4)^2+(z-4)^2=48\). 4. Therefore, \(M_1=(4,4,4)\) and \(r_1=4\sqrt{3}\). 5. Sphere \(K_2\) has center \(M_2=(2,2,2)\) and radius \(2\). The center distance is \(M_1M_2=2\sqrt{3}\). 6. Since \(2\sqrt{3}+2<4\sqrt{3}\), every point of \(K_2\) is strictly inside \(K_1\).

a) \(K_1: (x-4)^2+(y-4)^2+(z-4)^2=48\); center \((4,4,4)\); radius \(4\sqrt{3}\) b) \(M_1M_2+r_2=2\sqrt{3}+2<4\sqrt{3}=r_1\), so \(K_2\) lies completely inside \(K_1\).

- Use a point on the base rim and the highest point of the dome. - Both points are one sphere radius from the center. - Compare the instrument's squared distance from the center with \(R^2\).

1. A point on the base rim is \(A(24,0,0)\), and the dome's top is \(H(0,0,18)\). 2. Let the sphere center be \(M(0,0,c)\). Equal distances from \(M\) to \(A\) and \(H\) give \(24^2+c^2=(18-c)^2\). 3. Solving gives \(c=-7\). Then \(R^2=24^2+7^2=625\), so \(R=25\,\text{m}\). 4. The sphere equation is \(x^2+y^2+(z+7)^2=625\). 5. For \(P\), the squared distance to the center is \(15^2+10^2+(25+7)^2=1349\). 6. Since \(1349>625\), the instrument lies outside the sphere.

a) Radius \(25\,\text{m}\); equation \(x^2+y^2+(z+7)^2=625\) b) The instrument lies outside the sphere.

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

1. The radius equals the distance from \(M\) to the tangent plane: \(r=\frac{|1-10|}{\sqrt{1^2+(-2)^2+2^2}}=\frac93=3\). 2. The point of tangency is the perpendicular projection of \(M\) onto \(E\). Write \(B=(1,0,0)+t\langle1,-2,2\rangle\). 3. Substitution into the plane equation gives \(1+9t-10=0\), so \(t=1\). Therefore, \(B=(2,-2,2)\). 4. The reflected center lies the same distance beyond \(B\), so \(M'=M+2\overrightarrow{MB}\). 5. Since \(\overrightarrow{MB}=\langle1,-2,2\rangle\), \(M'=(1,0,0)+2\langle1,-2,2\rangle=(3,-4,4)\).

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

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

1. Two direction vectors in the plane are \(\overrightarrow{AB}=\langle-2,2,0\rangle\) and \(\overrightarrow{AC}=\langle0,2,-2\rangle\). 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. Equating coordinates produces an inconsistent system, so the lines are skew. 2. A vector perpendicular to both directions is \(\mathbf{n}=\langle 2,1,-2\rangle\), since both dot products are \(0\). Its magnitude is \(3\). 3. The vector between the given points is \(\mathbf{w}=\langle 3,4,6\rangle\). Thus, \(d(g,h)=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{\lvert 6+4-12\rvert}{3}=\frac{2}{3}\). 4. Let \(F_g=(2,0,1)+r\langle 1,2,2\rangle\) and \(F_h=(5,4,7)+s\langle 2,-2,1\rangle\). The shortest connecting vector \(F_h-F_g\) is perpendicular to both line directions. Solving \((F_h-F_g)\cdot\langle 1,2,2\rangle=0\) and \((F_h-F_g)\cdot\langle 2,-2,1\rangle=0\) gives \(r=\frac{23}{9}\) and \(s=-\frac{4}{9}\). 5. Therefore, \(F_g=\left(\frac{41}{9},\frac{46}{9},\frac{55}{9}\right)\) and \(F_h=\left(\frac{37}{9},\frac{44}{9},\frac{59}{9}\right)\). 6. For the tangent sphere, let \(G=(2+r,2r,1+2r)\) be a point on \(g\). The perpendicular-foot condition is \((G-M)\cdot\langle 1,2,2\rangle=0\), which gives \(9r-15=0\), so \(r=\frac{5}{3}\). Then \(G-M=\left\langle-\frac{4}{3},\frac{1}{3},\frac{1}{3}\right\rangle\), and the sphere's radius is \(\sqrt{\frac{16}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{2}\).

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

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

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

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

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

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

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

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

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

- Compare the two distances from \(C_t\) to the fixed endpoints. - Locate the midpoint of the base. - Use base times height for the area and account for the absolute value of \(t\).

1. \(C_t=(2+2t, 2-t, 4)\). 2. \(\overrightarrow{AC_t}=(1+2t, 2-t, 2)\), so \(\|\overrightarrow{AC_t}\|^2=(1+2t)^2+(2-t)^2+2^2=5t^2+9\). 3. \(\overrightarrow{BC_t}=(-1+2t, -2-t, -2)\), so \(\|\overrightarrow{BC_t}\|^2=(-1+2t)^2+(-2-t)^2+(-2)^2=5t^2+9\). Thus, \(AC_t=BC_t\). When \(t=0\), \(C_t\) is the midpoint of \(\overline{AB}\), so the triangle is degenerate. 4. The base length is \(AB=\sqrt{2^2+4^2+4^2}=6\). 5. Its midpoint is \(M(2, 2, 4)\), and \(\overrightarrow{MC_t}=(2t, -t, 0)\). Since \(\overrightarrow{AB}\cdot\overrightarrow{MC_t}=2\cdot2t+4\cdot(-t)+4\cdot0=0\), this vector is perpendicular to the base, so the height is \(|t|\sqrt5\). 6. The area is \(\frac12\cdot6\cdot|t|\sqrt5=3|t|\sqrt5\). 7. Setting this equal to \(3\sqrt5\) gives \(|t|=1\), so \(t=\pm1\).

a) \(AC_t=BC_t=\sqrt{5t^2+9}\) for every \(t\ne0\). b) \(t=-1\) or \(t=1\)

- Compare the squared lengths of \(AC_t\) and \(BC_t\). - For an equilateral triangle, set a leg length equal to \(AB\). - Use a dot product to verify the altitude. - Use the base and altitude to find the area.

1. \(\overrightarrow{AC_t}=\begin{pmatrix}1+2t\\2-2t\\2+t\end{pmatrix}\) and \(\overrightarrow{BC_t}=\begin{pmatrix}-1+2t\\-2-2t\\-2+t\end{pmatrix}\). Both squared lengths simplify to \(9t^2+9\). Thus, \(AC_t=BC_t\). When \(t=0\), \(C_t\) is the midpoint of \(AB\), so no triangle is formed. 2. \(\overrightarrow{AB}=\begin{pmatrix}2\\4\\4\end{pmatrix}\), so \(AB^2=36\). For an equilateral triangle, \(9t^2+9=36\), giving \(t=\pm\sqrt{3}\). 3. The midpoint is \(M=(3, 3, 1)\), and \(\overrightarrow{MC_t}=\begin{pmatrix}2t\\-2t\\t\end{pmatrix}\). Then \(\overrightarrow{MC_t}\cdot\overrightarrow{AB}=4t-8t+4t=0\). 4. The base length is \(6\), and the altitude is \(\|\overrightarrow{MC_t}\|=3|t|\). Thus, the area is \(\frac{1}{2}\cdot6\cdot3|t|=9|t|\). Setting \(9|t|=18\) gives \(t=\pm2\).

a) \(AC_t=BC_t=\sqrt{9t^2+9}\) for \(t\ne0\) b) \(t=\pm\sqrt{3}\) c) \(M(3, 3, 1)\) and \(\overrightarrow{MC_t}\cdot\overrightarrow{AB}=0\) d) \(t=\pm2\)

- Compare the squared side lengths. - For the right-angle condition, compare dot products of side vectors at the triangle's vertices. - Parameterize segment \(PQ\). - Use the squared-length equation for the final part.

1. \(\overrightarrow{PS_k}=\begin{pmatrix}k+1\\1-2k\\k+1\end{pmatrix}\) and \(\overrightarrow{QS_k}=\begin{pmatrix}k-1\\-1-2k\\k-1\end{pmatrix}\). Both squared lengths simplify to \(6k^2+3\), so they are equal. 2. In this isosceles triangle, a right angle must occur at \(S_k\). The dot product \(\overrightarrow{S_kP}\cdot\overrightarrow{S_kQ}=6k^2-3\). Setting it equal to \(0\) gives \(k=\pm\frac{\sqrt{2}}{2}\). 3. A point on \(PQ\) has the form \(P+r(Q-P)\), where \(0\le r\le1\). Solving \(S_k=P+r(Q-P)\) gives \(k=0\) and \(r=\frac{1}{2}\). Thus, \(S_0\) is the midpoint of \(PQ\). 4. Set \(6k^2+3=21\). Then \(k^2=3\), and the positive value is \(k=\sqrt{3}\).

a) \(PS_k=QS_k=\sqrt{6k^2+3}\) b) \(k=\pm\frac{\sqrt{2}}{2}\); the right angle is at \(S_k\). c) Yes. For \(k=0\), \(S_k\) is the midpoint of \(PQ\). d) \(k=\sqrt{3}\)

- Write a general point on each line. - The shortest connecting segment is perpendicular to both line directions. - Use two dot-product equations to solve for the parameters.

1. Write \(P=(1+r,r,1)\) and \(Q=(0,1+s,4+s)\). Then \(\overrightarrow{PQ}=\langle -1-r,1+s-r,3+s\rangle\). 2. For the shortest connection, \(\overrightarrow{PQ}\) must be perpendicular to both line directions. 3. From \(\overrightarrow{PQ}\cdot\langle 1,1,0\rangle=0\), obtain \(-2r+s=0\). 4. From \(\overrightarrow{PQ}\cdot\langle 0,1,1\rangle=0\), obtain \(-r+2s=-4\). 5. Solving gives \(r=-\frac{4}{3}\) and \(s=-\frac{8}{3}\). Therefore, \(P=\left(-\frac{1}{3},-\frac{4}{3},1\right)\) and \(Q=\left(0,-\frac{5}{3},\frac{4}{3}\right)\). 6. The connecting vector is \(\left\langle\frac{1}{3},-\frac{1}{3},\frac{1}{3}\right\rangle\), so the minimum distance is \(\frac{\sqrt{3}}{3}\approx0.577\).

\(P=\left(-\frac{1}{3},-\frac{4}{3},1\right)\), \(Q=\left(0,-\frac{5}{3},\frac{4}{3}\right)\), and \(d=\frac{\sqrt{3}}{3}\approx0.577\) unit

- Find a vector perpendicular to both line directions. - Form a vector between one point on each line. - Project that vector onto the common perpendicular direction.

1. A vector perpendicular to both direction vectors \(\langle 1,1,0\rangle\) and \(\langle 0,1,1\rangle\) is \(\mathbf{n}=\langle 1,-1,1\rangle\), with magnitude \(\sqrt{3}\). 2. The vector between the given points is \(\mathbf{w}=\langle 1,-2,1\rangle\). 3. The shortest distance is the magnitude of the scalar projection of \(\mathbf{w}\) onto \(\mathbf{n}\): \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\approx2.31\).

\(\frac{4\sqrt{3}}{3}\approx2.31\) units

- Find a direction perpendicular to both laser paths. - Project the vector between the paths onto that direction. - For the closest points, require the connecting vector to be perpendicular to both paths.

1. A vector perpendicular to both direction vectors is \(\mathbf{n}=\langle 1,-1,1\rangle\), with magnitude \(\sqrt{3}\). 2. The vector between the given points is \(\langle 2,0,2\rangle\). Therefore, the distance is \(d=\frac{\lvert\langle 2,0,2\rangle\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\approx2.31\,\text{m}\). 3. Let \(P=(r,r,0)\) and \(Q=(2,s,2+s)\). Then \(\overrightarrow{PQ}=\langle 2-r,s-r,2+s\rangle\). 4. The shortest connecting vector is perpendicular to both line directions. Solving \(\overrightarrow{PQ}\cdot\langle 1,1,0\rangle=0\) and \(\overrightarrow{PQ}\cdot\langle 0,1,1\rangle=0\) gives \(r=\frac{2}{3}\) and \(s=-\frac{2}{3}\). 5. Thus, \(P=\left(\frac{2}{3},\frac{2}{3},0\right)\) and \(Q=\left(2,-\frac{2}{3},\frac{4}{3}\right)\).

a) \(\frac{4\sqrt{3}}{3}\approx2.31\,\text{m}\) b) \(P=\left(\frac{2}{3},\frac{2}{3},0\right)\), \(Q=\left(2,-\frac{2}{3},\frac{4}{3}\right)\)

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

1. \(\overrightarrow{AB}=\langle 6,0,0\rangle\) and \(\overrightarrow{AD}=\langle 0,6,0\rangle\). They have equal magnitude \(6\) and dot product \(0\). Also, opposite side vectors are equal, so \(ABCD\) is a square with side length \(6\,\text{m}\). 2. A general point on \(AB\) is \(F(t)=(2+6t,2,0)\). The condition \(\overrightarrow{SF}\cdot\overrightarrow{AB}=0\) gives \(t=\frac{1}{2}\), so \(F=(5,2,0)\). Thus, the distance is \(SF=\sqrt{0^2+(-3)^2+(-10)^2}=\sqrt{109}\approx 10.44\,\text{m}\). 3. A general point on \(CS\) is \(G(r)=(8-3r,8-3r,10r)\). Its direction vector is \(\langle -3,-3,10\rangle\). 4. Require \(\overrightarrow{AG}\cdot\langle -3,-3,10\rangle=0\). This gives \(r=\frac{18}{59}\), so \(G=\left(\frac{418}{59},\frac{418}{59},\frac{180}{59}\right)\). 5. Therefore, \(AG=\frac{60}{\sqrt{59}}\approx 7.81\,\text{m}\).

a) Adjacent side vectors have length \(6\,\text{m}\) and dot product \(0\), and opposite sides are parallel and equal. b) \(\sqrt{109}\approx 10.44\,\text{m}\) c) \(\frac{60}{\sqrt{59}}\approx 7.81\,\text{m}\)

- Determine whether the three vectors from \(O\) can lie in one plane. - Use a dot product to find the perpendicular foot on \(QR\). - For the face area, use \(\frac{1}{2}\times\text{base}\times\text{height}\).

1. The vectors \(\overrightarrow{OP}=\langle 0,0,4\rangle\), \(\overrightarrow{OQ}=\langle 6,0,0\rangle\), and \(\overrightarrow{OR}=\langle 0,8,0\rangle\) point along three different coordinate axes. They do not lie in one plane through \(O\), so the four points are noncoplanar and form a tetrahedron. 2. A general point on \(QR\) is \(F(t)=(6-6t,8t,0)\). For the perpendicular from \(O\), require \(\overrightarrow{OF}\cdot\langle -6,8,0\rangle=0\). This gives \(t=\frac{9}{25}\), so \(F=\left(\frac{96}{25},\frac{72}{25},0\right)\). 3. Thus, the distance from \(O\) to \(QR\) is \(OF=\frac{24}{5}=4.8\). 4. The same parameter gives the perpendicular foot from \(P\) because \(P\) differs from \(O\) only in the z-coordinate, while line \(QR\) lies in the xy-plane. The altitude from \(P\) to \(QR\) is \(PF=\frac{4\sqrt{61}}{5}\). 5. Since \(QR=10\), the area is \(\frac{1}{2}(10)\left(\frac{4\sqrt{61}}{5}\right)=4\sqrt{61}\approx 31.24\) square units.

a) The three vectors from \(O\) to \(P\), \(Q\), and \(R\) lie along different coordinate axes, so the four points are noncoplanar. b) \(\frac{24}{5}=4.8\) units c) \(4\sqrt{61}\approx 31.24\) square units

- Use \(B-A\) for the direction of \(g\). - Test perpendicularity with a dot product. - Find a vector perpendicular to both line directions for the distance. - Use the Gram determinant for the triangle area.

1. \(\overrightarrow{AB}=\langle 1,2,-2\rangle\), so \(g: (x,y,z)=(2,1,3)+r\langle 1,2,-2\rangle\). 2. The direction vectors have dot product \(\langle 1,2,-2\rangle\cdot\langle 2,2,1\rangle=4\), so they are not perpendicular. 3. The direction vectors are not scalar multiples, and equating the line coordinates produces an inconsistent system. Thus, the lines are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle 6,-5,-2\rangle\), with magnitude \(\sqrt{65}\). The vector from the point on \(g\) to the point on \(h\) is \(\mathbf{w}=\langle -2,3,-2\rangle\). Therefore, \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{23}{\sqrt{65}}=\frac{23\sqrt{65}}{65}\approx2.85\). 5. For triangle \(ABC\), \(\overrightarrow{AB}=\langle 1,2,-2\rangle\) and \(\overrightarrow{AC}=\langle 0,5,-1\rangle\). Its area is \(\frac{1}{2}\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=\frac{3\sqrt{10}}{2}\approx4.74\).

a) \(g: (x,y,z)=(2,1,3)+r\langle 1,2,-2\rangle\) b) No; the dot product is \(4\). c) The lines are skew, and \(d=\frac{23\sqrt{65}}{65}\approx2.85\) units. d) \(\frac{3\sqrt{10}}{2}\approx4.74\) square units

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

1. \(\overrightarrow{P_1P_2}=\langle 2,1,-2\rangle\), so \(h: (x,y,z)=(1,2,2)+s\langle 2,1,-2\rangle\). 2. The direction vectors have dot product \(\langle 2,-2,1\rangle\cdot\langle 2,1,-2\rangle=0\), so they are perpendicular. 3. The direction vectors are not scalar multiples. Equating coordinates gives parameter values from the first two equations that fail the third, so the lines do not intersect. Therefore, they are skew. 4. A vector perpendicular to both directions is \(\mathbf{n}=\langle 1,2,2\rangle\), with magnitude \(3\). The vector between the given points is \(\mathbf{w}=\langle 1,2,-3\rangle\). 5. The distance is \(d=\frac{\lvert\mathbf{w}\cdot\mathbf{n}\rvert}{\lVert\mathbf{n}\rVert}=\frac{1}{3}\).

a) \(h: (x,y,z)=(1,2,2)+s\langle 2,1,-2\rangle\) b) The direction vectors are perpendicular. c) The lines are skew. d) \(\frac{1}{3}\) unit

- In three dimensions, a quadrilateral classification requires coplanarity first. - Try to express \(\overrightarrow{AD}\) as a linear combination of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). - Then compare pairs of adjacent side lengths. - A square must also have a right angle.

1. \(\overrightarrow{AB}=\langle0,-3,-6\rangle\), \(\overrightarrow{AC}=\langle4,2,-4\rangle\), and \(\overrightarrow{AD}=\langle k+2,2k+1,2k-2\rangle\). 2. The identity \(\overrightarrow{AD}=-\frac{k}{2}\overrightarrow{AB}+\left(\frac{k}{4}+\frac12\right)\overrightarrow{AC}\) holds for every real \(k\). Therefore, \(D\) lies in plane \(ABC\), so the four points are coplanar. 3. \(AB=BC=\sqrt{45}\), while \(AD=CD=3\sqrt{k^2+1}\). Thus, for a nondegenerate quadrilateral, \(ABCD\) is a kite. 4. For a rhombus, set \(3\sqrt{k^2+1}=\sqrt{45}\). Then \(k^2=4\), so \(k=\pm2\). The value \(k=-2\) is degenerate because \(D=B\), leaving \(k=2\). 5. At \(k=0\), \(D\) is the midpoint of \(AC\), so the quadrilateral is degenerate. At \(k=-2\), \(D=B\). 6. A square would need a right angle in addition to equal sides. The only nondegenerate rhombus value is \(k=2\), and at that value adjacent side vectors do not have dot product \(0\). Therefore, no value produces a square.

a) The four points are coplanar because \(\overrightarrow{AD}=-\frac{k}{2}\overrightarrow{AB}+\left(\frac{k}{4}+\frac12\right)\overrightarrow{AC}\). Also, \(AB=BC=\sqrt{45}\) and \(AD=CD=3\sqrt{k^2+1}\), so the nondegenerate cases are kites. b) \(k=2\) c) At \(k=-2\), \(D=B\); at \(k=0\), \(D\) is the midpoint of \(AC\). d) No value of \(k\) produces a square.

- Solve each pair of line equations to find a common point. - Form side vectors from the vertices. - Use the dot-product angle formula after the vertices are known. - The centroid is the coordinate average of the three vertices. - A parallel line can use a scalar multiple of the side vector.

1. Pairwise solution of the coordinate equations gives \(g_1\cap g_2=B(4,2,2)\), \(g_2\cap g_3=C(1,5,2)\), and \(g_3\cap g_1=A(1,2,-1)\). The three points are distinct, so the lines form a triangle. 2. The side vectors are \(\overrightarrow{AB}=\langle 3,0,3\rangle\), \(\overrightarrow{AC}=\langle 0,3,3\rangle\), and \(\overrightarrow{BC}=\langle -3,3,0\rangle\). 3. At \(A\), \(\cos A=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{\|\overrightarrow{AB}\|\|\overrightarrow{AC}\|}=\frac{9}{\sqrt{18}\sqrt{18}}=\frac12\), so \(A=60^\circ\). 4. The same calculation at \(B\) and \(C\) gives \(B=C=60^\circ\). Thus, the triangle is equilateral. 5. The centroid is the coordinate average: \(S=\frac13(A+B+C)=(2,3,1)\). 6. A line through \(S\) parallel to \(AB\) may use \(\langle 1,0,1\rangle\) as its direction: \(h:\mathbf{r}(u)=\langle 2,3,1\rangle+u\langle 1,0,1\rangle\).

a) The lines intersect pairwise in three distinct points. b) \(A(1,2,-1)\), \(B(4,2,2)\), \(C(1,5,2)\) c) \(A=B=C=60^\circ\); the triangle is equilateral. d) \(S(2,3,1)\) e) \(h:\mathbf{r}(u)=\langle 2,3,1\rangle+u\langle 1,0,1\rangle\)

- 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.

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

- Identify the two side faces crossed by a line parallel to the x-axis. - Find a normal vector to each face using two directions in the face. - Intersect the line with each face. - Compute the distance between the two boundary points.

1. The line crosses the side faces \(ADS\) and \(BCS\). 2. For face \(ADS\), directions are \(\overrightarrow{AD}=\langle 0,4,0\rangle\) and \(\overrightarrow{AS}=\langle 2,2,8\rangle\). A normal vector orthogonal to both is \(\langle -4,0,1\rangle\), giving the plane equation \(-4x+z=-8\). 3. On \(h\), \(x=s\) and \(z=2\). Thus, \(-4s+2=-8\), so \(s=\frac52\). The first boundary point is \(P_1\left(\frac52,4,2\right)\). 4. For face \(BCS\), a normal vector is \(\langle 4,0,1\rangle\), giving \(4x+z=24\). 5. Substitution gives \(4s+2=24\), so \(s=\frac{11}{2}\). The second boundary point is \(P_2\left(\frac{11}{2},4,2\right)\). 6. The distance between the points is \(\frac{11}{2}-\frac52=3\).

\(3\) units

- 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 \(g\).

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

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

\(8\) cubic units

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

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

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

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

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

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

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

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

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

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

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

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

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

1. Triangle \(OAB\) is right with legs of length \(6\), so its area is \(\frac{1}{2}(6)(6)=18\). 2. For plane \(ABC\), \(\overrightarrow{AB}=\langle -6,6,0\rangle\) and \(\overrightarrow{AC}=\langle 0,6,6\rangle\). A vector \(\langle a,b,c\rangle\) orthogonal to both satisfies \(-6a+6b=0\) and \(6b+6c=0\). Choose \(\mathbf{n}=\langle 1,1,-1\rangle\). 3. Using point \(A\), the plane equation is \(x+y-z=6\). 4. The distance from the origin is \(\frac{\lvert -6\rvert}{\sqrt{1^2+1^2+(-1)^2}}=2\sqrt{3}\approx3.46\). 5. Using base \(OAB\), the height from \(C\) is \(6\), so \(V=\frac{1}{3}(18)(6)=36\). 6. For a second method, the area of triangle \(ABC\) is \(\frac{1}{2}\sqrt{(\overrightarrow{AB}\cdot\overrightarrow{AB})(\overrightarrow{AC}\cdot\overrightarrow{AC})-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2}=18\sqrt{3}\). Using the height from \(O\), \(V=\frac{1}{3}(18\sqrt{3})(2\sqrt{3})=36\).

a) \(18\) square units b) \(E_{ABC}: x+y-z=6\) c) \(2\sqrt{3}\approx3.46\) units d) \(36\) cubic units by both methods

- Notice that \(A\), \(B\), and \(C\) have the same z-coordinate. - Use \(V=\frac{1}{3}Gh\) 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 \(\frac{1}{2}(4)(2\sqrt{5})=4\sqrt{5}\approx8.94\) square units. 5. Let \(h\) be the distance from \(C\) to plane \(ABD\). From \(16=\frac{1}{3}(4\sqrt{5})h\), \(h=\frac{12}{\sqrt{5}}=\frac{12\sqrt{5}}{5}\approx5.37\) units.

a) Area of \(ABC\): \(12\) square units; distance from \(D\) to plane \(ABC\): \(4\) units b) \(V=16\) cubic units c) Area of \(ABD\): \(4\sqrt{5}\approx8.94\) square units d) Distance from \(C\) to plane \(ABD\): \(\frac{12\sqrt{5}}{5}\approx5.37\) units

- 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) \(S\left(\frac{29}{20},\frac{11}{10},\frac{24}{5}\right)\)

- 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) \(A'=(-1, 6, 1)\)