Free Math Worksheets

- 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 \(\lVert\overrightarrow{AB}\rVert=7\)

- 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

- 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.

- 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: \(AB<AC<BC\).

- 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 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.

- 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 point set is the sphere centered at \(A\) with radius \(7\): \((x-2)^2+(y+3)^2+(z-5)^2=49\). 2. For \(P\), the left side is \(0^2+7^2+0^2=49\), so \(P\) satisfies the condition. 3. For \(Q\), the left side is \(4^2+3^2+0^2=25\), so \(Q\) does not satisfy the condition. 4. Set \(x=2\) and \(z=5\). Then \((y+3)^2=49\), so \(y=4\) or \(y=-10\). 5. The 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 \((2,-10,5)\)

- 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.

- 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.

- 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 \(\vec{v}=\begin{pmatrix}0\\0\\6\end{pmatrix}\). 3. \(\vec{s}=\begin{pmatrix}3\\4\\12\end{pmatrix}\), so \(\|\vec{s}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

1. Third component \(0\); magnitude \(8\) 2. \(\begin{pmatrix}0\\0\\6\end{pmatrix}\) 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 \(A'\): \(\vec{v}=\overrightarrow{AA'}=\begin{pmatrix}2-2\\1-1\\9-1\end{pmatrix}=\begin{pmatrix}0\\0\\8\end{pmatrix}\). 2. Translate \(B\) and \(C\) by \(\vec{v}\): \(B'=(6, 1, 1)+(0, 0, 8)=(6, 1, 9)\) and \(C'=(2, 4, 1)+(0, 0, 8)=(2, 4, 9)\). 3. A lateral edge has length \(\|\vec{v}\|=\sqrt{0^2+0^2+8^2}=8\).

a) \(\vec{v}=\begin{pmatrix}0\\0\\8\end{pmatrix}\) 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: \(\vec{r}_1=\begin{pmatrix}4\\-3\\1\end{pmatrix}\). 2. Reflection across the \(yz\)-plane changes the sign of the \(x\)-component: \(\vec{r}_2=\begin{pmatrix}-4\\-3\\1\end{pmatrix}\). 3. Relative to \(\vec{r}\), 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) \(\vec{r}_1=\begin{pmatrix}4\\-3\\1\end{pmatrix}\) b) \(\vec{r}_2=\begin{pmatrix}-4\\-3\\1\end{pmatrix}\) 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: \(\vec{r}_1=\begin{pmatrix}4\\-1\\-3\end{pmatrix}\). 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, \(\vec{r}_2=\begin{pmatrix}-4\\-1\\3\end{pmatrix}\).

a) \(\vec{r}_1=\begin{pmatrix}4\\-1\\-3\end{pmatrix}\) b) \(\vec{r}_2=\begin{pmatrix}-4\\-1\\3\end{pmatrix}\)

- 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.

- 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.

- 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.

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

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

- 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 \(g: \vec{r}(t)=\langle 2,-3,4\rangle+t\langle 3,3,-6\rangle\). 2. a) A point in the xy-plane has \(z=0\). Thus \(4-6t=0\), so \(t=\frac23\). 3. Substitution gives \(S=\left(2+3\left(\frac23\right), -3+3\left(\frac23\right), 0\right)=(4, -1, 0)\). 4. b) The displacement from \(S\) to \(Q\) is \(\overrightarrow{SQ}=\langle -4,2,0\rangle\). 5. The distance is \(\sqrt{(-4)^2+2^2+0^2}=\sqrt{20}=2\sqrt5\).

a) \(S(4, -1, 0)\) b) \(2\sqrt5\)

- 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 \((0, 0, 4.5)\)

- 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 \(\vec{r}(t)=\langle 2,4,9\rangle+t\langle 1,2,-2\rangle\). 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 \(\sqrt{(-8)^2+(-16)^2+20^2}=\sqrt{720}=12\sqrt5\,\text{m}\).

\(12\sqrt5\,\text{m}\)

- 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.

- The center is the midpoint of a diameter. - Compare \(MR^2\) with the squared radius. - In a circle, an inscribed angle that intercepts a diameter is a right angle.

1. Because \(M\) is the midpoint of the diameter, \(Q=2M-P=(-1,7,-1)\). 2. The squared radius is \(MP^2=3^2+(-4)^2+0^2=25\). 3. For \(R\), \(MR^2=0^2+3^2+4^2=25\), so \(R\) lies on the sphere. 4. The three points lie in one plane, and that plane intersects the sphere in a circle with diameter \(PQ\). By the Inscribed Angle Theorem, an angle that intercepts diameter \(PQ\) is a right angle. Thus, \(\angle PRQ=90^\circ\).

a) \(Q=(-1,7,-1)\) b) \(MR=MP=5\), so \(R\) lies on the sphere. c) Triangle \(PQR\) is right at \(R\).

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

- 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. - Points on the sphere are one radius from the center.

1. The center is \(M(5,-2,1)\), and the radius is \(7\). 2. The direction from \(M\) to \(Q\) is \(\overrightarrow{MQ}=\langle 2,3,6\rangle\), whose magnitude is \(7\). 3. The line is \((x,y,z)=(5,-2,1)+t\langle 2,3,6\rangle\). 4. Substitution into the sphere equation gives \(49t^2=49\), so \(t=1\) or \(t=-1\). 5. The 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 \(\overrightarrow{AB}=\begin{pmatrix}-6\\6\\0\end{pmatrix}\) and \(\overrightarrow{AS}=\begin{pmatrix}-6\\0\\8\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(-6a+6b=0\) and \(-6a+8c=0\). One solution is \(\vec{n}=\begin{pmatrix}4\\4\\3\end{pmatrix}\). 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

- 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.

- 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 \(a=-4\)

- 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 \(P_2(0, 7, 0)\)

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

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

- 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.

- Write a general point on the \(y\)-axis. - Use the squared distance formula. - Keep both solutions of the quadratic equation.

1. A point on the \(y\)-axis has the form \(P=(0, y, 0)\). 2. The distance condition gives \((0-1)^2+(y-3)^2+(0-4)^2=26\). 3. Thus, \(17+(y-3)^2=26\), so \((y-3)^2=9\). 4. Therefore, \(y=6\) or \(y=0\).

\((0, 6, 0)\) and \((0, 0, 0)\)

- 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) \(z=0\)

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

- 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.

- 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.

- 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) \(Q''=(0, 0, -3)\)

- 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 \(z=1\)

- 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 \(\overrightarrow{PS}=\begin{pmatrix}3-1\\3-1\\5-0\end{pmatrix}=\begin{pmatrix}2\\2\\5\end{pmatrix}\). 2. Applying this vector gives \(Q'=(5, 1, 0)+(2, 2, 5)=(7, 3, 5)\) and \(R'=(1, 4, 0)+(2, 2, 5)=(3, 6, 5)\). 3. In part b, the translation vector is \(\overrightarrow{QS}=\begin{pmatrix}3-5\\3-1\\5-0\end{pmatrix}=\begin{pmatrix}-2\\2\\5\end{pmatrix}\). 4. Applying this vector gives \(P'=(1, 1, 0)+(-2, 2, 5)=(-1, 3, 5)\) and \(R'= (1, 4, 0)+(-2, 2, 5)=(-1, 6, 5)\). 5. 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. \(\overrightarrow{PQ}=Q-P=(2, 3, -2)\). 4. \(\overrightarrow{P'Q'}=Q'-P'=(-2, -3, 2)=-\overrightarrow{PQ}\). 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) \(A''=(-3, 6, 5)\), \(B''=(2, 0, 9)\), \(C''=(5, 3, 4)\)

- 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

- 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 \(t=3\), \(\vec{x}=\begin{pmatrix}-11\\25\\10\end{pmatrix}\). 4. For \(t=-3\), \(\vec{x}=\begin{pmatrix}13\\-17\\-14\end{pmatrix}\).

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

- 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 \(P_2(0, 0, 24)\)

- 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 \(t=-2\)

- 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: \(P=2\sqrt{29}+2\sqrt{61}\approx 26.39\)

- What form do points on the x-axis have in three dimensions? - Use the three-dimensional distance formula. - Square both sides to remove the radical. - Check both solutions of the resulting squared equation.

1. The distance from \(A\) to the origin is \(\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\). 2. A point on the x-axis has the form \(P(x, 0, 0)\). 3. Set its distance from \(A\) equal to \(13\): \(\sqrt{(x-3)^2+(0-4)^2+(0-12)^2}=13\). 4. Squaring and simplifying gives \((x-3)^2+160=169\), so \((x-3)^2=9\). 5. Therefore, \(x-3=3\) or \(x-3=-3\), giving \(x=6\) or \(x=0\). The points are \((6, 0, 0)\) and \((0, 0, 0)\).

a) \(13\) units b) \((6, 0, 0)\) and \((0, 0, 0)\)

- 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 \(d=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}\).

- 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 \(64\).

- 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 \((0, 0, -10)\)

- 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 \(t\in\left[\frac12,\frac54\right]\).

- 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 \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\-1\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-1\\2\\2\end{pmatrix}\) span the base plane. One parametric equation is \(E:\vec{x}=\begin{pmatrix}4\\1\\1\end{pmatrix}+r\begin{pmatrix}2\\2\\-1\end{pmatrix}+s\begin{pmatrix}-1\\2\\2\end{pmatrix}\). 2. Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both spanning vectors. Then \(2a+2b-c=0\) and \(-a+2b+2c=0\). One solution is \(\vec{n}=\begin{pmatrix}2\\-1\\2\end{pmatrix}\). 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 \(\overline{AC}\): \(M=\left(\frac{9}{2},3,\frac{3}{2}\right)\). A unit normal vector is \(\frac{1}{3}\begin{pmatrix}2\\-1\\2\end{pmatrix}\). Therefore, \(S=M\pm3\begin{pmatrix}2\\-1\\2\end{pmatrix}\), giving \(S_1=\left(\frac{21}{2},0,\frac{15}{2}\right)\) or \(S_2=\left(-\frac{3}{2},6,-\frac{9}{2}\right)\).

a) \(E:\vec{x}=\begin{pmatrix}4\\1\\1\end{pmatrix}+r\begin{pmatrix}2\\2\\-1\end{pmatrix}+s\begin{pmatrix}-1\\2\\2\end{pmatrix}\); \(E:2x-y+2z=9\) b) \(AB=BC=3\) and \(\overrightarrow{AB}\perp\overrightarrow{BC}\); \(D=(3, 3, 3)\) c) \(S_1=\left(\frac{21}{2},0,\frac{15}{2}\right)\) or \(S_2=\left(-\frac{3}{2},6,-\frac{9}{2}\right)\)

- 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 \(\vec{r}=\langle 2,5,-1\rangle+r\langle 2,-1,0\rangle+s\langle 0,1,-2\rangle\). The two direction vectors are not scalar multiples, so they determine a plane. 2. Let \(\vec{n}=\langle a,b,c\rangle\). Orthogonality gives \(2a-b=0\) and \(b-2c=0\). Choose \(a=1\), so \(b=2\) and \(c=1\). Thus \(\vec{n}=\langle 1,2,1\rangle\). 3. Using \((2, 5, -1)\), the plane equation is \(x+2y+z=2+10-1=11\). 4. The perpendicular line through \(Q\) is \(\vec{\ell}(t)=\langle 4,1,8\rangle+t\langle 1,2,1\rangle\). 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 \(\lVert QP_0\rVert=\left\lVert-\frac12\langle 1,2,1\rangle\right\rVert=\frac12\sqrt{6}=\sqrt{\frac32}\approx1.22\).

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

a) \(E:\vec{x}=\begin{pmatrix}2\\2\\4\end{pmatrix}+r\begin{pmatrix}2\\0\\-2\end{pmatrix}+s\begin{pmatrix}0\\2\\-2\end{pmatrix}\); \(E:x+y+z=8\) b) The triangle is equilateral with side length \(2\sqrt{2}\), and \(G=\left(\frac{8}{3},\frac{8}{3},\frac{8}{3}\right)\).

- 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

- Find a normal vector by requiring zero dot products with two connecting vectors. - Compare the distances from \(N\) to \(L\) and \(M\). - Use a triangular base in the \(xy\)-plane and its perpendicular height. - Every point on the \(z\)-axis has the form \((0, 0, z)\).

1. Two vectors in \(H\) are \(\overrightarrow{LM}=\begin{pmatrix}-4\\4\\0\end{pmatrix}\) and \(\overrightarrow{LN}=\begin{pmatrix}-2\\2\\4\end{pmatrix}\). Let \(\vec{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\) be perpendicular to both. Then \(-4a+4b=0\) and \(-2a+2b+4c=0\), so one normal vector is \(\begin{pmatrix}1\\1\\0\end{pmatrix}\). Therefore, \(H:x+y=4\). 2. \(LN=\sqrt{(-2)^2+2^2+4^2}=\sqrt{24}\), and \(MN=\sqrt{2^2+(-2)^2+4^2}=\sqrt{24}\). Therefore, triangle \(LMN\) is isosceles. 3. Using triangle \(OLM\) as the base, its area is \(\frac{1}{2}\cdot4\cdot4=8\), and the perpendicular height from \(N\) to the \(xy\)-plane is \(4\). Thus \(V=\frac{1}{3}\cdot8\cdot4=\frac{32}{3}\) cubic units. 4. For tetrahedron \(OLMP\), the same base has area \(8\). If \(P=(0, 0, z)\), then the height is \(|z|\). Solve \(16=\frac{1}{3}\cdot8|z|\), giving \(|z|=6\). Therefore, \(P=(0, 0, 6)\) or \(P=(0, 0, -6)\).

a) \(H:x+y=4\) b) \(LN=MN=\sqrt{24}\) c) \(V=\frac{32}{3}\) cubic units d) \(P=(0, 0, 6)\) or \(P=(0, 0, -6)\)

- 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.

- 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 \(d>0\). - Equations that differ only by multiplication by \(-1\) describe the same plane.

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 are needed. 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. The remaining global sign choices describe the same four planes after the entire equation is multiplied by \(-1\), so 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

- 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. a) 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 \(\vec{r}=\langle 1,1,5\rangle+\lambda\langle 3,8,-5\rangle+\mu\langle 9,-2,-2\rangle\). 3. Let \(\vec{n}=\langle a,b,c\rangle\). Solving \(\vec{n}\cdot\overrightarrow{AB}=0\) and \(\vec{n}\cdot\overrightarrow{AC}=0\) gives \(\vec{n}=\langle 2,3,6\rangle\). 4. Using point \(A\), the standard equation is \(2x+3y+6z=35\). 5. b) The distance from the origin is \(\frac{|0-35|}{\sqrt{2^2+3^2+6^2}}=\frac{35}{7}=5\). 6. c) 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) \(\vec{r}=\langle 1,1,5\rangle+\lambda\langle 3,8,-5\rangle+\mu\langle 9,-2,-2\rangle\); \(2x+3y+6z=35\) b) \(5\) c) \(S\left(\frac{29}{20}, \frac{11}{10}, \frac{24}{5}\right)\)