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

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

- 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 cube has side length \(4\), 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: \(OP=\sqrt{4^2+4^2+4^2}=\sqrt{48}=4\sqrt{3}\).

\(P=(4,4,4)\) and \(OP=4\sqrt{3}\) 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 square base has side length \(6\), so each coordinate from the center to a side is \(3\). Thus \(C=(3,3,0)\). 2. The 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.

- Which axis is perpendicular to the \(yz\)-plane? - A point in a coordinate plane has one coordinate equal to zero.

1. Points in the \(yz\)-plane have an \(x\)-coordinate of \(0\). 2. Only \(A\) has first coordinate \(0\), so \(A\) lies in the plane.

\(A(0,4,-3)\), because its \(x\)-coordinate is \(0\).

- Compare the corresponding coordinates of the two points. - Which coordinate changes while the other two stay the same?

1. Subtract the starting coordinates from the ending coordinates: \(Q-P=(0,0,4)\). 2. Only the \(z\)-coordinate changes, so the motion is parallel to the \(z\)-axis.

\(\overrightarrow{PQ}=\langle0,0,4\rangle\), parallel to the \(z\)-axis.

- Which coordinate measures how far a point is above or below the \(xy\)-plane? - A distance is never negative, even for a point below the plane.

1. The \(xy\)-plane consists of points with \(z=0\). 2. The perpendicular distance is the absolute value of the \(z\)-coordinate: \(|-5|=5\) units.

\(5\) units

- A displacement describes the change in each coordinate, not the final position itself. - Add the corresponding entries of the starting point and displacement.

1. Add the displacement components to the starting coordinates. 2. \(B=(1+2,2-1,3-4)=(3,1,-1)\).

\(B(3,1,-1)\)

- The midpoint is halfway between the endpoints in each coordinate. - Work with each of the three coordinates separately.

1. Average corresponding coordinates of the endpoints. 2. The midpoint is \(M=(\frac{2+6}{2},\frac{-2+4}{2},\frac{4+2}{2})=(4,1,3)\).

\(M(4,1,3)\)

- Which coordinate is always zero for points on the \(xy\)-plane? - What remains unchanged when moving straight up or down?

1. Moving straight toward the \(xy\)-plane leaves the \(x\)- and \(y\)-coordinates unchanged. 2. Set the \(z\)-coordinate to \(0\), giving \(Q(2,3,0)\).

\(Q(2,3,0)\)

- Which coordinate measures the signed distance from the \(xy\)-plane? - A reflected point is the same distance on the opposite side of the plane.

1. Reflection across the \(xy\)-plane keeps the first two coordinates unchanged. 2. Reverse the sign of the third coordinate to get \(P'(4,-2,-3)\).

\(P'(4,-2,-3)\)

- A displacement gives signed changes, not just three lengths. - Does the final corner lie above or below the starting corner?

1. Traveling to the opposite bottom corner changes all three coordinates. 2. From the shown edge lengths, the horizontal changes are \(+2\,\text{cm}\) and \(+3\,\text{cm}\). Traveling downward makes the vertical change \(-5\,\text{cm}\). 3. The displacement is \(\langle2,3,-5\rangle\,\text{cm}\).

\(\langle2,3,-5\rangle\,\text{cm}\)

- 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

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

The side lengths are \(AB=5\), \(AC=6\), and \(BC=\sqrt{61}\). 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 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 \(\overrightarrow{PQ}=\langle3,0,4\rangle\), \(\overrightarrow{QR}=\langle-3,4,-1\rangle\), and \(\overrightarrow{PR}=\langle0,4,3\rangle\). 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) Isosceles, not equilateral: \(PQ=PR=5\), but \(QR=\sqrt{26}\ne5\).

- 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.

- 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.

- 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.

- 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 halfway between each vertex and its image. - Express the midpoint relationship for one original vertex and its unknown image. - Check that the midpoint of each original-image pair is \(Z\).

1. For a point reflection through \(Z\), 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) \(\mathbf{r}_2=\langle -4,-1,3\rangle\)

- 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 \(P_2\): \(\overrightarrow{P_1P_2}=\langle14-10,17-20,5-5\rangle=\langle4,-3,0\rangle\). 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 \(\sqrt{10^2+20^2+5^2}=\sqrt{525}=5\sqrt{21}\approx22.91\,\text{m}\). 4. The distance from \(P_2\) to the origin is \(\sqrt{14^2+17^2+5^2}=\sqrt{510}\approx22.58\,\text{m}\). 5. Since \(525>510\), drone \(P_1\) is farther from the origin.

a) \(5\,\text{m}\) b) \(P_1\) is \(5\sqrt{21}\approx22.91\,\text{m}\) from the origin, and \(P_2\) is \(\sqrt{510}\approx22.58\,\text{m}\) 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 remove the square root from the equation? - A squared expression can produce two possible values for its base.

1. The coordinate differences from \(A\) to \(B_t\) are \(\langle2-2,t-(-1),1-5\rangle=\langle0,t+1,-4\rangle\). 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: \(13+\sqrt{194}+5=18+\sqrt{194}\).

\(18+\sqrt{194}\) 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 \(AC\): \(AC=\sqrt{(1-5)^2+(2-2)^2+(5-1)^2}=\sqrt{32}=4\sqrt2\). 4. Since \(AB=BC\), the triangle is isosceles. It is not equilateral because \(AC\ne5\).

a) \(AB=5\), \(BC=5\), and \(AC=4\sqrt2\) 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.

- A space diagonal travels across the inside of the prism, not along one face. - Use a right triangle in the base, then a second right triangle involving the height. - Think about how all three edge lengths contribute to the diagonal.

1. The face diagonal across the \(2\,\text{cm}\) and \(3\,\text{cm}\) edges has squared length \(2^2+3^2=13\). 2. Add the squared height: \(d^2=13+6^2=49\). 3. Hence the space diagonal has length \(d=7\,\text{cm}\).

\(7\,\text{cm}\)

- The midpoint is the average of the two endpoints in each coordinate. - What does the midpoint equation become when one endpoint is unknown? - Check your result by averaging the two endpoints.

1. The midpoint equation is \(M=\frac{A+B}{2}\), so \(B=2M-A\). 2. Calculate each coordinate: \(B=(-2-3,4-(-2),10-1)=(-5,6,9)\). 3. Check that the average of \(A\) and \(B\) is \(M\).

\(B(-5,6,9)\)

- Which coordinate is zero for every point in the \(xz\)-plane? - Reflection changes the sign of the coordinate perpendicular to the mirror plane. - Both points have two coordinates in common. How does that simplify their distance?

1. Reflection across the \(xz\)-plane changes the sign of the \(y\)-coordinate only: \(P'(2,4,3)\). 2. The points differ by \(8\) units along the \(y\)-axis, so \(PP'=8\) units.

\(P'(2,4,3)\); \(PP'=8\) units.

- Find the squared distances between each pair of vertices; you do not need square roots. - Which relationship must hold among the three squared side lengths for a right triangle? - The largest squared side length belongs to the side opposite the right angle.

1. \(PQ^2=2^2+1^2+2^2=9\). 2. \(PR^2=(-2)^2+4^2+0^2=20\), and \(QR^2=(-4)^2+3^2+(-2)^2=29\). 3. Since \(9+20=29\), the Pythagorean theorem holds, so the right angle is at \(P\).

Yes. \(PQ^2=9\), \(PR^2=20\), and \(QR^2=29\). Since \(9+20=29\), the right angle is at \(P\).

- 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 displacement vector from \(A\) to \(B\). - Use the given length to write an equation involving the unknown component. - Remember that a squared expression can equal a positive number in two ways.

1. \(\overrightarrow{AB}=\langle2,1-k,-4\rangle\). 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. \(\overrightarrow{AB}=\langle0,4,3\rangle\) and \(\overrightarrow{AD}=\langle-5,0,0\rangle\). 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 \(DCGH\), because their edges in the \(AD\) direction are perpendicular to the faces. Since \(\overrightarrow{AD}=\langle-5,0,0\rangle\) is parallel to the \(x\)-axis, both faces are parallel to the \(yz\)-plane.

- 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: \(AB=AC\) and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), so the equal sides meet at a right angle.

- 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 \(KL=4\), \(KM=\sqrt{2^2+1^2+1^2}=\sqrt6\), and \(LM=\sqrt{(-2)^2+1^2+1^2}=\sqrt6\). 2. Since \(KM=LM\), the triangle is isosceles. Its longest side is \(KL\), so the largest angle is at \(M\). 3. Use \(\overrightarrow{MK}=\langle-2,-1,-1\rangle\) and \(\overrightarrow{ML}=\langle2,-1,-1\rangle\). 4. Then \(\cos\angle M=\frac{-4+1+1}{\sqrt6\sqrt6}=-\frac{1}{3}\). 5. Therefore, \(\angle M=\cos^{-1}(-\frac{1}{3})\approx109.47^\circ\). 6. The triangle is isosceles and obtuse.

The side lengths are \(KL=4\) and \(KM=LM=\sqrt6\). 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. \(\overrightarrow{PQ}=\langle2,1,2\rangle\) and \(\overrightarrow{PS}=\langle1,2,2\rangle\). Both vectors have length \(3\). 2. \(\overrightarrow{SR}=\langle2,1,2\rangle=\overrightarrow{PQ}\), 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) \(\overrightarrow{PQ}=\langle2,1,2\rangle\), \(\overrightarrow{PS}=\langle1,2,2\rangle\); both lengths are \(3\). b) \(\overrightarrow{SR}=\overrightarrow{PQ}\) makes the quadrilateral a parallelogram, and \(PQ=PS=3\) makes it 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, \(\overrightarrow{AB}=\langle2-5,3-(-1)\rangle=\langle-3,4\rangle\). In three dimensions, \(\overrightarrow{AB}=\langle2-5,3-(-1),0-0\rangle=\langle-3,4,0\rangle\). The three-dimensional vector has the same first two components and an added zero component. 3. \(M=(\frac{5+2}{2},\frac{-1+3}{2},\frac{0+0}{2})=(3.5,1,0)\). 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. \(\overrightarrow{AB}_{2D}=\langle-3,4\rangle\), \(\overrightarrow{AB}_{3D}=\langle-3,4,0\rangle\). The 3D vector preserves the 2D components and adds \(0\) for the unchanged \(z\)-coordinate. 3. \(M=(3.5,1,0)\) 4. \(P=(3.5,1,4)\)

- 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 \(\overline{PP'}\): \(M=(\frac{3+1}{2},\frac{-2+6}{2},\frac{4+0}{2})=(2,2,2)\). 2. Use \(Q'=2M-Q\): \(Q'=2(2,2,2)-(5,1,2)=(-1,3,2)\). 3. \(\overrightarrow{PQ}=Q-P=\langle2,3,-2\rangle\). 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. - Check the images of the three vertices one at a time using the midpoint relationship.

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 through 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 through 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 reflected directed vector points the opposite way, while the new line stays parallel to the old one. - 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 reverses a directed vector, but the reflected line remains parallel to the original. Reversing the line parameter gives the equivalent direction \(\langle0,0,1\rangle\), so \(h_r: \mathbf{x}=\langle7,9,1\rangle+t\langle0,0,1\rangle\). 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) \(h_r: \mathbf{x}=\langle7,9,1\rangle+t\langle0,0,1\rangle\) 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. \(RP=\sqrt{(1.2-1.2)^2+(0-4)^2+(3.5-7.5)^2}=\sqrt{32}=4\sqrt2\). 4. The perimeter is \(5+5+4\sqrt2=10+4\sqrt2\). 5. Because \(PQ=QR=5\), the triangle has two congruent sides and is isosceles.

a) \(10+4\sqrt2\) 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: \(AC=\sqrt{(5-2)^2+(7-1)^2+(0-(-3))^2}=\sqrt{54}=3\sqrt6\), and \(BD=\sqrt{(1-6)^2+(5-3)^2+(-1-(-2))^2}=\sqrt{30}\). 3. Add the four side lengths: \(P=4\sqrt{21}\). 4. Opposite side vectors are equal: \(\overrightarrow{AB}=\overrightarrow{DC}=\langle4,2,1\rangle\) and \(\overrightarrow{BC}=\overrightarrow{AD}=\langle-1,4,2\rangle\). The quadrilateral is a parallelogram, and its four equal sides make it a rhombus.

Sides: \(AB=BC=CD=DA=\sqrt{21}\) Diagonals: \(AC=3\sqrt6\), \(BD=\sqrt{30}\) Perimeter: \(P=4\sqrt{21}\) Classification: \(\overrightarrow{AB}=\overrightarrow{DC}=\langle4,2,1\rangle\) and \(\overrightarrow{BC}=\overrightarrow{AD}=\langle-1,4,2\rangle\). All sides have length \(\sqrt{21}\), so it is a rhombus.

- 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: \(P=2\sqrt{29}+2\sqrt{61}\). 3. Find the diagonals: \(PR=\sqrt{(4-3)^2+(7-0)^2+(10-2)^2}=\sqrt{114}\), and \(QS=\sqrt{(0-7)^2+(4-3)^2+(8-4)^2}=\sqrt{66}\). 4. The opposite side vectors agree: \(\overrightarrow{PQ}=\overrightarrow{SR}=\langle4,3,2\rangle\) and \(\overrightarrow{QR}=\overrightarrow{PS}=\langle-3,4,6\rangle\). Thus \(PQRS\) is a parallelogram, but \(PQ\ne QR\), so it is not a rhombus.

Sides: \(PQ=RS=\sqrt{29}\), \(QR=SP=\sqrt{61}\) Diagonals: \(PR=\sqrt{114}\), \(QS=\sqrt{66}\) Perimeter: \(P=2\sqrt{29}+2\sqrt{61}\) Classification: \(\overrightarrow{PQ}=\overrightarrow{SR}=\langle4,3,2\rangle\) and \(\overrightarrow{QR}=\overrightarrow{PS}=\langle-3,4,6\rangle\). It is a parallelogram, but not a rhombus because \(PQ=\sqrt{29}\ne\sqrt{61}=QR\).

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

- 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 \(\overrightarrow{AB}=\langle3,4,0\rangle\), \(\overrightarrow{BC}=\langle-3,0,4\rangle\), and \(\overrightarrow{AC}=\langle0,4,4\rangle\). 2. Thus, \(AB=5\), \(BC=5\), and \(AC=\sqrt{32}=4\sqrt2\). 3. For the angle at \(B\), use \(\overrightarrow{BA}=\langle-3,-4,0\rangle\) and \(\overrightarrow{BC}=\langle-3,0,4\rangle\). 4. Their dot product is \(9\), so \(\cos\beta=\frac9{5\cdot5}=\frac9{25}\). 5. Therefore, \(\beta=\cos^{-1}(\frac9{25})\approx68.90^\circ\). 6. Using the two sides adjacent to \(\beta\), \(K=\frac12\cdot5\cdot5\sin\beta=2\sqrt{34}\).

a) \(AB=5\), \(BC=5\), and \(AC=4\sqrt2\) b) \(\beta\approx68.90^\circ\) c) The area is \(2\sqrt{34}\) 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 \(64\).

- 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. \(\overrightarrow{BF}=F-B=\langle-8,6,10\rangle\). Therefore, \(\|\overrightarrow{BF}\|=\sqrt{(-8)^2+6^2+10^2}=\sqrt{200}=10\sqrt{2}\) units. 3. The base is a right triangle with area \(\frac{1}{2}\cdot8\cdot6=24\) square units. The volume is \(24\cdot10=240\) cubic units.

a) \(D=(0,0,10)\), \(E=(8,0,10)\), \(F=(0,6,10)\) b) \(10\sqrt{2}\) units c) \(240\) cubic units

- 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) \(k=\sqrt{3}\)

- 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.