Free math worksheets

- Compare the given equality with the dot-product angle formula. - Use the fact that both magnitudes are nonzero. - Interpret the resulting cosine value geometrically.

1. By the dot-product angle formula, \(\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha\). 2. Compare this with the given equation: \(\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha=-\|\mathbf{u}\|\|\mathbf{v}\|\). 3. Since both vectors are nonzero, divide by their positive magnitudes to get \(\cos\alpha=-1\). 4. Therefore, \(\alpha=180^\circ\). 5. The vectors are parallel but point in opposite directions; equivalently, one is a negative scalar multiple of the other.

\(\alpha=180^\circ\). The vectors are parallel and point in opposite directions.

- Read the horizontal and vertical components from point \(P\). - Treat those components as the legs of a right triangle. - Which trigonometric ratio connects the vertical component, horizontal component, and direction angle?

1. The endpoint shows an eastward component of \(4\) and a northward component of \(3\). 2. Therefore, \(\tan\theta=\frac{3}{4}\). 3. Thus, \(\theta=\tan^{-1}\left(\frac{3}{4}\right)\approx36.9^\circ\).

\(36.9^\circ\) north of east

- Resolve the force into horizontal and vertical components. - Decide which trigonometric ratio connects the vertical component to the given magnitude and angle. - Keep the exact trigonometric value rather than converting to a decimal.

The vertical component is opposite the \(60^\circ\) angle, so \(F_y=12\sin 60^\circ=12\cdot\frac{\sqrt{3}}{2}=6\sqrt{3}\,\text{N}\).

\(6\sqrt{3}\,\text{N}\)

- Identify which component is opposite the given angle. - Choose the trigonometric ratio that relates the opposite side to the hypotenuse. - The force magnitude is the hypotenuse of the component triangle.

1. The vertical component is opposite the \(30^\circ\) angle, so use \(F_y=F\sin\theta\). 2. \(F_y=20\sin 30^\circ=20\cdot\frac12=10\,\text{N}\).

\(10\,\text{N}\)

- Read each force's horizontal and vertical components from its terminal point. - Add corresponding components. - Use the magnitude formula on the resultant.

1. From the diagram, \(\mathbf{F}_1=\langle3,0\rangle\) and \(\mathbf{F}_2=\langle0,4\rangle\). 2. Add components: \(\mathbf{R}=\langle3,4\rangle\). 3. Its magnitude is \(\|\mathbf{R}\|=\sqrt{3^2+4^2}=5\,\text{N}\).

\(\mathbf{R}=\langle3,4\rangle\,\text{N}\) and \(\|\mathbf{R}\|=5\,\text{N}\).

- Form two vectors that both begin at \(B\). - Find their dot product and magnitudes. - Apply the dot-product angle formula.

1. Use the vectors that begin at \(B\): \(\overrightarrow{BA}=\begin{pmatrix}-3\\-2\\0\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}0\\4\\-3\end{pmatrix}\). 2. Their dot product is \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-8\). 3. Their magnitudes are \(\|\overrightarrow{BA}\|=\sqrt{13}\) and \(\|\overrightarrow{BC}\|=5\). 4. Therefore, \(\cos\beta=\frac{-8}{5\sqrt{13}}\). 5. Thus, \(\beta=\cos^{-1}\left(\frac{-8}{5\sqrt{13}}\right)\approx116.34^\circ\). The required intermediate dot product is \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-8\).

\(\beta\approx116.34^\circ\) \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-8\).

- Find the squared magnitude of each vector. - Find the dot product of the vectors. - Use the dot-product identity for the area of a parallelogram.

1. Compute the squared magnitudes: \(\|\mathbf{u}\|^2=6^2+(-2)^2=40\) and \(\|\mathbf{v}\|^2=3^2+4^2=25\). 2. Compute the dot product: \(\mathbf{u}\cdot\mathbf{v}=6\cdot3+(-2)\cdot4=10\). 3. The squared area of the parallelogram is \(\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2=40\cdot25-10^2=900\). 4. Therefore, the area is \(\sqrt{900}=30\).

\(30\) square units

- Write two side vectors with a common initial point. - Use their magnitudes and dot product to find the area of the parallelogram they form. - The triangle has half the area of that parallelogram.

1. Two side vectors are \(\overrightarrow{AB}=\begin{pmatrix}-2\\2\\0\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}-2\\0\\2\end{pmatrix}\). 2. Their squared magnitudes are both \(8\), and their dot product is \(4\). 3. The squared area of the parallelogram formed by the vectors is \(8\cdot8-4^2=48\). 4. The triangle has half the parallelogram's area, so its area is \(\frac{1}{2}\sqrt{48}=2\sqrt{3}\) square units.

\(2\sqrt{3}\) square units \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4\); area \(=2\sqrt3\).

- Read each directed segment by subtracting its initial point from its endpoint. - Head-to-tail vectors add to the displacement from the first initial point to the last endpoint. - Use the resultant components to find both magnitude and direction.

1. Reading the endpoints from the coordinate grid gives \(\overrightarrow{OA}=\langle 4,0\rangle\) and \(\overrightarrow{AB}=\langle 0,3\rangle\), in meters per second. 2. The resultant is \(\overrightarrow{OB}=\langle 4,3\rangle\,\text{m/s}\). 3. Its speed is \(\sqrt{4^2+3^2}=5\,\text{m/s}\). 4. Its direction angle satisfies \(\tan\theta=\frac{3}{4}\), so \(\theta\approx36.9^\circ\).

a) \(\langle 4,3\rangle\,\text{m/s}\) b) \(5\,\text{m/s}\) c) \(36.9^\circ\)

- Represent the boat velocity and current with perpendicular component vectors. - Add the component vectors before finding the speed. - Pay attention to whether the requested angle is measured from east or from north.

1. The boat's water-relative velocity is \(\langle 0,12\rangle\) and the current is \(\langle 5,0\rangle\), so the shore-relative velocity is \(\langle 5,12\rangle\,\text{mph}\). 2. The speed is \(\sqrt{5^2+12^2}=13\,\text{mph}\). 3. If \(\alpha\) is measured east of north, then \(\tan\alpha=\frac{5}{12}\). 4. Therefore, \(\alpha\approx22.6^\circ\).

a) \(\langle 5,12\rangle\,\text{mph}\) b) \(13\,\text{mph}\) c) \(22.6^\circ\) east of north

- Use the two side vectors that begin at \(A\). - A zero dot product proves a right angle. - For part b), use two vectors that both begin at \(B\). - Apply the dot-product angle formula.

1. The vectors meeting at \(A\) are \(\overrightarrow{AB}=\begin{pmatrix}3\\0\\4\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\4\\0\end{pmatrix}\). 2. Their dot product is \(0\), so the angle at \(A\) is \(90^\circ\). 3. For the angle at \(B\), use \(\overrightarrow{BA}=\begin{pmatrix}-3\\0\\-4\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-3\\4\\-4\end{pmatrix}\). 4. Their magnitudes are \(\|\overrightarrow{BA}\|=5\) and \(\|\overrightarrow{BC}\|=\sqrt{41}\), while their dot product is \(25\). 5. Therefore, \(\cos\beta=\frac{25}{5\sqrt{41}}=\frac{5}{\sqrt{41}}\). 6. Thus, \(\beta=\cos^{-1}\left(\frac{5}{\sqrt{41}}\right)\approx38.66^\circ\). The two required reported dot products are \(0\) at \(A\) and \(25\) at \(B\).

a) Since \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), the triangle has a right angle at \(A\). b) \(\beta\approx38.66^\circ\) Required dot products: \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\) and \(\overrightarrow{BA}\cdot\overrightarrow{BC}=25\).

- Add the three side lengths for the perimeter. - For the angle at \(Q\), use two vectors that begin at \(Q\). - Apply the given translation to \(P\) in part c. - To prove a line is perpendicular to a plane, relate its direction to two nonparallel directions in that plane.

1. The side lengths are \(PQ=4\), \(PR=3\), and \(QR=\sqrt{(-4)^2+3^2}=5\). 2. Therefore, the base perimeter is \(4+3+5=12\). 3. At \(Q\), \(\overrightarrow{QP}=\langle -4,0,0\rangle\) and \(\overrightarrow{QR}=\langle -4,0,3\rangle\). 4. Thus, \(\cos\beta=\frac{\overrightarrow{QP}\cdot\overrightarrow{QR}}{\|\overrightarrow{QP}\|\|\overrightarrow{QR}\|}=\frac{16}{20}=\frac45\), so \(\beta\approx36.87^\circ\). 5. Translating \(P\) gives \(P'=(2, 7, 0)\), and \(\overrightarrow{PP'}=\langle 0,4,0\rangle\). 6. Since \(\overrightarrow{PP'}\cdot\overrightarrow{PQ}=0\) and \(\overrightarrow{PP'}\cdot\overrightarrow{PR}=0\), the edge is perpendicular to two nonparallel vectors in the base plane. Therefore, \(PP'\) is perpendicular to the base plane.

a) \(12\) units b) \(\beta\approx36.87^\circ\) c) \(P'=(2, 7, 0)\), and \(PP'\) is perpendicular to the base plane because its direction vector has dot product \(0\) with both \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\). Required dot products: \(\overrightarrow{QP}\cdot\overrightarrow{QR}=16\), \(\overrightarrow{PP'}\cdot\overrightarrow{PQ}=0\), and \(\overrightarrow{PP'}\cdot\overrightarrow{PR}=0\).

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

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

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

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

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

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

- Expand the squared magnitude of a difference vector. - Replace the dot product with the angle formula. - Use the equal magnitudes to eliminate the unknown length.

1. Let \(L=\|\mathbf{a}\|=\|\mathbf{b}\|>0\). Then \(\|\mathbf{a}-\mathbf{b}\|=\sqrt3L\). 2. Square the magnitude: \(\|\mathbf{a}-\mathbf{b}\|^2=\|\mathbf{a}\|^2+\|\mathbf{b}\|^2-2\mathbf{a}\cdot\mathbf{b}\). 3. Using \(\mathbf{a}\cdot\mathbf{b}=L^2\cos\alpha\), obtain \(3L^2=2L^2-2L^2\cos\alpha\). 4. Since \(L>0\), divide by \(L^2\): \(3=2-2\cos\alpha\). 5. Therefore, \(\cos\alpha=-\frac12\), so \(\alpha=120^\circ\).

\(\alpha=120^\circ\)

- Represent the shorter magnitude by one variable. - Expand the squared magnitude of the sum. - Replace the dot product with the angle formula.

1. Let \(L=\|\mathbf{u}\|>0\). Then \(\|\mathbf{v}\|=2L\) and \(\|\mathbf{u}+\mathbf{v}\|=\sqrt7L\). 2. Use \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+\|\mathbf{v}\|^2+2\mathbf{u}\cdot\mathbf{v}\). 3. Since \(\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\alpha\), substitution gives \(7L^2=L^2+4L^2+4L^2\cos\alpha\). 4. Divide by \(L^2\): \(7=5+4\cos\alpha\). 5. Thus, \(\cos\alpha=\frac12\), so \(\alpha=60^\circ\).

\(\alpha=60^\circ\)

- Write the face diagonal with a zero \(z\)-component. - Use the dot-product angle formula. - Make sure both vectors begin at the origin.

1. The face-diagonal vector is \(\mathbf{f}=\langle 4,3,0\rangle\). 2. The dot product is \(\mathbf{d}\cdot\mathbf{f}=4\cdot4+3\cdot3+12\cdot0=25\). 3. The magnitudes are \(\|\mathbf{d}\|=13\) and \(\|\mathbf{f}\|=5\). 4. Therefore, \(\cos\alpha=\frac{25}{13\cdot5}=\frac{5}{13}\). 5. Thus, \(\alpha=\cos^{-1}\left(\frac{5}{13}\right)\approx67.38^\circ\).

\(\alpha\approx67.38^\circ\) \(\mathbf{d}\cdot\mathbf{f}=25\).

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

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

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

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

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

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

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

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

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

- Combine the swimming velocity and current as vectors. - Crossing time depends on the component perpendicular to the banks. - For no drift, the two components parallel to the bank must cancel. - Use the fixed swimming speed to determine the remaining component.

1. Add the swimmer and current velocities: \(\mathbf{v}_r=\mathbf{v}_s+\mathbf{v}_w=\langle 0.6,1.0,0\rangle\,\text{m/s}\). 2. Its magnitude is \(\|\mathbf{v}_r\|=\sqrt{0.6^2+1.0^2}=\sqrt{1.36}\approx1.17\,\text{m/s}\). 3. The across-river component is \(1.0\,\text{m/s}\), so the crossing time is \(\frac{100}{1.0}=100\,\text{s}\). The drift is \(0.6\cdot100=60\,\text{m}\). 4. To eliminate drift, the swimmer's \(x\)-component must be \(-0.6\,\text{m/s}\). With total speed \(1.0\,\text{m/s}\), the across-river component is \(\sqrt{1-0.6^2}=0.8\,\text{m/s}\). 5. Thus, \(\mathbf{v}_h=\langle -0.6,0.8,0\rangle\,\text{m/s}\). 6. Let \(\mathbf{i}=\langle 1,0,0\rangle\). Then \(\cos\alpha=\frac{\mathbf{v}_h\cdot\mathbf{i}}{\|\mathbf{v}_h\|\|\mathbf{i}\|}=-0.6\), so \(\alpha\approx126.87^\circ\).

a) \(\mathbf{v}_r=\langle 0.6,1.0,0\rangle\,\text{m/s}\), with magnitude approximately \(1.17\,\text{m/s}\) b) Crossing time: \(100\,\text{s}\); downstream drift: \(60\,\text{m}\) c) \(\mathbf{v}_h=\langle -0.6,0.8,0\rangle\,\text{m/s}\) d) \(\alpha\approx126.87^\circ\) measured counterclockwise from the positive \(x\)-axis

- Use the dot-product angle relationship to connect the dot product, magnitudes, and angle. - Relate the sine and cosine of the same angle. - A triangle generated by two vectors has half the corresponding parallelogram area.

1. The dot product is \(\mathbf{a}\cdot\mathbf{b}=1\cdot3+2\cdot0+(-1)\cdot4=-1\). 2. The magnitudes are \(\|\mathbf{a}\|=\sqrt6\) and \(\|\mathbf{b}\|=5\). 3. Therefore, \(\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}=-\frac{1}{5\sqrt6}\). 4. Use \(\sin^2\theta=1-\cos^2\theta\): \(\sin\theta=\sqrt{1-\frac{1}{150}}=\sqrt{\frac{149}{150}}\). 5. The triangle's area is \(A=\frac12\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta=\frac{\sqrt{149}}2\approx6.10\).

a) \(\mathbf{a}\cdot\mathbf{b}=-1\), \(\cos\theta=-\frac{1}{5\sqrt6}\) b) \(A=\frac{\sqrt{149}}2\approx6.10\) square units

- 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) \(k=\pm\frac{3\sqrt2}{2}\)

- Apply the translation to \(C\) first. - Use two side vectors that begin at \(A\). - Relate the side-vector magnitudes and dot product to the parallelogram area. - The triangle has half the parallelogram area.

1. \(F=C+\mathbf{v}=(4, 5, 6)\). 2. \(\overrightarrow{AB}=\langle 3,1,0\rangle\) and \(\overrightarrow{AF}=\langle 2,5,6\rangle\). 3. Their squared magnitudes are \(\|\overrightarrow{AB}\|^2=10\) and \(\|\overrightarrow{AF}\|^2=65\). 4. Their dot product is \(\overrightarrow{AB}\cdot\overrightarrow{AF}=3\cdot2+1\cdot5+0\cdot6=11\). 5. The squared parallelogram area is \(10\cdot65-11^2=529\), so the parallelogram area is \(23\). 6. Therefore, the triangle's area is \(\frac{23}{2}=11.5\).

\(11.5\) square units \(\mathbf{u}\cdot\mathbf{w}=11\); area \(=\frac{23}{2}\) square units.

- Find the coordinates of the three midpoints first. - Write two vectors that form sides of the triangle. - Use the dot-product area identity for the parallelogram formed by those vectors. - Remember that the triangle has half the area of the parallelogram.

1. The midpoints are \(M_1=(2, 0, 0)\), \(M_2=(4, 4, 2)\), and \(M_3=(0, 2, 4)\). 2. Two side vectors are \(\overrightarrow{M_1M_2}=\begin{pmatrix}2\\4\\2\end{pmatrix}\) and \(\overrightarrow{M_1M_3}=\begin{pmatrix}-2\\2\\4\end{pmatrix}\). 3. Their squared magnitudes and dot product are \(\|\overrightarrow{M_1M_2}\|^2=24\), \(\|\overrightarrow{M_1M_3}\|^2=24\), and \(\overrightarrow{M_1M_2}\cdot\overrightarrow{M_1M_3}=12\). 4. The squared area of the parallelogram formed by the vectors is \(24\cdot24-12^2=432\). 5. Therefore, the area of the triangle is \(\frac{1}{2}\sqrt{432}=6\sqrt{3}\).

\(6\sqrt{3}\) square units \(\mathbf{u}\cdot\mathbf{v}=12\); area \(=6\sqrt3\) square units.

- Use the relationship between opposite sides of a parallelogram to find \(D\). - Use the two adjacent side vectors at \(A\). - The dot product can be used both in a vector area identity and in the angle formula. - Be sure to distinguish the parallelogram area from a triangle area.

1. In a parallelogram, \(\overrightarrow{AD}=\overrightarrow{BC}\). Since \(\overrightarrow{BC}=\begin{pmatrix}1\\2\\1\end{pmatrix}\), \(D=A+\overrightarrow{BC}=(2, 3, 3)\). 2. The adjacent side vectors at \(A\) are \(\overrightarrow{AB}=\begin{pmatrix}3\\1\\0\end{pmatrix}\) and \(\overrightarrow{AD}=\begin{pmatrix}1\\2\\1\end{pmatrix}\). Their squared magnitudes are \(10\) and \(6\), and their dot product is \(5\). Therefore, the squared area is \(10\cdot6-5^2=35\), so the exact area is \(\sqrt{35}\) square units. 3. The angle formula gives \(\cos\alpha=\frac{\overrightarrow{AB}\cdot\overrightarrow{AD}}{\|\overrightarrow{AB}\|\|\overrightarrow{AD}\|}=\frac{5}{\sqrt{60}}\). Thus, \(\alpha=\cos^{-1}\left(\frac{5}{\sqrt{60}}\right)\approx49.80^\circ\).

1. \(D(2, 3, 3)\) 2. \(\sqrt{35}\) square units 3. \(\alpha\approx49.80^\circ\) \(\overrightarrow{AB}\cdot\overrightarrow{AD}=5\).

- Compare the equality with the absolute-value form of the dot-product bound. - Interpret the equality case geometrically before introducing a scale factor. - Use the required magnitude to determine the possible scale factors.

1. By the dot-product angle formula, \(\left|\mathbf{a}\cdot\mathbf{x}\right|=\|\mathbf{a}\|\|\mathbf{x}\||\cos\theta|\). 2. The given equality requires \(|\cos\theta|=1\), so \(\mathbf{x}\) must be parallel to \(\mathbf{a}\). Write \(\mathbf{x}=k\mathbf{a}\). 3. Since \(\|\mathbf{a}\|=\sqrt{2^2+(-1)^2+2^2}=3\), the magnitude condition gives \(9=\|\mathbf{x}\|=3|k|\). 4. Thus, \(|k|=3\), so \(k=3\) or \(k=-3\). 5. Therefore, \(\mathbf{x}=\langle6,-3,6\rangle\) or \(\mathbf{x}=\langle-6,3,-6\rangle\).

\(\mathbf{x}=\langle6,-3,6\rangle\) or \(\mathbf{x}=\langle-6,3,-6\rangle\)

- Place one vertex at the origin and align the cube's edges with the axes. - Represent two space diagonals using sign variations of \(a\). - Use the absolute dot product when finding the acute angle between lines.

1. Place the cube so its edges are parallel to the coordinate axes. 2. Two distinct space diagonals can be represented by direction vectors \(\mathbf{d}_1=\langle a,a,a\rangle\) and \(\mathbf{d}_2=\langle -a,a,a\rangle\). 3. Their dot product is \(\mathbf{d}_1\cdot\mathbf{d}_2=-a^2+a^2+a^2=a^2\). 4. Each vector has magnitude \(a\sqrt{3}\). 5. Therefore, \(\cos\alpha=\frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{\|\mathbf{d}_1\|\|\mathbf{d}_2\|}=\frac{a^2}{3a^2}=\frac{1}{3}\). 6. Any pair of distinct space-diagonal lines has direction vectors differing in the sign of one or two components, so the absolute dot-product calculation gives the same acute-angle cosine.

\(\cos\alpha=\frac{1}{3}\)

- Write two side vectors from \(A\). - Use the dot-product area identity, then divide the parallelogram area by \(2\). - Use equal opposite-side vectors to locate \(D\). - Relate the triangle's area to the base \(AC\) and its altitude.

1. The side vectors from \(A\) are \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\1\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}0\\4\\4\end{pmatrix}\). 2. Their squared magnitudes and dot product are \(\|\overrightarrow{AB}\|^2=9\), \(\|\overrightarrow{AC}\|^2=32\), and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=12\). Therefore, the triangle's area is \(\frac{1}{2}\sqrt{9\cdot32-12^2}=\frac{1}{2}\sqrt{144}=6\) square units. 3. Since \(\overrightarrow{AD}=\overrightarrow{BC}\), \(\overrightarrow{BC}=\begin{pmatrix}-2\\2\\3\end{pmatrix}\), so \(D=A+\overrightarrow{BC}=(-1, 2, 5)\). 4. Side \(AC\) has length \(\sqrt{32}=4\sqrt{2}\). Using \(6=\frac{1}{2}\cdot4\sqrt{2}\cdot h\), the altitude is \(h=\frac{12}{4\sqrt{2}}=\frac{3\sqrt{2}}{2}\).

a) \(6\) square units b) \(D(-1, 2, 5)\) c) \(\frac{3\sqrt{2}}{2}\) units \(\overrightarrow{AB}\cdot\overrightarrow{AC}=12\).

- Audit the dot-product arithmetic before checking the angle formula. - Compare the denominator with the exact dot-product angle formula. - Check the inverse-cosine value independently of the earlier steps.

1. The first error is arithmetic: \(0\cdot5=0\), not \(5\). Thus, \(\mathbf{u}\cdot\mathbf{v}=0\). 2. The second error is in the angle formula. The denominator is the product of the magnitudes, not their sum: \(\|\mathbf{u}\|\|\mathbf{v}\|=5\cdot5\). 3. The third error is the inverse-cosine value: \(\cos\phi=0.5\) would give \(\phi=60^\circ\), not \(30^\circ\). 4. Using the correct dot product, \(\cos\phi=\frac{0}{5\cdot5}=0\). 5. Therefore, \(\phi=90^\circ\).

Error 1: \(0\cdot5=0\), so the dot product is \(0\), not \(5\). Error 2: The vector magnitudes must be multiplied in the denominator, not added. Error 3: \(\cos^{-1}(0.5)=60^\circ\), not \(30^\circ\). The correct angle is \(\phi=90^\circ\).

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

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

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

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

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

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

- First verify that opposite sides are parallel. - Use a dot product on adjacent side vectors. - Use the absolute dot product to find the acute angle between diagonal lines. - A rectangle's circumcenter is the common midpoint of its diagonals.

1. The side vectors are \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\0\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-2\\2\\2\end{pmatrix}\), \(\overrightarrow{CD}=\begin{pmatrix}-2\\-2\\0\end{pmatrix}\), and \(\overrightarrow{DA}=\begin{pmatrix}2\\-2\\-2\end{pmatrix}\). 2. Since \(\overrightarrow{AB}=-\overrightarrow{CD}\) and \(\overrightarrow{BC}=-\overrightarrow{DA}\), the quadrilateral is a parallelogram. 3. Also, \(\overrightarrow{AB}\cdot\overrightarrow{BC} =2\cdot(-2)+2\cdot2+0\cdot2=0\). Thus, adjacent sides are perpendicular, so the parallelogram is a rectangle. 4. The diagonal vectors are \(\overrightarrow{AC}=\begin{pmatrix}0\\4\\2\end{pmatrix}\) and \(\overrightarrow{BD}=\begin{pmatrix}-4\\0\\2\end{pmatrix}\). 5. Their dot product is \(4\), and each has magnitude \(\sqrt{20}\). Therefore, \(\cos\phi=\frac{|4|}{\sqrt{20}\sqrt{20}}=\frac{1}{5}\). 6. Thus, \(\phi=\cos^{-1}\left(\frac{1}{5}\right)\approx78.46^\circ\). 7. The circle's center is the common midpoint of the diagonals: \(M=\frac{A+C}{2}=(1, 3, 3)\). 8. Its radius is half a diagonal: \(r=\frac{\sqrt{20}}{2}=\sqrt{5}\).

a) \(ABCD\) is a rectangle because its opposite-side vectors are negatives of each other and \(\overrightarrow{AB}\cdot\overrightarrow{BC}=0\). b) \(\phi\approx78.46^\circ\) c) \(M=(1, 3, 3)\) and \(r=\sqrt{5}\approx2.24\) units

- Use equal opposite-side vectors to find \(C\). - The angle at \(B\) is formed by \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\). - Use the dot-product area identity with two adjacent side vectors. - The diagonals of a parallelogram bisect each other.

1. Since \(\overrightarrow{DC}=\overrightarrow{AB}\), \(\overrightarrow{AB}=\begin{pmatrix}3\\0\\4\end{pmatrix}\), so \(C=D+\overrightarrow{AB}=(6, 4, 5)\). 2. At \(B\), \(\overrightarrow{BA}=\begin{pmatrix}-3\\0\\-4\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}1\\3\\0\end{pmatrix}\). Thus, \(\cos\beta=\frac{-3}{5\sqrt{10}}\), so \(\beta\approx100.94^\circ\). 3. The adjacent vectors at \(A\) are \(\overrightarrow{AB}=\begin{pmatrix}3\\0\\4\end{pmatrix}\) and \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\0\end{pmatrix}\). Their squared magnitudes are \(25\) and \(10\), and their dot product is \(3\). Therefore, the squared area is \(25\cdot10-3^2=241\), so the area is \(\sqrt{241}\) square units. 4. The diagonals bisect each other, so \(M=\frac{A+C}{2}=(4, 2.5, 3)\).

a) \(C(6, 4, 5)\) b) \(\beta\approx100.94^\circ\) c) \(\sqrt{241}\) square units d) \(M(4, 2.5, 3)\) Required dot products: \(\overrightarrow{BA}\cdot\overrightarrow{BC}=-3\) and \(\overrightarrow{AB}\cdot\overrightarrow{AD}=3\).

- Use equal opposite-side vectors to locate \(S\). - Use the dot-product area identity with the adjacent side vectors. - The diagonals of a parallelogram bisect each other. - Use the absolute value of the dot product when finding the acute angle between two lines.

1. Since \(\overrightarrow{PS}=\overrightarrow{QR}\), \(\overrightarrow{QR}=\begin{pmatrix}1\\2\\3\end{pmatrix}\), so \(S=P+\overrightarrow{QR}=(2, 3, 4)\). 2. The adjacent side vectors are \(\overrightarrow{PQ}=\begin{pmatrix}4\\0\\0\end{pmatrix}\) and \(\overrightarrow{PS}=\begin{pmatrix}1\\2\\3\end{pmatrix}\). Their squared magnitudes are \(16\) and \(14\), and their dot product is \(4\). Thus, the squared area is \(16\cdot14-4^2=208\), so the area is \(\sqrt{208}=4\sqrt{13}\) square units. 3. The diagonals bisect each other, so \(T=\frac{P+R}{2}=(3.5, 2, 2.5)\). 4. Direction vectors for the diagonals are \(\overrightarrow{PR}=\begin{pmatrix}5\\2\\3\end{pmatrix}\) and \(\overrightarrow{QS}=\begin{pmatrix}-3\\2\\3\end{pmatrix}\). Their dot product is \(-2\). For the acute angle between the diagonals, \(\cos\phi=\frac{|-2|}{\sqrt{38}\sqrt{22}}=\frac{2}{\sqrt{836}}\). Therefore, \(\phi\approx86.03^\circ\).

a) \(S(2, 3, 4)\) b) \(4\sqrt{13}\) square units c) \(T(3.5, 2, 2.5)\) and \(\phi\approx86.03^\circ\) Required dot products: \(\overrightarrow{PQ}\cdot\overrightarrow{PS}=4\) and \(\overrightarrow{PR}\cdot\overrightarrow{QS}=-2\).

- Convert each leg of the trip into horizontal and vertical components. - Add corresponding components before finding the overall displacement. - Use the final component signs to identify the correct direction reference.

1. The first displacement is \(\langle 120,0\rangle\). 2. The second displacement is \(\langle 80\cos30^\circ,80\sin30^\circ\rangle=\langle 40\sqrt3,40\rangle\). 3. The total displacement is \(\langle 120+40\sqrt3,40\rangle\). 4. Its magnitude is \(\sqrt{(120+40\sqrt3)^2+40^2}\approx193.5\,\text{mi}\). 5. Its direction is \(\theta=\tan^{-1}\left(\frac{40}{120+40\sqrt3}\right)\approx11.9^\circ\) north of east.

\(193.5\,\text{mi}\) at \(11.9^\circ\) north of east

- Resolve the angled force into x- and y-components. - Add the two force vectors componentwise. - Use the signs of the resultant components before selecting the direction angle.

1. The first force is \(\langle 50,0\rangle\). 2. The second force is \(\langle 40\cos120^\circ,40\sin120^\circ\rangle=\langle -20,20\sqrt3\rangle\). 3. The resultant is \(\langle 30,20\sqrt3\rangle\). 4. Its magnitude is \(\sqrt{30^2+(20\sqrt3)^2}=10\sqrt{21}\approx45.8\,\text{N}\). 5. Both components are positive, so the resultant is in quadrant I. Its direction angle is \(\tan^{-1}\left(\frac{20\sqrt3}{30}\right)\approx49.1^\circ\).

\(45.8\,\text{N}\) at a direction angle of \(49.1^\circ\)

- The ground track is due north only if the east-west components cancel. - Use the wind speed to determine the required westward component of the air-velocity vector. - The airspeed is the magnitude of the aircraft's velocity relative to the air.

1. For the ground track to have no east-west component, the aircraft's air-velocity vector must have a westward component of \(20\,\text{mph}\) to cancel the wind. 2. If \(\theta\) is the heading angle west of north, then \(100\sin\theta=20\), so \(\theta=\sin^{-1}(0.2)\approx11.5^\circ\). 3. The northward air-velocity component is \(\sqrt{100^2-20^2}=40\sqrt6\approx98.0\,\text{mph}\). 4. The wind has no north-south component, so the ground speed is approximately \(98.0\,\text{mph}\).

a) \(11.5^\circ\) west of north b) \(98.0\,\text{mph}\)

- 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 \(t=1\)

- Find \(G\) by adding the prism's height to \(C\)'s \(z\)-coordinate. - Use coordinate-axis unit vectors as normals to the coordinate planes. - Relate a line-plane angle to the angle between the line direction and a plane normal.

1. Since \(G\) is directly above \(C\), \(G=(0, 5, 16)\). 2. Therefore, \(\overrightarrow{AG}=\langle -12,5,16\rangle\). 3. Its magnitude is \(\|\overrightarrow{AG}\|=\sqrt{144+25+256}=\sqrt{425}=5\sqrt{17}\). 4. For the \(xy\)-plane, use normal vector \(\mathbf{n}_{xy}=\langle 0,0,1\rangle\). The line-plane angle \(\alpha_{xy}\) satisfies \(\sin\alpha_{xy}=\frac{16}{\sqrt{425}}\), so \(\alpha_{xy}\approx50.91^\circ\). 5. For the \(yz\)-plane, \(\sin\alpha_{yz}=\frac{12}{\sqrt{425}}\), so \(\alpha_{yz}\approx35.60^\circ\). 6. For the \(xz\)-plane, \(\sin\alpha_{xz}=\frac{5}{\sqrt{425}}\), so \(\alpha_{xz}\approx14.04^\circ\). The required reported dot products with \(\mathbf{n}_{xy}=\langle0,0,1\rangle\), \(\mathbf{n}_{yz}=\langle1,0,0\rangle\), and \(\mathbf{n}_{xz}=\langle0,1,0\rangle\) are \(16,-12,5\).

a) \(\overrightarrow{AG}=\langle -12,5,16\rangle\), and \(\|\overrightarrow{AG}\|=5\sqrt{17}\approx20.62\) units b) With the \(xy\)-plane: \(50.91^\circ\) With the \(yz\)-plane: \(35.60^\circ\) With the \(xz\)-plane: \(14.04^\circ\) Required dot products: \(16\), \(-12\), and \(5\), respectively.

- Form vectors from the base vertices to the apex. - Use the base plane's normal direction for the line-plane angle. - Use the dot-product angle formula for the angle between two lateral edges.

1. The edge vector is \(\overrightarrow{AS}=\langle -4,4,12\rangle\). 2. Its magnitude is \(\|\overrightarrow{AS}\|=\sqrt{16+16+144}=4\sqrt{11}\). 3. The base is the \(xy\)-plane, with unit normal \(\mathbf{n}=\langle 0,0,1\rangle\). If \(\beta\) is the line-plane angle, then \(\sin\beta=\frac{12}{4\sqrt{11}}=\frac{3}{\sqrt{11}}\), so \(\beta\approx64.76^\circ\). 4. Also, \(\overrightarrow{BS}=\langle -4,-4,12\rangle\), with magnitude \(4\sqrt{11}\). 5. The angle \(\gamma\) between the lateral edges satisfies \(\cos\gamma=\frac{\overrightarrow{AS}\cdot\overrightarrow{BS}}{\|\overrightarrow{AS}\|\|\overrightarrow{BS}\|}=\frac{144}{176}=\frac{9}{11}\). 6. Therefore, \(\gamma\approx35.10^\circ\).

a) \(AS=4\sqrt{11}\approx13.27\) units b) The angle with the base plane is approximately \(64.76^\circ\). c) The angle between \(AS\) and \(BS\) is approximately \(35.10^\circ\). Required dot products: \(\overrightarrow{AS}\cdot\langle0,0,1\rangle=12\) and \(\overrightarrow{SA}\cdot\overrightarrow{SB}=144\).

- A due-north ground velocity has zero \(x\)-component. - Relate ground velocity, air velocity, and wind velocity with vector addition. - Use the fixed airspeed to determine the unknown northward component. - Use distance divided by ground speed for the flight time.

1. Write the required ground velocity as \(\mathbf{v}_g=\langle 0,v_y,0\rangle\), where \(v_y>0\). 2. Since \(\mathbf{v}_g=\mathbf{v}_h+\mathbf{v}_w\), the air-velocity vector must be \(\mathbf{v}_h=\langle -80,v_y+60,0\rangle\). 3. Use the airspeed: \((-80)^2+(v_y+60)^2=170^2\). 4. This gives \((v_y+60)^2=22{,}500\). The northward solution is \(v_y+60=150\), so \(v_y=90\). 5. Therefore, \(\mathbf{v}_g=\langle 0,90,0\rangle\,\text{km/h}\) and \(\mathbf{v}_h=\langle -80,150,0\rangle\,\text{km/h}\). 6. For the angle \(\phi\), \(\cos\phi=\frac{\mathbf{v}_h\cdot\mathbf{v}_g}{\|\mathbf{v}_h\|\|\mathbf{v}_g\|}=\frac{15}{17}\). Thus, \(\phi\approx28.07^\circ\). 7. The flight time is \(\frac{540\,\text{km}}{90\,\text{km/h}}=6\,\text{h}\).

a) \(\mathbf{v}_g=\langle 0,90,0\rangle\,\text{km/h}\) b) \(\mathbf{v}_h=\langle -80,150,0\rangle\,\text{km/h}\) c) \(\phi\approx28.07^\circ\) d) \(6\,\text{h}\)

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

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

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

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

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

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

- Write the two side vectors that start at \(A\). - Use the dot-product area identity for the parallelogram formed by the vectors. - A triangle has half the corresponding parallelogram's area. - Remember that an absolute-value equation can produce two cases.

1. The side vectors from \(A\) are \(\overrightarrow{AB}=\begin{pmatrix}4\\2\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}x\\5\end{pmatrix}\). 2. Their squared magnitudes and dot product are \(\|\overrightarrow{AB}\|^2=20\), \(\|\overrightarrow{AC}\|^2=x^2+25\), and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4x+10\). 3. The square of twice the triangle's area is \(20(x^2+25)-(4x+10)^2=4(x-10)^2\). 4. Since the triangle's area is \(13\), \(\frac{1}{2}\sqrt{4(x-10)^2}=13\), so \(|x-10|=13\). 5. Thus, \(x=-3\) or \(x=23\). Because \(x>0\), the required value is \(x=23\).

\(x=23\) \(\overrightarrow{AB}\cdot\overrightarrow{AC}=4x+10\); \(x=23\).

- Check whether the two base vectors are perpendicular. - The height is the component of \(\mathbf{w}\) in the normal direction. - Use \(V=Gh\). - Think about why changing the base of a fixed solid cannot change its volume.

1. The base vectors are perpendicular because \(\mathbf{u}\cdot\mathbf{v}=0\). Their lengths are \(\|\mathbf{u}\|=5\) and \(\|\mathbf{v}\|=5\), so the base area is \(G=5\cdot5=25\) square units. 2. The component of \(\mathbf{w}\) perpendicular to the base is \(h=\|\mathbf{w}\|\cos30^\circ=6\left(\frac{\sqrt{3}}{2}\right)=3\sqrt{3}\). 3. Therefore, the volume is \(V=Gh=25\cdot3\sqrt{3}=75\sqrt{3}\) cubic units. 4. Choosing a different face as the base changes both the base area and the corresponding perpendicular height, but their product still measures the same three-dimensional region. Therefore, each valid base-height choice gives the same volume.

a) \(25\) square units b) \(75\sqrt{3}\) cubic units c) The product of a face's area and its corresponding perpendicular height measures the same fixed solid, regardless of the chosen base.

- Find each connecting vector and use its magnitude for the corresponding length. - Check whether opposite sides have equal or opposite vectors. - Use the dot-product angle formula at \(A\). - Use the dot-product area identity with the two adjacent side vectors.

1. The side vectors have lengths \(\|\overrightarrow{AB}\|=\sqrt{29}\), \(\|\overrightarrow{BC}\|=\sqrt{22}\), \(\|\overrightarrow{CD}\|=\sqrt{29}\), and \(\|\overrightarrow{DA}\|=\sqrt{22}\). Also, \(\overrightarrow{AB}=-\overrightarrow{CD}\), so the quadrilateral is a parallelogram. 2. The diagonals have vectors \(\overrightarrow{AC}=\begin{pmatrix}3\\3\\5\end{pmatrix}\) and \(\overrightarrow{BD}=\begin{pmatrix}-7\\3\\1\end{pmatrix}\). Thus, their lengths are \(\sqrt{43}\) and \(\sqrt{59}\), respectively. 3. At \(A\), the adjacent vectors are \(\overrightarrow{AB}=\begin{pmatrix}5\\0\\2\end{pmatrix}\) and \(\overrightarrow{AD}=\begin{pmatrix}-2\\3\\3\end{pmatrix}\). Their dot product is \(-4\), so \(\cos\alpha=\frac{-4}{\sqrt{29}\sqrt{22}}=\frac{-4}{\sqrt{638}}\). Therefore, \(\alpha\approx99.11^\circ\). 4. The squared area of the parallelogram is \(\|\overrightarrow{AB}\|^2\|\overrightarrow{AD}\|^2-(\overrightarrow{AB}\cdot\overrightarrow{AD})^2 =29\cdot22-(-4)^2=622\). Therefore, its area is \(\sqrt{622}\) square units.

a) Side lengths: \(\sqrt{29},\sqrt{22},\sqrt{29},\sqrt{22}\); diagonal lengths: \(\sqrt{43}\) and \(\sqrt{59}\) b) \(\alpha\approx99.11^\circ\) c) \(\sqrt{622}\) square units \(\overrightarrow{AB}\cdot\overrightarrow{AD}=-4\).

- Use diagonals from one vertex to divide the pentagon into nonoverlapping triangles. - For each triangle, use two side vectors with a common initial point. - Find each triangle's area with the dot-product area identity. - Add the three areas.

1. Use diagonals \(AC\) and \(AD\) to divide the pentagon into triangles \(ABC\), \(ACD\), and \(ADE\). 2. For triangle \(ABC\), \(\overrightarrow{AB}=\begin{pmatrix}2\\0\\2\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}3\\2\\5\end{pmatrix}\). Using the dot-product area identity gives an area of \(2\sqrt{3}\). 3. For triangle \(ACD\), \(\overrightarrow{AC}=\begin{pmatrix}3\\2\\5\end{pmatrix}\) and \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\4\end{pmatrix}\). Using the same identity gives an area of \(\frac{7\sqrt{3}}{2}\). 4. For triangle \(ADE\), \(\overrightarrow{AD}=\begin{pmatrix}1\\3\\4\end{pmatrix}\) and \(\overrightarrow{AE}=\begin{pmatrix}-1\\1\\0\end{pmatrix}\). Its area is \(2\sqrt{3}\). 5. Therefore, the pentagon's area is \(2\sqrt{3}+\frac{7\sqrt{3}}{2}+2\sqrt{3}=\frac{15\sqrt{3}}{2}\) square units.

\(\frac{15\sqrt{3}}{2}\) square units Required dot products: \(16\) for \(ABC\), \(29\) for \(ACD\), and \(2\) for \(ADE\).

- Use the given vector relationship to locate \(D\). - Find each side length from the magnitude of its connecting vector. - Use the dot-product angle formula at \(A\). - Divide the trapezoid into two triangles and use the dot-product area identity for each.

1. \(\overrightarrow{AB}=\begin{pmatrix}6\\2\\-2\end{pmatrix}\), so \(\overrightarrow{DC}=\frac{1}{2}\overrightarrow{AB}=\begin{pmatrix}3\\1\\-1\end{pmatrix}\). Therefore, \(D=C-\overrightarrow{DC}=(2, 4, 4)\). 2. The side lengths are \(\| \overrightarrow{AB}\|=\sqrt{44}=2\sqrt{11}\), \(\|\overrightarrow{BC}\|=\sqrt{17}\), \(\|\overrightarrow{CD}\|=\sqrt{11}\), and \(\|\overrightarrow{DA}\|=\sqrt{14}\). 3. At \(A\), \(\overrightarrow{AB}\cdot\overrightarrow{AD}=8\), so \(\cos\alpha=\frac{8}{\sqrt{44}\sqrt{14}}=\frac{8}{\sqrt{616}}\). Thus, \(\alpha\approx71.20^\circ\). 4. Divide the trapezoid along diagonal \(BD\). The dot-product area identity gives \(\text{Area}(\triangle ABD)=\sqrt{138}\) and \(\text{Area}(\triangle BCD)=\frac{\sqrt{138}}{2}\). 5. The trapezoid's area is \(\sqrt{138}+\frac{\sqrt{138}}{2}=\frac{3\sqrt{138}}{2}\) square units.

a) \(D(2, 4, 4)\) b) \(AB=2\sqrt{11}\), \(BC=\sqrt{17}\), \(CD=\sqrt{11}\), \(DA=\sqrt{14}\) c) \(\alpha\approx71.20^\circ\); area \(=\frac{3\sqrt{138}}{2}\) square units \(\overrightarrow{AB}\cdot\overrightarrow{AD}=8\).

- Use equal opposite-side vectors to find \(D\). - In a parallelogram, equal adjacent sides make the figure a rhombus. - The diagonals of a parallelogram bisect each other. - Use a dot product to test whether the diagonal vectors are perpendicular. - A rhombus with perpendicular diagonals has area equal to half the product of the diagonal lengths.

1. Since \(\overrightarrow{AD}=\overrightarrow{BC}\), \(\overrightarrow{BC}=\begin{pmatrix}0\\3\\4\end{pmatrix}\), so \(D=A+\overrightarrow{BC}=(2, 2, 7)\). 2. \(\|\overrightarrow{AB}\|=5\) and \(\|\overrightarrow{BC}\|=5\). A parallelogram with equal adjacent side lengths is a rhombus. 3. The diagonals bisect each other, so \(M=\frac{A+C}{2}=(3.5, 2.5, 5)\). 4. The diagonal vectors are \(\overrightarrow{AC}=\begin{pmatrix}3\\7\\4\end{pmatrix}\) and \(\overrightarrow{BD}=\begin{pmatrix}-3\\-1\\4\end{pmatrix}\). Their dot product is \(0\), so the diagonals intersect at \(90^\circ\). This confirms that the diagonals of a rhombus are perpendicular. 5. Their lengths are \(\sqrt{74}\) and \(\sqrt{26}\). Therefore, the area is \(\frac{1}{2}\sqrt{74}\sqrt{26}=\sqrt{481}\) square units.

a) \(D(2, 2, 7)\); \(AB=BC=5\), so \(ABCD\) is a rhombus. b) \(M(3.5, 2.5, 5)\) c) \(90^\circ\); the diagonals of a rhombus are perpendicular. d) \(\sqrt{481}\) square units \(\overrightarrow{AC}\cdot\overrightarrow{BD}=0\), so the diagonals meet at \(90^\circ\).

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

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

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

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

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

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

- Because the sign is motionless, the horizontal and vertical force components each balance separately. - Resolve each cable tension into horizontal and vertical components. - Use the horizontal balance equation to express one tension in terms of the other before using the vertical balance.

1. Let \(T_L\) and \(T_R\) be the left- and right-cable tensions. 2. Horizontal equilibrium gives \(T_L\cos30^\circ=T_R\cos45^\circ\). 3. Vertical equilibrium gives \(T_L\sin30^\circ+T_R\sin45^\circ=500\). 4. From the horizontal equation, \(T_R=T_L\frac{\cos30^\circ}{\cos45^\circ}=T_L\frac{\sqrt6}{2}\). 5. Substitute into the vertical equation: \(\frac12T_L+\frac{\sqrt2}{2}\left(T_L\frac{\sqrt6}{2}\right)=500\), so \(T_L=500(\sqrt3-1)\approx366.0\,\text{lb}\). 6. Then \(T_R\approx448.3\,\text{lb}\).

Left cable: \(366.0\,\text{lb}\) Right cable: \(448.3\,\text{lb}\)

- Read each force vector from its initial point and endpoint on the coordinate grid. - Add the two forces componentwise to obtain the resultant. - A balancing force must have the same magnitude as the resultant and the opposite direction.

1. Reading the endpoint coordinates from the diagram gives the two forces \(\langle 6,2\rangle\) and \(\langle -1,4\rangle\), in newtons. 2. Their resultant is \(\langle 6,2\rangle+\langle -1,4\rangle=\langle 5,6\rangle\,\text{N}\). 3. Its magnitude is \(\sqrt{5^2+6^2}=\sqrt{61}\approx7.8\,\text{N}\). 4. The equilibrant is the opposite of the resultant, so it is \(\langle -5,-6\rangle\,\text{N}\).

a) \(\langle 5,6\rangle\,\text{N}\) b) \(7.8\,\text{N}\) c) \(\langle -5,-6\rangle\,\text{N}\)