**Problem 9:** Find two unit vectors parallel to the yz-plane and orthogonal to $\mathbf{v} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$. 1. Vectors parallel to the yz-plane have zero $x$-component, so they are of the form $\mathbf{u} = 0\mathbf{i} + y\mathbf{j} + z\mathbf{k}$. 2. Orthogonality to $\mathbf{v}$ means $\mathbf{u} \cdot \mathbf{v} = 0$. Calculate: $$0 \cdot 3 + y \cdot (-1) + z \cdot 2 = -y + 2z = 0.$$ 3. Solve for $y$: $y = 2z$. 4. Choose $z=1$, then $y=2$, so $\mathbf{u}_1 = 0\mathbf{i} + 2\mathbf{j} + 1\mathbf{k}$. 5. Normalize $\mathbf{u}_1$: $$||\mathbf{u}_1|| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}.$$ 6. Unit vector: $$\mathbf{u}_1 = \frac{1}{\sqrt{5}}(0,2,1).$$ 7. Another vector orthogonal to $\mathbf{v}$ and parallel to yz-plane can be $\mathbf{u}_2 = 0\mathbf{i} -1\mathbf{j} + 2\mathbf{k}$ (negate $\mathbf{u}_1$'s $y$ and $z$ components). 8. Normalize $\mathbf{u}_2$: $$||\mathbf{u}_2|| = \sqrt{0^2 + (-1)^2 + 2^2} = \sqrt{5}.$$ 9. Unit vector: $$\mathbf{u}_2 = \frac{1}{\sqrt{5}}(0,-1,2).$$ **Problem 10:** Find area, sine, and cosine of angles of triangle with vertices $A(0,0,0)$, $B(4,-1,3)$, $C(1,2,3)$. 1. Compute vectors: $$\mathbf{AB} = B - A = (4,-1,3), \quad \mathbf{AC} = C - A = (1,2,3).$$ 2. Compute lengths: $$||\mathbf{AB}|| = \sqrt{4^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}.$$ $$||\mathbf{AC}|| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}.$$ 3. Compute $\mathbf{BC} = C - B = (1-4, 2+1, 3-3) = (-3,3,0)$ and its length: $$||\mathbf{BC}|| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.$$ 4. Area of triangle: $$\text{Area} = \frac{1}{2} ||\mathbf{AB} \times \mathbf{AC}||.$$ 5. Compute cross product: $$\mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 3 \\ 1 & 2 & 3 \end{vmatrix} = \mathbf{i}((-1)(3) - 3(2)) - \mathbf{j}(4(3) - 3(1)) + \mathbf{k}(4(2) - (-1)(1)) = \mathbf{i}(-3 - 6) - \mathbf{j}(12 - 3) + \mathbf{k}(8 + 1) = -9\mathbf{i} - 9\mathbf{j} + 9\mathbf{k}.$$ 6. Magnitude: $$||\mathbf{AB} \times \mathbf{AC}|| = \sqrt{(-9)^2 + (-9)^2 + 9^2} = \sqrt{81 + 81 + 81} = \sqrt{243} = 9\sqrt{3}.$$ 7. Area: $$\frac{1}{2} \times 9\sqrt{3} = \frac{9\sqrt{3}}{2}.$$ 8. Compute cosines of angles using dot products: $$\cos \angle BAC = \frac{\mathbf{AB} \cdot \mathbf{AC}}{||\mathbf{AB}|| ||\mathbf{AC}||} = \frac{4 \cdot 1 + (-1) \cdot 2 + 3 \cdot 3}{\sqrt{26} \sqrt{14}} = \frac{4 - 2 + 9}{\sqrt{364}} = \frac{11}{\sqrt{364}} = \frac{11}{2\sqrt{91}}.$$ 9. Similarly, $$\cos \angle ABC = \frac{\mathbf{BA} \cdot \mathbf{BC}}{||\mathbf{BA}|| ||\mathbf{BC}||} = \frac{(-4)(-3) + 1(3) + (-3)(0)}{\sqrt{26} \times 3\sqrt{2}} = \frac{12 + 3 + 0}{3\sqrt{52}} = \frac{15}{3 \times 2\sqrt{13}} = \frac{15}{6\sqrt{13}} = \frac{5}{2\sqrt{13}}.$$ 10. And $$\cos \angle ACB = \frac{\mathbf{CA} \cdot \mathbf{CB}}{||\mathbf{CA}|| ||\mathbf{CB}||} = \frac{(-1)(-3) + (-2)(3) + 0(0)}{\sqrt{14} \times 3\sqrt{2}} = \frac{3 - 6 + 0}{3\sqrt{28}} = \frac{-3}{3 \times 2\sqrt{7}} = \frac{-3}{6\sqrt{7}} = -\frac{1}{2\sqrt{7}}.$$ 11. Sine of angles: $$\sin \theta = \sqrt{1 - \cos^2 \theta}.$$ **Problem 11:** Check if quadrilateral $PQRS$ with $P(1,-2,3)$, $Q(4,3,-1)$, $R(2,2,1)$, $S(5,7,-3)$ is a parallelogram and find its area. 1. Compute vectors: $$\mathbf{PQ} = Q - P = (3,5,-4), \quad \mathbf{RS} = S - R = (3,5,-4).$$ 2. Since $\mathbf{PQ} = \mathbf{RS}$, opposite sides are equal and parallel. 3. Compute vectors: $$\mathbf{QR} = R - Q = (-2,-1,2), \quad \mathbf{SP} = P - S = (-4,-9,6).$$ 4. Check if $\mathbf{QR} = \mathbf{SP}$: no, so check if $\mathbf{QR} + \mathbf{SP} = 0$: $$(-2,-1,2) + (-4,-9,6) = (-6,-10,8) \neq 0.$$ 5. So $PQRS$ is not a parallelogram by this test. 6. Area of parallelogram formed by $\mathbf{PQ}$ and $\mathbf{QR}$: $$\text{Area} = ||\mathbf{PQ} \times \mathbf{QR}||.$$ 7. Compute cross product: $$\mathbf{PQ} \times \mathbf{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 5 & -4 \\ -2 & -1 & 2 \end{vmatrix} = \mathbf{i}(5 \cdot 2 - (-4)(-1)) - \mathbf{j}(3 \cdot 2 - (-4)(-2)) + \mathbf{k}(3 \cdot (-1) - 5(-2)) = \mathbf{i}(10 - 4) - \mathbf{j}(6 - 8) + \mathbf{k}(-3 + 10) = 6\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}.$$ 8. Magnitude: $$\sqrt{6^2 + 2^2 + 7^2} = \sqrt{36 + 4 + 49} = \sqrt{89}.$$ 9. Area of quadrilateral (assuming it is a parallelogram) is $\sqrt{89}$. **Problem 12:** Find volume and total surface area of parallelepiped with edges $$\mathbf{A} = \mathbf{i} - \mathbf{j} - 2\mathbf{k}, \quad \mathbf{B} = 3\mathbf{i} - 4\mathbf{j} + \mathbf{k}, \quad \mathbf{C} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k}.$$ 1. Volume is absolute value of scalar triple product: $$V = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})|.$$ 2. Compute $\mathbf{B} \times \mathbf{C}$: $$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -4 & 1 \\ 3 & 1 & 2 \end{vmatrix} = \mathbf{i}((-4)(2) - 1(1)) - \mathbf{j}(3(2) - 1(3)) + \mathbf{k}(3(1) - (-4)(3)) = \mathbf{i}(-8 - 1) - \mathbf{j}(6 - 3) + \mathbf{k}(3 + 12) = -9\mathbf{i} - 3\mathbf{j} + 15\mathbf{k}.$$ 3. Compute dot product: $$\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (1)(-9) + (-1)(-3) + (-2)(15) = -9 + 3 - 30 = -36.$$ 4. Volume: $$| -36 | = 36.$$ 5. Surface area is sum of areas of parallelograms on each pair of edges: $$S = 2(||\mathbf{A} \times \mathbf{B}|| + ||\mathbf{B} \times \mathbf{C}|| + ||\mathbf{C} \times \mathbf{A}||).$$ 6. Compute $||\mathbf{A} \times \mathbf{B}||$: $$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -2 \\ 3 & -4 & 1 \end{vmatrix} = \mathbf{i}((-1)(1) - (-2)(-4)) - \mathbf{j}(1(1) - (-2)(3)) + \mathbf{k}(1(-4) - (-1)(3)) = \mathbf{i}(-1 - 8) - \mathbf{j}(1 + 6) + \mathbf{k}(-4 + 3) = -9\mathbf{i} - 7\mathbf{j} - \mathbf{k}.$$ $$||\mathbf{A} \times \mathbf{B}|| = \sqrt{(-9)^2 + (-7)^2 + (-1)^2} = \sqrt{81 + 49 + 1} = \sqrt{131}.$$ 7. Compute $||\mathbf{C} \times \mathbf{A}||$: $$\mathbf{C} \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ 1 & -1 & -2 \end{vmatrix} = \mathbf{i}(1(-2) - 2(-1)) - \mathbf{j}(3(-2) - 2(1)) + \mathbf{k}(3(-1) - 1(1)) = \mathbf{i}(-2 + 2) - \mathbf{j}(-6 - 2) + \mathbf{k}(-3 - 1) = 0\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}.$$ $$||\mathbf{C} \times \mathbf{A}|| = \sqrt{0^2 + 8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}.$$ 8. Recall $||\mathbf{B} \times \mathbf{C}|| = \sqrt{(-9)^2 + (-3)^2 + 15^2} = \sqrt{81 + 9 + 225} = \sqrt{315} = 3\sqrt{35}.$ 9. Total surface area: $$2(\sqrt{131} + 3\sqrt{35} + 4\sqrt{5}).$$ **Problem 13:** Determine if points $P(\frac{1}{2}, \frac{1}{3}, 0)$, $Q(1,1,-1)$, $R(-2,-3,5)$ are collinear or not. 1. Compute vectors: $$\mathbf{PQ} = Q - P = \left(1 - \frac{1}{2}, 1 - \frac{1}{3}, -1 - 0\right) = \left(\frac{1}{2}, \frac{2}{3}, -1\right).$$ $$\mathbf{PR} = R - P = \left(-2 - \frac{1}{2}, -3 - \frac{1}{3}, 5 - 0\right) = \left(-\frac{5}{2}, -\frac{10}{3}, 5\right).$$ 2. Check if $\mathbf{PR} = \lambda \mathbf{PQ}$ for some scalar $\lambda$. 3. Compare components: $$-\frac{5}{2} = \lambda \cdot \frac{1}{2} \Rightarrow \lambda = -5,$$ $$-\frac{10}{3} = \lambda \cdot \frac{2}{3} = -5 \cdot \frac{2}{3} = -\frac{10}{3},$$ $$5 = \lambda \cdot (-1) = -5 \cdot (-1) = 5.$$ 4. All components satisfy $\mathbf{PR} = -5 \mathbf{PQ}$, so points are collinear. 5. Symmetric equation of line through $P$ and $Q$: $$\frac{x - \frac{1}{2}}{\frac{1}{2}} = \frac{y - \frac{1}{3}}{\frac{2}{3}} = \frac{z - 0}{-1}.$$ **Summary:** - Problem 9: Unit vectors $\frac{1}{\sqrt{5}}(0,2,1)$ and $\frac{1}{\sqrt{5}}(0,-1,2)$. - Problem 10: Area $= \frac{9\sqrt{3}}{2}$, cosines of angles $\frac{11}{2\sqrt{91}}$, $\frac{5}{2\sqrt{13}}$, $-\frac{1}{2\sqrt{7}}$. - Problem 11: Quadrilateral is not parallelogram; area $= \sqrt{89}$. - Problem 12: Volume $= 36$, surface area $= 2(\sqrt{131} + 3\sqrt{35} + 4\sqrt{5})$. - Problem 13: Points are collinear; symmetric line equation $\frac{x - \frac{1}{2}}{\frac{1}{2}} = \frac{y - \frac{1}{3}}{\frac{2}{3}} = \frac{z}{-1}$.