1. **Find a vector of length 10 units along the line through A(0,−1,1) and B(2,−2,3):** Step 1: Find the vector \( \overrightarrow{AB} = B - A = (2-0, -2+1, 3-1) = (2, -1, 2) \). Step 2: Find the length of \( \overrightarrow{AB} \): $$\|\overrightarrow{AB}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$$ Step 3: Find the unit vector along \( \overrightarrow{AB} \): $$\hat{u} = \frac{1}{3}(2, -1, 2) = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right).$$ Step 4: Multiply by 10 to get vector of length 10: $$\vec{v} = 10 \times \hat{u} = \left(\frac{20}{3}, -\frac{10}{3}, \frac{20}{3}\right).$$ 2. **Find the direction cosines of \( \vec{V} = -6\mathbf{i} + 7\mathbf{j} + 2\mathbf{k} \):** Step 1: Find the magnitude of \( \vec{V} \): $$\|\vec{V}\| = \sqrt{(-6)^2 + 7^2 + 2^2} = \sqrt{36 + 49 + 4} = \sqrt{89}.$$ Step 2: Direction cosines are cosines of angles with axes: $$\cos \alpha = \frac{-6}{\sqrt{89}}, \quad \cos \beta = \frac{7}{\sqrt{89}}, \quad \cos \gamma = \frac{2}{\sqrt{89}}.$$ 3. Given \( A=(3,-1,1), B=(2,1,-1), C=(1,0,-2) \), and \( p=\overrightarrow{OA}, q=\overrightarrow{OB}, w=\overrightarrow{OC} \): a) **Find the angle between \( p \) and \( w \):** Step 1: Compute dot product: $$p \cdot w = 3 \times 1 + (-1) \times 0 + 1 \times (-2) = 3 + 0 - 2 = 1.$$ Step 2: Compute magnitudes: $$\|p\| = \sqrt{3^2 + (-1)^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11},$$ $$\|w\| = \sqrt{1^2 + 0^2 + (-2)^2} = \sqrt{1 + 0 + 4} = \sqrt{5}.$$ Step 3: Use formula for angle \( \theta \): $$\cos \theta = \frac{p \cdot w}{\|p\| \|w\|} = \frac{1}{\sqrt{11} \sqrt{5}} = \frac{1}{\sqrt{55}}.$$ Step 4: Angle: $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{55}}\right).$$ b) **Find a vector \( v \) orthogonal to both \( q \) and \( w \):** Step 1: Compute \( q = (2,1,-1), w = (1,0,-2) \). Step 2: Use cross product: $$v = q \times w = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & 0 & -2 \end{vmatrix} = \mathbf{i}(1 \times -2 - (-1) \times 0) - \mathbf{j}(2 \times -2 - (-1) \times 1) + \mathbf{k}(2 \times 0 - 1 \times 1)$$ $$= \mathbf{i}(-2 - 0) - \mathbf{j}(-4 + 1) + \mathbf{k}(0 - 1) = (-2)\mathbf{i} - (-3)\mathbf{j} - 1\mathbf{k} = (-2, 3, -1).$$ c) **Find the vector component (orthogonal projection) of \( p \) along \( q \) and orthogonal to \( q \):** Step 1: Projection of \( p \) on \( q \): $$\text{proj}_q p = \frac{p \cdot q}{\|q\|^2} q.$$ Calculate dot product: $$p \cdot q = 3 \times 2 + (-1) \times 1 + 1 \times (-1) = 6 - 1 - 1 = 4.$$ Calculate \( \|q\|^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6.\) So, $$\text{proj}_q p = \frac{4}{6} (2,1,-1) = \left(\frac{8}{6}, \frac{4}{6}, -\frac{4}{6}\right) = \left(\frac{4}{3}, \frac{2}{3}, -\frac{2}{3}\right).$$ Step 2: Vector orthogonal to \( q \) is: $$p - \text{proj}_q p = \left(3 - \frac{4}{3}, -1 - \frac{2}{3}, 1 + \frac{2}{3}\right) = \left(\frac{5}{3}, -\frac{5}{3}, \frac{5}{3}\right).$$ d) **Find the area of triangle ABC:** Step 1: Compute vectors: $$\overrightarrow{AB} = B - A = (2-3, 1+1, -1-1) = (-1, 2, -2),$$ $$\overrightarrow{AC} = C - A = (1-3, 0+1, -2-1) = (-2, 1, -3).$$ Step 2: Area = \( \frac{1}{2} \| \overrightarrow{AB} \times \overrightarrow{AC} \| \). Calculate cross product: $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & -2 \\ -2 & 1 & -3 \end{vmatrix} = \mathbf{i}(2 \times -3 - (-2) \times 1) - \mathbf{j}(-1 \times -3 - (-2) \times -2) + \mathbf{k}(-1 \times 1 - 2 \times -2)$$ $$= \mathbf{i}(-6 + 2) - \mathbf{j}(3 - 4) + \mathbf{k}(-1 + 4) = (-4, 1, 3).$$ Step 3: Magnitude: $$\sqrt{(-4)^2 + 1^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}.$$ Step 4: Area: $$\frac{1}{2} \sqrt{26}.$$ e) **Do vectors \( p, q, w \) lie in the same plane?** Step 1: Check if scalar triple product \( p \cdot (q \times w) = 0 \). Calculate \( q \times w = (-2, 3, -1) \) from part b. Calculate dot product: $$p \cdot (q \times w) = 3 \times (-2) + (-1) \times 3 + 1 \times (-1) = -6 - 3 - 1 = -10 \neq 0.$$ Step 2: Since scalar triple product \( \neq 0 \), vectors are not coplanar. f) **Find the equation of the plane through points A, B, C:** Step 1: Use normal vector \( \vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = (-4, 1, 3) \). Step 2: Equation of plane: $$-4(x - 3) + 1(y + 1) + 3(z - 1) = 0,$$ Simplify: $$-4x + 12 + y + 1 + 3z - 3 = 0,$$ $$-4x + y + 3z + 10 = 0.$$ g) **Find the equation of the plane through C where vector \( \overrightarrow{AB} \) is normal:** Step 1: Normal vector is \( \overrightarrow{AB} = (-1, 2, -2) \). Step 2: Equation: $$-1(x - 1) + 2(y - 0) - 2(z + 2) = 0,$$ Simplify: $$-x + 1 + 2y - 2z - 4 = 0,$$ $$-x + 2y - 2z - 3 = 0.$$