Let's solve each part step-by-step: --- ## Given Vectors \[ \vec{a} = 3\hat{i} + 2\hat{j} - 2\hat{k} \] \[ \vec{b} = -2\hat{i} + 3\hat{j} + 4\hat{k} \] \[ \vec{c} = 1\hat{i} - 2\hat{j} - 1\hat{k} \] --- ## (a) \(\vec{d} = 2\vec{a} - \vec{b} - 2\vec{c}\) Calculate each term: **\(2\vec{a}\):** \[ 2\vec{a} = 2(3\hat{i} + 2\hat{j} - 2\hat{k}) = 6\hat{i} + 4\hat{j} - 4\hat{k} \] **\(-\vec{b}\):** \[ -\vec{b} = -1(-2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\hat{i} - 3\hat{j} - 4\hat{k} \] **\(-2\vec{c}\):** \[ -2\vec{c} = -2(1\hat{i} - 2\hat{j} - 1\hat{k}) = -2\hat{i} + 4\hat{j} + 2\hat{k} \] Add them together: \[ \vec{d} = (6\hat{i} + 4\hat{j} - 4\hat{k}) + (2\hat{i} - 3\hat{j} - 4\hat{k}) + (-2\hat{i} + 4\hat{j} + 2\hat{k}) \] Combine like terms: - \(i\): \(6 + 2 - 2 = 6\) - \(j\): \(4 - 3 + 4 = 5\) - \(k\): \(-4 - 4 + 2 = -6\) **Final answer:** \[ \boxed{\vec{d} = 6\hat{i} + 5\hat{j} - 6\hat{k}} \] --- ## (b) \(|\vec{d}|\) (Magnitude) \[ |\vec{d}| = \sqrt{6^2 + 5^2 + (-6)^2} = \sqrt{36 + 25 + 36} = \sqrt{97} \] \[ \boxed{|\vec{d}| = \sqrt{97}} \] --- ## (c) \(\vec{a} \cdot \vec{b}\) (Dot Product) \[ \vec{a} \cdot \vec{b} = (3)(-2) + (2)(3) + (-2)(4) \] \[ = -6 + 6 - 8 = -8 \] \[ \boxed{\vec{a} \cdot \vec{b} = -8} \] --- ## (d) \(\vec{b} \times \vec{c}\) (Cross Product) Set up determinant: \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 4 \\ 1 & -2 & -1 \end{vmatrix} \] \[ = \hat{i} \left(3 \times -1 - 4 \times -2 \right) - \hat{j} \left(-2 \times -1 - 4 \times 1 \right) + \hat{k} \left(-2 \times -2 - 3 \times 1 \right) \] \[ = \hat{i} (-3 + 8) - \hat{j} (2 - 4) + \hat{k} (4 - 3) \] \[ = \hat{i}(5) - \hat{j}(-2) + \hat{k}(1) \] \[ = 5\hat{i} + 2\hat{j} + \hat{k} \] \[ \boxed{\vec{b} \times \vec{c} = 5\hat{i} + 2\hat{j} + 1\hat{k}} \] --- ## (e) \(\angle(\vec{a}, \vec{c})\) (Angle Between Vectors) \[ \cos\theta = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} \] First, calculate \(\vec{a} \cdot \vec{c}\): \[ \vec{a} \cdot \vec{c} = (3)(1) + (2)(-2) + (-2)(-1) = 3 - 4 + 2 = 1 \] \[ |\vec{a}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} \] \[ |\vec{c}| = \sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \] \[ \cos\theta = \frac{1}{\sqrt{17}\sqrt{6}} = \frac{1}{\sqrt{102}} \] \[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{102}} \right ) \] **Final answer:** \[ \boxed{\theta = \cos^{-1} \left( \frac{1}{\sqrt{102}} \right )} \] --- ## (f) \(\operatorname{proj}_{\vec{b}}\vec{a}\) (Projection) \[ \operatorname{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b} \] From part (c): \( \vec{a} \cdot \vec{b} = -8 \) \[ |\vec{b}|^2 = (-2)^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29 \] \[ \operatorname{proj}_{\vec{b}}\vec{a} = \frac{-8}{29} \vec{b} = \frac{-8}{29}(-2\hat{i} + 3\hat{j} + 4\hat{k}) \] \[ = \left(\frac{16}{29}\right)\hat{i} - \left(\frac{24}{29}\right)\hat{j} - \left(\frac{32}{29}\right)\hat{k} \] **Final answer:** \[ \boxed{ \operatorname{proj}_{\vec{b}}\vec{a} = \frac{16}{29}\hat{i} - \frac{24}{29}\hat{j} - \frac{32}{29}\hat{k} } \] --- Let me know if you need further explanations or calculations!