1. **Problem (a):** Find the cross product $\mathbf{A} \times \mathbf{B}$ and the angle $\theta$ between vectors $\mathbf{A} = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$ and $\mathbf{B} = 2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$. 2. **Cross product formula:** $$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$ 3. **Calculate $\mathbf{A} \times \mathbf{B}$:** $$= \mathbf{i} \begin{vmatrix} -2 & 4 \\ 3 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 4 \\ 2 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -2 \\ 2 & 3 \end{vmatrix}$$ $$= \mathbf{i}((-2)(-2) - 4 \times 3) - \mathbf{j}(3 \times (-2) - 4 \times 2) + \mathbf{k}(3 \times 3 - (-2) \times 2)$$ $$= \mathbf{i}(4 - 12) - \mathbf{j}(-6 - 8) + \mathbf{k}(9 + 4)$$ $$= -8\mathbf{i} + 14\mathbf{j} + 13\mathbf{k}$$ 4. **Magnitude of $\mathbf{A}$ and $\mathbf{B}$:** $$|\mathbf{A}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$$ $$|\mathbf{B}| = \sqrt{2^2 + 3^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$$ 5. **Dot product $\mathbf{A} \cdot \mathbf{B}$:** $$= 3 \times 2 + (-2) \times 3 + 4 \times (-2) = 6 - 6 - 8 = -8$$ 6. **Angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$:** Using formula $$\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|} = \frac{-8}{\sqrt{29} \times \sqrt{17}} = \frac{-8}{\sqrt{493}}$$ $$\theta = \cos^{-1} \left( \frac{-8}{\sqrt{493}} \right)$$ --- 7. **Problem (b):** Find the directional derivative of $$\phi = x^2 z + 2 x y^2 + y z^2$$ at point $(1,2,-1)$ in direction of vector $\mathbf{A} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$. 8. **Directional derivative formula:** $$D_{\mathbf{u}} \phi = \nabla \phi \cdot \mathbf{u}$$ where $\mathbf{u}$ is the unit vector in direction of $\mathbf{A}$. 9. **Gradient $\nabla \phi$ components:** $$\frac{\partial \phi}{\partial x} = 2 x z + 2 y^2$$ $$\frac{\partial \phi}{\partial y} = 4 x y + z^2$$ $$\frac{\partial \phi}{\partial z} = x^2 + 2 y z$$ 10. **Evaluate gradient at $(1,2,-1)$:** $$\frac{\partial \phi}{\partial x} = 2 \times 1 \times (-1) + 2 \times 2^2 = -2 + 8 = 6$$ $$\frac{\partial \phi}{\partial y} = 4 \times 1 \times 2 + (-1)^2 = 8 + 1 = 9$$ $$\frac{\partial \phi}{\partial z} = 1^2 + 2 \times 2 \times (-1) = 1 - 4 = -3$$ 11. **Unit vector $\mathbf{u}$:** $$|\mathbf{A}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$$ $$\mathbf{u} = \frac{1}{\sqrt{29}} (2, 3, -4)$$ 12. **Directional derivative:** $$D_{\mathbf{u}} \phi = (6, 9, -3) \cdot \frac{1}{\sqrt{29}} (2, 3, -4) = \frac{1}{\sqrt{29}} (6 \times 2 + 9 \times 3 + (-3) \times (-4))$$ $$= \frac{1}{\sqrt{29}} (12 + 27 + 12) = \frac{51}{\sqrt{29}}$$ --- 13. **Problem (c):** Find vector triple product $$\mathbf{A} \times (\mathbf{B} \times \mathbf{C})$$ where $$\mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}, \quad \mathbf{B} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}, \quad \mathbf{C} = 3\mathbf{i} + \mathbf{j} + 3\mathbf{k}$$ 14. **Use vector triple product identity:** $$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C} (\mathbf{A} \cdot \mathbf{B})$$ 15. **Calculate dot products:** $$\mathbf{A} \cdot \mathbf{C} = 2 \times 3 + (-3) \times 1 + 1 \times 3 = 6 - 3 + 3 = 6$$ $$\mathbf{A} \cdot \mathbf{B} = 2 \times 1 + (-3) \times 2 + 1 \times (-1) = 2 - 6 - 1 = -5$$ 16. **Calculate vector triple product:** $$= 6 \mathbf{B} - (-5) \mathbf{C} = 6 (1, 2, -1) + 5 (3, 1, 3)$$ $$= (6, 12, -6) + (15, 5, 15) = (21, 17, 9)$$ **Final answers:** (a) $\mathbf{A} \times \mathbf{B} = -8\mathbf{i} + 14\mathbf{j} + 13\mathbf{k}$, $\theta = \cos^{-1} \left( \frac{-8}{\sqrt{493}} \right)$ (b) Directional derivative $= \frac{51}{\sqrt{29}}$ (c) Vector triple product $= 21\mathbf{i} + 17\mathbf{j} + 9\mathbf{k}$