1. Stating the problem: (a) Determine $A\times B$ and angle $\theta$ between the two vectors where $A = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$ and $B = 2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$. 2. Formula and important rules: Use the determinant formula for the cross product and the dot-product formula $\cos\theta = \dfrac{A\cdot B}{|A||B|}$ to get the angle. 3. Compute $A\times B$ using the determinant formula. $$A \times B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 4 \\ 2 & 3 & -2 \end{vmatrix}$$ 4. Expand the determinant and simplify. $$= \mathbf{i}((-2)(-2)-4\cdot3)-\mathbf{j}(3(-2)-4\cdot2)+\mathbf{k}(3\cdot3-(-2)\cdot2)$$ $$= \mathbf{i}(4-12)-\mathbf{j}(-6-8)+\mathbf{k}(9+4)$$ $$= -8\mathbf{i}+14\mathbf{j}+13\mathbf{k}$$ 5. Final cross product for part (a): $A\times B = \langle -8,\,14,\,13 \rangle$. 6. Compute the dot product and magnitudes to find the angle. $$A\cdot B = 3\cdot2 + (-2)\cdot3 + 4\cdot(-2) = 6 -6 -8 = -8$$ $$|A| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{29}$$ $$|B| = \sqrt{2^2 + 3^2 + (-2)^2} = \sqrt{17}$$ 7. Use the cosine formula and simplify. $$\cos\theta = \frac{A\cdot B}{|A||B|} = \frac{-8}{\sqrt{29}\sqrt{17}} = \frac{-8}{\sqrt{493}}$$ 8. Numerical value of the angle (approximate): $\theta = \arccos\left(\dfrac{-8}{\sqrt{493}}\right) \approx 111.07^{\circ} \approx 1.938\ \text{rad}$. 1. Stating the problem (b): Find the directional derivative of $\Phi(x,y,z)=x^{2}z+2xy^{2}+yz^{2}$ at the point $(1,2,-1)$ in the direction of $A=2\mathbf{i}+3\mathbf{j}-4\mathbf{k}$. 2. Formula and important rules: The directional derivative in direction of a unit vector $\mathbf{u}$ is $D_{\mathbf{u}}\Phi=\nabla\Phi\cdot\mathbf{u}$, and $\mathbf{u}=\dfrac{A}{|A|}$. 3. Compute the gradient components symbolically. $$\nabla\Phi = \left\langle \frac{\partial\Phi}{\partial x},\,\frac{\partial\Phi}{\partial y},\,\frac{\partial\Phi}{\partial z} \right\rangle$$ $$\frac{\partial\Phi}{\partial x}=2xz+2y^{2},\quad \frac{\partial\Phi}{\partial y}=4xy+z^{2},\quad \frac{\partial\Phi}{\partial z}=x^{2}+2yz$$ 4. Evaluate the gradient at $(1,2,-1)$. $$\nabla\Phi(1,2,-1)=\langle 2(1)(-1)+2(2)^{2},\,4(1)(2)+(-1)^{2},\,1^{2}+2(2)(-1)\rangle$$ $$=\langle -2+8,\,8+1,\,1-4\rangle = \langle 6,\,9,\,-3\rangle$$ 5. Compute the unit vector in the direction of $A$. $$|A|=\sqrt{2^{2}+3^{2}+(-4)^{2}}=\sqrt{29}$$ $$\mathbf{u}=\frac{1}{\sqrt{29}}\langle 2,\,3,\,-4\rangle$$ 6. Compute the directional derivative by dot product. $$D_{\mathbf{u}}\Phi = \nabla\Phi(1,2,-1)\cdot\mathbf{u} = \frac{\nabla\Phi(1,2,-1)\cdot A}{|A|}$$ $$\nabla\Phi(1,2,-1)\cdot A = 6\cdot2 + 9\cdot3 + (-3)\cdot(-4) = 12 +27 +12 = 51$$ $$D_{\mathbf{u}}\Phi = \frac{51}{\sqrt{29}}$$ 7. Numerical approximation for part (b): $D_{\mathbf{u}}\Phi \approx 9.472$ (approximately). 1. Stating the problem (c): For $A=2\mathbf{i}-3\mathbf{j}+\mathbf{k}$, $B=\mathbf{i}+2\mathbf{j}-\mathbf{k}$, $C=3\mathbf{i}+\mathbf{j}+3\mathbf{k}$, determine $A\times(B\times C)$. 2. Formula and important rule: Use the cross product determinant or the vector triple product identity $A\times(B\times C)=B(A\cdot C)-C(A\cdot B)$ if desired. 3. Compute $B\times C$ by determinant. $$B\times C = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & 1 & 3 \end{vmatrix}$$ $$=\mathbf{i}(2\cdot3 - (-1)\cdot1)-\mathbf{j}(1\cdot3 - (-1)\cdot3)+\mathbf{k}(1\cdot1 - 2\cdot3)$$ $$=\mathbf{i}(6+1)-\mathbf{j}(3+3)+\mathbf{k}(1-6)$$ $$=7\mathbf{i}-6\mathbf{j}-5\mathbf{k}$$ 4. Now compute $A\times(B\times C)=A\times\langle 7,-6,-5\rangle$. $$A\times(B\times C)=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 7 & -6 & -5 \end{vmatrix}$$ $$=\mathbf{i}((-3)(-5)-1(-6))-\mathbf{j}(2(-5)-1\cdot7)+\mathbf{k}(2(-6)-(-3)\cdot7)$$ $$=\mathbf{i}(15+6)-\mathbf{j}(-10-7)+\mathbf{k}(-12+21)$$ $$=21\mathbf{i}+17\mathbf{j}+9\mathbf{k}$$ 5. Final answer for part (c): $A\times(B\times C)=\langle 21,\,17,\,9 \rangle$.