91Ó°ÊÓ

To find the cross product \( \mathbf{a} \times \mathbf{b} \), we use the determinant formula for vectors. The cross product of two vectors \( \mathbf{u} = u_1 \mathbf{i} + u_2 \mathbf{j} + u_3 \mathbf{k} \) and \( \mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \) is given by:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3\end{vmatrix}\]Substitute \( \mathbf{a} = -3 \mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{b} = -\mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k} \) into the determinant:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & 2 & -2 \-1 & 2 & -4\end{vmatrix}\]Calculating this, we get:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(2(-4) - (-2)(2)) - \mathbf{j}(-3(-4) - (-2)(-1)) + \mathbf{k}(-3(2) - 2(-1))\]\[= \mathbf{i}(-8 + 4) - \mathbf{j}(12 - 2) + \mathbf{k}(-6 + 2)\]\[= -4\mathbf{i} - 10\mathbf{j} - 4\mathbf{k}\]

First, find \( \mathbf{b} + \mathbf{c} \). Given \( \mathbf{b} = -\mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k} \) and \( \mathbf{c} = 7 \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k} \), we add them component-wise:\[\mathbf{b} + \mathbf{c} = (-1 + 7)\mathbf{i} + (2 + 3)\mathbf{j} + (-4 - 4)\mathbf{k} = 6 \mathbf{i} + 5 \mathbf{j} - 8 \mathbf{k}\]Now find \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) \) using the cross product formula:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & 2 & -2 \6 & 5 & -8\end{vmatrix}\]Calculate as follows:\[\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{i}(2(-8) - (-2)(5)) - \mathbf{j}(-3(-8) - (-2)(6)) + \mathbf{k}(-3(5) - 2(6))\]\[= \mathbf{i}(-16 + 10) - \mathbf{j}(24 - 12) + \mathbf{k}(-15 - 12)\]\[= -6 \mathbf{i} - 12 \mathbf{j} - 27 \mathbf{k}\]

Using the dot product formula, \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \), calculate for \( \mathbf{a} \) and \( \mathbf{b} + \mathbf{c} = 6 \mathbf{i} + 5 \mathbf{j} - 8 \mathbf{k} \):\[\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = (-3)(6) + (2)(5) + (-2)(-8)\]\[= -18 + 10 + 16\]\[= 8\]

First, find \( \mathbf{b} \times \mathbf{c} \). Use the determinant formula for vectors, given \( \mathbf{b} = -\mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k} \) and \( \mathbf{c} = 7 \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k} \):\[\mathbf{b} \times \mathbf{c} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 2 & -4 \7 & 3 & -4\end{vmatrix}\]Calculate:\[\mathbf{b} \times \mathbf{c} = \mathbf{i}(2(-4) - (-4)(3)) - \mathbf{j}((-1)(-4) - (-4)(7)) + \mathbf{k}((-1)(3) - 2(7))\]\[= \mathbf{i}(-8 + 12) - \mathbf{j}(4 + 28) + \mathbf{k}(-3 - 14)\]\[= 4\mathbf{i} - 32\mathbf{j} - 17\mathbf{k}\]Now, calculate \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \):\[\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & 2 & -2 \4 & -32 & -17\end{vmatrix}\]Calculate:\[= \mathbf{i}(2(-17) - (-2)(-32)) - \mathbf{j}((-3)(-17) - (-2)(4)) + \mathbf{k}((-3)(-32) - 2(4))\]\[= \mathbf{i}(-34 - 64) - \mathbf{j}(51 - (-8)) + \mathbf{k}(96 - 8)\]\[= -98\mathbf{i} - 59\mathbf{j} + 88\mathbf{k}\]