91Ó°ÊÓ

We will use the following formula for the cross product of two vectors: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix} \] where \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\). Let's use this formula to find \(u \times v\) and \(u \times w\).

Using the formula, we set up the determinant for the cross product of \(u\) and \(v\): \[ u \times v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 4 \\ 3 & 1 & -2 \\ \end{vmatrix} \] Now, we compute the determinant: \[ u \times v = (\mathbf{i}(-3)(-2) + \mathbf{j}(4)(3) + \mathbf{k}(2)(1)) - (\mathbf{k}(-3)(3) + \mathbf{i}(4)(1) + \mathbf{j}(2)(-2)) \] Simplifying and combining terms, we have: \[ u \times v = (6\mathbf{i} + 12\mathbf{j} + 2\mathbf{k}) - (-9\mathbf{k} + 4\mathbf{i} - 4\mathbf{j}) \] Finally, we get: \[ u \times v = 2\mathbf{i} + 16\mathbf{j} + 11\mathbf{k} \]

Using the formula, we set up the determinant for the cross product of \(u\) and \(w\): \[ u \times w = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 4 \\ 1 & 5 & 3 \\ \end{vmatrix} \] Now, we compute the determinant: \[ u \times w = (\mathbf{i}(-3)(3) + \mathbf{j}(4)(1) + \mathbf{k}(2)(5)) - (\mathbf{k}(-3)(1) + \mathbf{i}(4)(5) + \mathbf{j}(2)(3)) \] Simplifying and combining terms, we have: \[ u \times w = (-9\mathbf{i} + 4\mathbf{j} + 10\mathbf{k}) - (-3\mathbf{k} + 20\mathbf{i} - 6\mathbf{j}) \] Finally, we get: \[ u \times w = -29\mathbf{i} + 10\mathbf{j} + 7\mathbf{k} \] In conclusion, \((a) \: u \times v = 2\mathbf{i} + 16\mathbf{j} + 11\mathbf{k}\) and \((b) \: u \times w = -29\mathbf{i} + 10\mathbf{j} + 7\mathbf{k}\).