Given the vectors below, find the scalar triple product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \) \( \begin{array}{l}\mathbf{u}=[0,-5,3] \\ \mathbf{v}=[0,-9,2] \\ \mathbf{w}=[6,-3,8] \\ \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=\text { Ex: } 5\end{array} \)

To find the scalar triple product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \), we first need to compute the cross product \( \mathbf{v} \times \mathbf{w} \). Let's calculate \( \mathbf{v} \times \mathbf{w} \): \[ \mathbf{v} = \begin{bmatrix} 0 \\ -9 \\ 2 \end{bmatrix}, \quad \mathbf{w} = \begin{bmatrix} 6 \\ -3 \\ 8 \end{bmatrix} \] Using the determinant of a matrix to find the cross product, we get: \[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -9 & 2 \\ 6 & -3 & 8 \end{vmatrix} \] Calculating this determinant, we find: \[ = \mathbf{i} \begin{vmatrix} -9 & 2 \\ -3 & 8 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & 2 \\ 6 & 8 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & -9 \\ 6 & -3 \end{vmatrix} \] Calculating each 2x2 determinant: 1. \(\begin{vmatrix} -9 & 2 \\ -3 & 8 \end{vmatrix} = (-9)(8) - (2)(-3) = -72 + 6 = -66\) 2. \(\begin{vmatrix} 0 & 2 \\ 6 & 8 \end{vmatrix} = (0)(8) - (2)(6) = 0 - 12 = -12\) 3. \(\begin{vmatrix} 0 & -9 \\ 6 & -3 \end{vmatrix} = (0)(-3) - (-9)(6) = 0 + 54 = 54\) Now, substituting back into the cross product: \[ \mathbf{v} \times \mathbf{w} = -66 \mathbf{i} + 12 \mathbf{j} + 54 \mathbf{k} = \begin{bmatrix} -66 \\ 12 \\ 54 \end{bmatrix} \] Next, compute \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \): \[ \mathbf{u} = \begin{bmatrix} 0 \\ -5 \\ 3 \end{bmatrix} \] Perform the dot product: \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{bmatrix} 0 \\ -5 \\ 3 \end{bmatrix} \cdot \begin{bmatrix} -66 \\ 12 \\ 54 \end{bmatrix} = (0)(-66) + (-5)(12) + (3)(54) \] Calculating this: \[ = 0 - 60 + 162 = 102 \] Therefore, the scalar triple product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \) is: \[ \boxed{102} \]