In Exercises 21-24, compute the scalar triple product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \) \( 21 . \mathbf{u}=(-2,0,6), \mathbf{v}=(1,-3,1), \mathbf{w}=(-5,-1,1) \)

To compute the scalar triple product \( \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) \), we first need to calculate the cross product \( \mathbf{v} \times \mathbf{w} \), and then take the dot product of \( \mathbf{u} \) with that result. Let's calculate \( \mathbf{v} \times \mathbf{w} \): \[ \mathbf{v} = (1, -3, 1), \quad \mathbf{w} = (-5, -1, 1) \] The cross product \( \mathbf{v} \times \mathbf{w} \) is given by the determinant: \[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 1 \\ -5 & -1 & 1 \end{vmatrix} \] Calculating this determinant: \[ = \mathbf{i} \left((-3)(1) - (1)(-1)\right) - \mathbf{j} \left((1)(1) - (1)(-5)\right) + \mathbf{k} \left((1)(-1) - (-3)(-5)\right) \] This simplifies to: \[ = \mathbf{i} (-3 + 1) - \mathbf{j} (1 + 5) + \mathbf{k} (-1 - 15) \] \[ = \mathbf{i} (-2) - \mathbf{j} (6) + \mathbf{k} (-16) \] \[ = (-2, -6, -16) \] Now we take the dot product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \): \[ \mathbf{u} = (-2, 0, 6) \] Calculating the dot product: \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (-2)(-2) + (0)(-6) + (6)(-16) \] \[ = 4 + 0 - 96 \] \[ = 4 - 96 = -92 \] Thus, the scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) is \( \boxed{-92} \).