Find the volume of the parallelepiped spanned by the vectors below. \( \begin{array}{l}\mathbf{u}=[2,8,-6] \\ \mathbf{v}=[-5,0,-2] \\ \mathbf{w}=[7,6,5] \\ \text { Volume of parallelepiped }=\text { Ex: } 5\end{array} \)

To find the volume of the parallelepiped spanned by the vectors \(\mathbf{u}=[2,8,-6]\), \(\mathbf{v}=[-5,0,-2]\), and \(\mathbf{w}=[7,6,5]\), you can use the scalar triple product formula. The volume \(V\) is given by the absolute value of the dot product of one vector with the cross product of the other two: \[ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \] First, calculate the cross product \(\mathbf{v} \times \mathbf{w}\): \[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 0 & -2 \\ 7 & 6 & 5 \end{vmatrix} = \mathbf{i}(0 \cdot 5 - (-2) \cdot 6) - \mathbf{j}(-5 \cdot 5 - (-2) \cdot 7) + \mathbf{k}(-5 \cdot 6 - 0 \cdot 7) \] Calculating this step-by-step: 1. For \(\mathbf{i}\): \(0 \cdot 5 - (-2) \cdot 6 = 0 + 12 = 12\) 2. For \(\mathbf{j}\): \(-5 \cdot 5 - (-2) \cdot 7 = -25 + 14 = -11\) (notice that we subtract this component) 3. For \(\mathbf{k}\): \(-5 \cdot 6 - 0 \cdot 7 = -30\) Thus, \(\mathbf{v} \times \mathbf{w} = [12, 11, -30]\). Now, calculate \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\): \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = [2, 8, -6] \cdot [12, 11, -30] = 2 \cdot 12 + 8 \cdot 11 - 6 \cdot 30 \] Calculating this: - \(2 \cdot 12 = 24\) - \(8 \cdot 11 = 88\) - \(-6 \cdot 30 = -180\) Thus: \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 24 + 88 - 180 = 112 - 180 = -68 \] Finally, the volume of the parallelepiped is: \[ V = |-68| = 68 \] So the volume of the parallelepiped is \(68\).