Find the cross product \( \vec{u} \times \vec{v} \) where \( \vec{u}=6 \mathbf{i}-7 \mathbf{j}-6 \mathrm{k} \) and \( \vec{v}=\langle 7,-3,8) \) \( \vec{u} \times \vec{v}= \)

To find the cross product \( \vec{u} \times \vec{v} \), we can use the determinant of a matrix composed of the unit vectors and the components of the vectors \( \vec{u} \) and \( \vec{v} \). We can write it as: \[ \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & -7 & -6 \\ 7 & -3 & 8 \end{vmatrix} \] Calculating the determinant: \[ \vec{u} \times \vec{v} = \mathbf{i} \begin{vmatrix} -7 & -6 \\ -3 & 8 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 6 & -6 \\ 7 & 8 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 6 & -7 \\ 7 & -3 \end{vmatrix} \] Calculating each of the 2x2 determinants: 1. For \( \mathbf{i} \): \[ (-7)(8) - (-6)(-3) = -56 - 18 = -74 \] 2. For \( \mathbf{j} \): \[ (6)(8) - (-6)(7) = 48 + 42 = 90 \] 3. For \( \mathbf{k} \): \[ (6)(-3) - (-7)(7) = -18 + 49 = 31 \] Putting it all together, we have: \[ \vec{u} \times \vec{v} = -74 \mathbf{i} - 90 \mathbf{j} + 31 \mathbf{k} \] Therefore, the final result is: \[ \vec{u} \times \vec{v} = -74 \mathbf{i} - 90 \mathbf{j} + 31 \mathbf{k} \]