1. **State the problem:** We need to compute the cross product $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = \langle 3, 0, -1 \rangle$ and $\mathbf{b} = \langle 1, 2, 2 \rangle$. 2. **Recall the formula for the cross product:** $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \mathbf{i}(a_2 b_3 - a_3 b_2) - \mathbf{j}(a_1 b_3 - a_3 b_1) + \mathbf{k}(a_1 b_2 - a_2 b_1)$$ 3. **Substitute the components:** $$a_1=3, a_2=0, a_3=-1, b_1=1, b_2=2, b_3=2$$ 4. **Calculate each component:** - $\mathbf{i}$ component: $0 \times 2 - (-1) \times 2 = 0 + 2 = 2$ - $\mathbf{j}$ component: $3 \times 2 - (-1) \times 1 = 6 + 1 = 7$ - $\mathbf{k}$ component: $3 \times 2 - 0 \times 1 = 6 - 0 = 6$ 5. **Apply the sign for the $\mathbf{j}$ component:** $$\mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 7\mathbf{j} + 6\mathbf{k}$$ 6. **Write the final vector:** $$\mathbf{a} \times \mathbf{b} = \langle 2, -7, 6 \rangle$$ This is the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$.