1. **Problem:** Find the cross product $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{b} = 3\mathbf{i} - \mathbf{k}$. 2. **Formula:** The cross product of two vectors $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$ is given by the determinant: $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$ 3. **Step-by-step calculation:** - Identify components: $a_1=2$, $a_2=1$, $a_3=0$ (since $\mathbf{a}$ has no $\mathbf{k}$ component). - $b_1=3$, $b_2=0$, $b_3=-1$. - Compute determinant: $$\mathbf{a} \times \mathbf{b} = \mathbf{i}(a_2 b_3 - a_3 b_2) - \mathbf{j}(a_1 b_3 - a_3 b_1) + \mathbf{k}(a_1 b_2 - a_2 b_1)$$ $$= \mathbf{i}(1 \times (-1) - 0 \times 0) - \mathbf{j}(2 \times (-1) - 0 \times 3) + \mathbf{k}(2 \times 0 - 1 \times 3)$$ $$= \mathbf{i}(-1) - \mathbf{j}(-2) + \mathbf{k}(-3)$$ $$= -\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$$ 4. **Answer:** The cross product is $-\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$. 5. **Explanation:** The cross product results in a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. The determinant method helps calculate each component systematically. **Final answer:** (a) $-\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$