The teacher hasn't uploaded a solution for this question yet.
The area of a parallelogram whose diagonals are given by vectors $\vec{a}$ and $\vec{b}$ is given by half the magnitude of the cross product of the diagonals: $$Area = \frac{1}{2} |\vec{a} \times \vec{b}|$$
Given $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$, we compute the cross product as follows: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} $$ $$ = \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1)) $$ $$ = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) $$ $$ = -2\hat{i} + 3\hat{j} + 7\hat{k} $$
The magnitude of $\vec{a} \times \vec{b} = -2\hat{i} + 3\hat{j} + 7\hat{k}$ is: $$ |\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62} $$
The area of the parallelogram is half the magnitude of the cross product: $$ Area = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{62} $$
Final Answer: $\frac{\sqrt{62}}{2}$ square units
AI generated content. Review strictly for academic accuracy.