Find the area of a parallelogram whose adjacent side are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.

$15\sqrt{2}$ sq. units

The correct answer is Option (2) → $15\sqrt{2}$ sq. units ##

The adjacent sides of the parallelogram are $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.

$ \vec{a} \times \vec{b} = (\hat{i} - \hat{j} + 3\hat{k}) \times (2\hat{i} - 7\hat{j} + \hat{k}) $

$ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & -1 & 3 \\2 & -7 & 1\end{vmatrix} $

$ = (-1 + 21)\hat{i} - (1 - 6)\hat{j} + (-7 + 2)\hat{k} $

$ = 20\hat{i} + 5\hat{j} - 5\hat{k} $

The area of the parallelogram is $|\vec{a} \times \vec{b}|$

$ = \sqrt{20^2 + 5^2 + 5^2} = 15\sqrt{2} \text{ sq. units} $