Find the area of a parallelogram whose adjacent sides are represented by the vectors $\overrightarrow{a} = \hat{i} -\hat{j} + 3\hat{k}$, $\overrightarrow{b} = 2\hat{i} -7\hat{j} + \hat{k}$.

The correct answer is $15\sqrt{2}$

Given $\overrightarrow{a} = \hat{i} -\hat{j} + 3\hat{k}$, $\overrightarrow{b} = 2\hat{i} -7\hat{j} + \hat{k}$

$\overrightarrow{a}$ $\times$ $\overrightarrow{b}$ =
$\left| \begin{matrix}


\hat{i} & \phantom{-}\hat{j} & \phantom{-}\hat{k} \\


1 & \phantom{-}-1 & \phantom{-}3\\


2 & \phantom{-}-7 & \phantom{-}1\\


\end{matrix} \right|$ = $20\hat{i} - 5\hat{j} -5\hat{k}$ = $5(4\hat{i} - \hat{j} -\hat{k})$


Area of parallelogram = $|\overrightarrow{a}$ $\times$ $\overrightarrow{b}|$ = $5\sqrt{16 +1 +1 }$ = $5\sqrt{18}$ = $15\sqrt{2}$ sq. units