Question

The area of the parallelogram whose adjacent sides are $3\hat{i}+2\hat{j}+\hat{k}$ and $\hat{j}+2\hat{k}$ is

$\left(a\right)3\sqrt{6}\left(b\right)4$
$\left(c\right)\frac{3\sqrt{6}}{2}\left(d\right)54$

Answer 1

Required area is the magnitude of the cross product of the two vectors.
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 1\\ 0& 1& 2\end{array}\right|=3\hat{i}-6\hat{j}+3\hat{k}$
Required Area $=\sqrt{3^2+(-6{)}^2+3^2}$
$=\sqrt{54}=3\sqrt{6}$

Option $\left(a\right)$ is correct.