If the diagonals of a parallelogram are represented by the vectors $3\hat i+\hat j-2\hat k$ and $\hat i +3\hat j-4\hat k$, then its area in square units, is

$\sqrt{42}$

Let $\vec a = 3\hat i+\hat j-2\hat k$ and $\vec b=\hat i +3\hat j-4\hat k$. Then,

$\vec a×\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\3&1&-2\\1&3&-4\end{vmatrix}=2\hat i+10\hat j+8\hat k$

$⇒|\vec a×\vec b|=\sqrt{4+100+64}=\sqrt{168}=2\sqrt{42}$

∴ Required area = $\frac{1}{2}|\vec a×\vec b|=\sqrt{42}$ sq. units