The two co-initial adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k}$ and $2\hat{i} + 2\hat{j} + 3\hat{k}$. Find its diagonals and use them to find the area of the parallelogram.

$2\sqrt{101}$ sq. units

The correct answer is Option (1) → $2\sqrt{101}$ sq. units ##

$\vec{d_1} = \vec{a} + \vec{b} = 4\hat{i} - 2\hat{j} - 2\hat{k}$, $\vec{d_2} = \vec{a} - \vec{b} = -6\hat{j} - 8\hat{k}$

Area of the parallelogram $= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$

$ = \frac{1}{2} \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\4 & -2 & -2 \\0 & -6 & -8\end{vmatrix} = |2\hat{i} + 8\hat{j} - 6\hat{k}|$

Area of the parallelogram $= 2\sqrt{101}$ sq. units.