Find the area of the parallelogram whose one side and a diagonal are represented by coinitial vectors $\hat{i} - \hat{j} + \hat{k}$ and $4\hat{i} + 5\hat{k}$ respectively.

$\sqrt{42}$ sq. units

The correct answer is Option (1) → $\sqrt{42}$ sq. units ##

Let $\vec{a} = \hat{i} - \hat{j} + \hat{k}$

$\vec{d} = 4\hat{i} + 5\hat{k}$

$∵\vec{a} + \vec{b} = \vec{d}$

$∴\vec{b} = \vec{d} - \vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 1 & 4 \end{vmatrix} = -5\hat{i} - 1\hat{j} + 4\hat{k}$

Area of parallelogram $= |\vec{a} \times \vec{b}| = \sqrt{25 + 1 + 16} = \sqrt{42}$ sq. units.