Q 11-03-022JEE MainMedium
The diagonals of a parallelogram are $\vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k}$. The area of the parallelogram is
Answer: (C) $5\sqrt{3}$
For a parallelogram in terms of its diagonals, area $= \dfrac{1}{2}|\vec{d}_1 \times \vec{d}_2|$.
$$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = (4 - 6)\hat{i} - (12 + 2)\hat{j} + (-9 - 1)\hat{k} = -2\hat{i} - 14\hat{j} - 10\hat{k}$$
$$|\vec{d}_1 \times \vec{d}_2| = \sqrt{4 + 196 + 100} = 10\sqrt{3} \;\Rightarrow\; \text{Area} = 5\sqrt{3}$$
Solution by Sreeraj P, M.Sc Physics