Question

Area of a parallelogram, whose diagonals are $3\hat{i}+\hat{j}-2\hat{k}and\hat{i}-3\hat{j}+4\hat{k}$ will be:

• A. $14unit$
• B. $5\sqrt{3}unit$
• C. $10\sqrt{3}unit$
• D. $20\sqrt{3}unit$

Answer 1

The correct option is B $5\sqrt{3}unit$

Area = $\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$ when diagonal vectors $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ are given
Hence we need to find the cross product of $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$
$\overrightarrow{d_1}\times \overrightarrow{d_2}=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& -3& 4\end{array}\right|$
=$\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)$
=$-2\hat{i}-14\hat{j}-10\hat{k}$
Area$=\frac{1}{2}\sqrt{4+196+100}=\frac{1}{2}\left(\sqrt{300}\right)=5\sqrt{3}$