Question

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

• A. $14$ unit
• 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

$\overrightarrow{d_1}=3\hat{i}+\hat{j}-2\hat{k}\overrightarrow{d_2}=\hat{i}-3\hat{j}+4\hat{k}$

Area of parallelogram $=\frac{1}{2}\Vert \overrightarrow{d_1}\times \overrightarrow{d_2}\Vert =\frac{1}{2}\times \Vert (-9\hat{k}-12\hat{j}+4\hat{i}-\hat{k}-2\hat{j}-6\hat{i})\Vert =\frac{1}{2}\times \Vert (-2\hat{i}-14\hat{j}-10\hat{k})\Vert =\Vert (-\hat{i}-7\hat{j}-5\hat{k})\Vert =\sqrt{75}=5\sqrt{3}$