Question

The area of parallelogram whose diagonals represent the vectors $3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$ and $\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}$ is

• A. $10\sqrt{3}$
• B. $5\sqrt{3}$
• C. $8$
• D. $4$

Answer 1

Answer: (B)

$5\sqrt{3}$

The area of a parallelogram is given by $\frac{1}{2}\times$ the cross product of its diagonals.

So, area $=\frac{1}{2}[(3\hat{i}+\hat{j}-2\hat{k})\times (\hat{i}-3\hat{j}+4\hat{k})]$

$=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& -3& 4\end{array}\right|$

$=\frac{1}{2}\left|\hat{i}\right(4-6)-\hat{j}(12+2)+\hat{k}(-9-1\left)\right|$

$=\frac{1}{2}\left|(-2\hat{i}-14\hat{j}-10\hat{k})\right|$

$=\sqrt{1+49+25}$

$=\sqrt{75}$