Question

If the adjacent sides of a parallelogram are $\hat{i}+2\hat{j}+3\hat{k}$, and $-3\hat{i}-2\hat{j}+\hat{k}$, then the area of the parallelogram is

• A. $6\sqrt{5}$
• B. $7\sqrt{5}$
• C. $8\sqrt{5}$
• D. $5\sqrt{7}$

Answer 1

Answer: (A)

$6\sqrt{5}$

$\overrightarrow{n}=(\hat{i}+2\hat{j}+3\hat{k})\times (-3\hat{i}-2\hat{j}+\hat{k})$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -3& -2& 1\end{array}\right|=8\hat{i}-10\hat{j}+4\hat{k}$

$\therefore$ Area $=|\overrightarrow{n}|=6\sqrt{5}$