Question

Find the area of the parallelogram whose adjacent side are $\overrightarrow{a}=3\hat{i}+4\hat{j}+3\hat{k}$, $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}.$

• A. $2\sqrt{3}$ sq. units.
• B. $5\sqrt{3}$ sq. units.
• C. $6\sqrt{3}$ sq. units.
• D. $8\sqrt{3}$ sq. units.

Answer 1

The correct option is D $8\sqrt{3}$ sq. units.

Area of parallelogram $=\left|(a\times b)\right|$

But $a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& -2& 1\end{array}\right|=8(i+j+k)$

$\therefore |a\times b|$

$=8\sqrt{1+1+1}$

$=8\sqrt{3}$ sq. units.