Question

The vectors from origin to the points A and B are $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k}and\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$ respectively,then the area of $△OAB$ is equal to

(a) 340 (b) $\sqrt{25}$ (c) $\sqrt{229}$ (d) $\frac{1}{2}\sqrt{229}$

Answer 1

(d) $Areaof△OAB=\frac{1}{2}|\overrightarrow{OA}\times \overrightarrow{OB}|=\frac{1}{2}|(2\hat{i}-3\hat{j}+2\hat{k})\times (2\hat{i}+3\hat{j}+\hat{k})|=\frac{1}{2}\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 2& -3& 2\\ 2& 3& 1\end{array}\right|=\frac{1}{2}|\left[\hat{i}\right(-3-6)-\hat{j}(2-4)+\hat{k}(6+6\left)\right]|=\frac{1}{2}|-\hat{i}+2\hat{j}+12\hat{k}|\therefore Areaof△OAB=\frac{1}{2}\sqrt{(81+4+144)}=\frac{1}{2}\sqrt{229}$