Question

Let $\overrightarrow{PR}=3\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{SQ}=\hat{i}-3\hat{j}-4\hat{k}$ determine diagonals of a parallelogram $PQRS$ and $\overrightarrow{PT}=\hat{i}+2\hat{j}+3\hat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{PT},\overrightarrow{PQ}$ and $\overrightarrow{PS}$ is

• A. $5$
• B. $20$
• C. $10$
• D. $30$

Answer 1

The correct option is C $10$

Area of base (PQRS)$=\frac{1}{2}|\overrightarrow{PR}\times \overrightarrow{SQ}|=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -2\\ 1& -3& -4\end{array}\right|$
$=\frac{1}{2}|-10\hat{i}+10\hat{j}-10\hat{k}|=5|\hat{i}-\hat{j}+\hat{k}|=5\sqrt{3}$

Height$=$ proj. of $PT$ on $\hat{i}-\hat{j}+\hat{k}=\left|\frac{1-2+3}{\sqrt{3}}\right|=\frac{2}{\sqrt{3}}$

Volume $=(5\sqrt{3})\left(\frac{2}{\sqrt{3}}\right)=10$ cu. units