Question

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

• A. 20
• B. 30
• C. 5
• D. 10

Answer 1

The correct option is D 10

$\overline{PR}=\overrightarrow{a}+\overrightarrow{b}=3\hat{i}+\hat{j}-2\hat{k}={\overline{d}}_1$
$\overline{SQ}=\overrightarrow{a}-\overrightarrow{b}=\hat{i}-3\hat{j}-4\hat{k}={\overline{d}}_2$

$\Rightarrow \overrightarrow{a}=2\hat{i}-\hat{j}-3\hat{k}$
$\Rightarrow \overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}$
$\therefore$ Volume of the required parallelepiped $=\left|\begin{array}{ccc}2& -1& -3\\ 1& 2& 1\\ 1& 2& 3\end{array}\right|$
$=|2(6-2)+1(3-1)-3(2-2)|$
= 10 cubic units
OR
Area of parallelogram $=\frac{1}{2}|\overline{d_1}\times {\overline{d}}_2|$
$\therefore$ Volume of parallelepiped $=\frac{1}{2}\Vert \begin{array}{ccc}3& 1& -2\\ 1& -3& -4\\ 1& 2& 3\end{array}\Vert =10\text{cubic units}$.