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

10

In Δ's PQR and PSQ, we have

$\vec{PQ}+\vec{QR} =\vec{PR}$ and, $\vec{PS}+\vec{SQ} =\vec{PQ}$

$⇒\vec{PQ}+\vec{QR} =\vec{PR}$ and $\vec{PQ}-\vec{PS}=\vec{SQ}$

$⇒\vec{PQ}+\vec{PS}=\vec{PR}$ and $\vec{PQ}-\vec{PS}=\vec{SQ}$

$⇒\vec{PQ}=\frac{\vec{PR}+\vec{SQ}}{2}$ and $\vec{PS}=\vec{PS}=\frac{\vec{PR}-\vec{SQ}}{2}$

$⇒\vec{PQ}=2\hat i-\hat j-3\hat k$ and $\vec{PS}=\hat i+2\hat j+\hat k$

Thus, we obtain $\vec{PQ}=2\hat i-\hat j-3\hat k, \vec{PS}=\hat i+2\hat j+\hat k$ and $\vec{PT}=\hat i+2\hat j+3\hat k$

$[\vec{PT}\, \vec{PQ}\,\vec{PS}]=\begin{vmatrix}1&2&3\\2&-1&-3\\1&2&1\end{vmatrix}=5-10+15=10$

Hence, volume of the parallelopiped determined by the vectors $\vec{PT}, \vec{PQ}$ and $\vec{PS}$ is 10 cubic units.