Let overrightarrow{PR}=3 hat{i}+hat{j}-2 hat{ | vedclass

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(C) In a parallelogram $PQRS$,the diagonals are $\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{PS}$ and $\overrightarrow{SQ} = \overrigh

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) In a parallelogram $PQRS$,the diagonals are $\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{PS}$ and $\overrightarrow{SQ} = \overrightarrow{PQ} - \overrightarrow{PS}$.Solving for $\overrightarrow{PQ}$ and $\overrightarrow{PS}$:$\overrightarrow{PQ} = \frac{\overrightarrow{PR} + \overrightarrow{SQ}}{2}$$\overrightarrow{PS} = \frac{\overrightarrow{PR} - \overrightarrow{SQ}}{2}$The volume $V$ of the parallelepiped determined by $\overrightarrow{PT}, \overrightarrow{PQ}, \overrightarrow{PS}$ is given by the scalar triple product $|[\overrightarrow{PT}, \overrightarrow{PQ}, \overrightarrow{PS}]|$.$V = |\overrightarrow{PT} \cdot (\overrightarrow{PQ} 	imes \overrightarrow{PS})|$$V = |\overrightarrow{PT} \cdot (\frac{\overrightarrow{PR} + \overrightarrow{SQ}}{2} 	imes \frac{\overrightarrow{PR} - \overrightarrow{SQ}}{2})|$$V = \frac{1}{4} |\overrightarrow{PT} \cdot (\overrightarrow{PR} 	imes \overrightarrow{PR} - \overrightarrow{PR} 	imes \overrightarrow{SQ} + \overrightarrow{SQ} 	imes \overrightarrow{PR} - \overrightarrow{SQ} 	imes \overrightarrow{SQ})|$Since $\overrightarrow{PR} 	imes \overrightarrow{PR} = 0$ and $\overrightarrow{SQ} 	imes \overrightarrow{SQ} = 0$,and $\overrightarrow{SQ} 	imes \overrightarrow{PR} = -(\overrightarrow{PR} 	imes \overrightarrow{SQ})$:$V = \frac{1}{4} |\overrightarrow{PT} \cdot (-2(\overrightarrow{PR} 	imes \overrightarrow{SQ}))| = \frac{1}{2} |[\overrightarrow{PT}, \overrightarrow{PR}, \overrightarrow{SQ}]|$$V = \frac{1}{2} |\begin{vmatrix} 1 & 2 & 3 \ 3 & 1 & -2 \ 1 & -3 & -4 \end{vmatrix}|$$V = \frac{1}{2} |1(-4 - 6) - 2(-12 + 2) + 3(-9 - 1)|$$V = \frac{1}{2} |-10 + 20 - 30| = \frac{1}{2} |-20| = 10$.

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