Let $O A, O B, O C$ be the co-terminal edges of a rectangular parallelopiped of volume $V$ and let $P$ be…

Let $O A, O B, O C$ be the co-terminal edges of a rectangular parallelopiped of volume $V$ and let $P$ be the vertex opposite to $O$. Then, $[\overrightarrow{\mathbf{A P}} \overrightarrow{\mathbf{B P}} \overrightarrow{\mathbf{C P}}]$ is equal to
  1. $2 \mathrm{~V}$
  2. $12 \mathrm{~V}$
  3. $3 \sqrt{3} \mathrm{~V}$
  4. 0

Solution

Here, $O A, O B, O C$ are the co-terminal edges of a rectangular parallelopiped of volume $V$.
Also, we know that the volume of rectangular parallelopiped $=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$ ie, $\quad V=[\overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}}]$ $\ldots$ (i) Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}$ then from figure $\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{BP}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{CP}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}$ $\because$ (By vector addition) Now, we find $[\overrightarrow{\mathrm{AP}} \overrightarrow{\mathrm{BP}} \overrightarrow{\mathrm{CP}}]=[(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})(\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}})]$ $=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}})]$ $=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}]$ $[\because \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}=0]$ $\left\{\begin{array}{l}\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}\end{array}\right.$ $=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+0+\overrightarrow{\mathbf{b}}]$ $=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}+\overrightarrow{\mathrm{b}}]$ $=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}+(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{b}}$ $=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}$ $=1+0-0-0+0+1$ $=2 \cdot 1$ $\because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}=1$ $=2\{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\}$ $=2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$ $=2[\overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}}]$ $=2 \mathrm{~V} \quad$ [from Eq. (i)] Hence, $[\overrightarrow{\mathrm{AP}} \overrightarrow{\mathrm{BP}} \overrightarrow{\mathrm{CP}}]=2 V$

Asked in: AP EAMCET 2010

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