Q. $\overrightarrow{\mathbf{a}} \bullet \lbrack\left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) \times \left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right)\rbrack$ equals

A) 0
B) $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}$
C) $\overrightarrow{\mathbf{a}}$
D) $\overrightarrow{\mathbf{a}} \bullet (\overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}})$
Explanation

(a)
$\overrightarrow{\mathbf{a}} \bullet \left\lbrack \left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) \times \left( \ \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) \right\rbrack$
$= \overrightarrow{\mathbf{a}} \bullet \lbrack\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}})\rbrack$
$= \overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} \right) + \overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}} \right)$
$= \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} \right\rbrack + \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{b}} \right\rbrack = 0\ \ \ \ \ \ \ \left\lbrack \because\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{b}} \right\rbrack = - \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} \right\rbrack \right\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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