Q. If $\overrightarrow{\mathbf{A}} \times \left( \overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}} \right) = \overrightarrow{\mathbf{B}} \times (\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{A}})$ and $\left\lbrack \overrightarrow{\mathbf{A}}\ \overrightarrow{\mathbf{B}}\ \overrightarrow{\mathbf{C}} \right\rbrack \neq 0,$ then $\overrightarrow{\mathbf{A}} \times (\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})$is equal to
Explanation
(a)
Given, $\overrightarrow{\mathbf{A}} \times \left( \overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}} \right) = \overrightarrow{\mathbf{B}} \times \left( \overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{A}} \right)\ \ldots.(i)$
Also, $\left\lbrack \overrightarrow{\mathbf{A}}\ \overrightarrow{\mathbf{B}}\ \overrightarrow{\mathbf{C}} \right\rbrack \neq 0ie.\overrightarrow{\mathbf{A}},\overrightarrow{\mathbf{B}},\overrightarrow{\mathbf{C}}\ \$are not coplanar.
$\therefore$From Eq. (i)
$\left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{C}} \right) - \left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{B}} \right)\overrightarrow{\mathbf{C}} = \left( \overrightarrow{\mathbf{B}} \bullet \overrightarrow{\mathbf{A}} \right)\overrightarrow{\mathbf{C}} - \left( \overrightarrow{\mathbf{B}} \bullet \overrightarrow{\mathbf{C}} \right)\overrightarrow{\mathbf{A}}$
$\Longrightarrow \ \left( \overrightarrow{\mathbf{B}} \bullet \overrightarrow{\mathbf{C}} \right)\overrightarrow{\mathbf{A}} + \left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{C}} \right)\overrightarrow{\mathbf{B}} - \left\lbrack \left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{B}} \right) + \left( \overrightarrow{\mathbf{B}} \bullet \overrightarrow{\mathbf{C}} \right) \right\rbrack\overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \ \overrightarrow{\mathbf{B}} \bullet \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{B}} = \overrightarrow{\mathbf{0}}$
$\left\lbrack \because\left\lbrack \overrightarrow{\mathbf{A}}\ \overrightarrow{\mathbf{B}}\ \overrightarrow{\mathbf{C}} \right\rbrack \neq 0 \right\rbrack$
Now, consider
$\overrightarrow{\mathbf{A}} \times \left( \overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}} \right) = \left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{C}} \right)\overrightarrow{\mathbf{B}} - \left( \overrightarrow{\mathbf{A}} \bullet \overrightarrow{\mathbf{B}} \right)\overrightarrow{\mathbf{C}}$
$= 0 \bullet \overrightarrow{\mathbf{B}} - 0 \bullet \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.