Q. If $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}},$ $\ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}},$ $\ \overrightarrow{\mathbf{c}} = 3\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}},$ then a vector perpendicular to $\overrightarrow{\mathbf{a}}$ and in the plane containing $\overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}$ is

A) $- 17\widehat{\dot{\mathbf{i}}} + 21\widehat{\dot{\mathbf{j}}} - 97\widehat{\mathbf{k}}$
B) $17\widehat{\dot{\mathbf{i}}} + 21\widehat{\dot{\mathbf{j}}} - 123\widehat{\mathbf{k}}$
C) $- 17\widehat{\dot{\mathbf{i}}} - 21\widehat{\dot{\mathbf{j}}} + 97\widehat{\mathbf{k}}$
D) $- 17\widehat{\dot{\mathbf{i}}} - 21\widehat{\dot{\mathbf{j}}} - 97\widehat{\mathbf{k}}$
Explanation

(d)
$\because\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & 2 & - 5 \\
3 & 5 & - 1
\end{matrix} \right| = 23\widehat{\dot{\mathbf{i}}} - 14\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}\$
$\therefore\overrightarrow{\mathbf{a}} \times \left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} \right) = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
2 & 3 & - 1 \\
23 & - 14 & - 1
\end{matrix} \right| = - 17\widehat{\dot{\mathbf{i}}} - 21\widehat{\dot{\mathbf{j}}} - 97\widehat{\mathbf{k}}\$

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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