Q. Which one of the following vectors is of magnitude 6 and perpendicular to both $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ and$\overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$?

A) $2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}$
B) $2(2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})$
C) $3(2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})$
D) $2(2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})$
Explanation

(d)
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
2 & 2 & 1 \\
1 & - 2 & 2
\end{matrix} \right|$
$= \widehat{\dot{\mathbf{i}}}(4 + 2) - \widehat{\dot{\mathbf{j}}}(4 - 1) + \widehat{\mathbf{k}}( - 4 - 2)$
$= 6\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right| = \sqrt{36 + 9 + 36} = \sqrt{81} = 9$
$\therefore\$Required vectors are
$\pm 6\left| \frac{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|} \right|$
$= \pm \frac{6}{9}\left( 6\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}} \right)$
$= \pm 2(2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})$

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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