Q. Let $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ and$\ \overrightarrow{\mathbf{c}} = \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$. Then the vectors $\ \overrightarrow{\mathbf{b}}$ satisfying $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$ and $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 3$ is

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

(a)
$\overrightarrow{\mathbf{a}} \times \ \overrightarrow{\mathbf{c}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
0 & 1 & - 1 \\
1 & - 1 & - 1
\end{matrix} \right|$
$= \widehat{\dot{\mathbf{i}}}( - 1 - 1) - \widehat{\dot{\mathbf{j}}}(0 + 1) + \widehat{\mathbf{k}}(0 - 1)$
$= - 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
Given,
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} \times \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right)\overrightarrow{\mathbf{a}} - \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{a}} \right)\overrightarrow{\mathbf{b}} = - \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}$
$\Longrightarrow 3\overrightarrow{\mathbf{a}} - 2\overrightarrow{\mathbf{b}} = - \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}$
$\Longrightarrow \overrightarrow{\mathbf{b}} = \frac{3\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}}{2}$
$\Longrightarrow \ \overrightarrow{\mathbf{b}} = \frac{3\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}} - 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}}{2}$
$= \frac{- 2\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 4\widehat{\mathbf{k}}}{2} = - \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}$

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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