Q. Let the unit vectors $\overrightarrow{\mathbf{a}}and\ \overrightarrow{\mathbf{b}}$ be perpendicular to each other and the unit vector $\overrightarrow{\mathbf{c}}$ be inclined at an angle $\theta$ to both $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$. If $\overrightarrow{\mathbf{c}} = \alpha,\overrightarrow{\mathbf{a}} + \beta,\ \overrightarrow{\mathbf{b}} + \gamma\left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right),$ where $\alpha,\beta,\gamma$ are scalars, then

A) $\alpha = \cot{\theta,\ \beta = \sin{\theta,\ \gamma^{2} = \cos{2\theta}}}$
B) $\alpha = \cos{\theta,\ \beta = \cos{\theta,\ \gamma^{2} = \cos{2\theta}}}$
C) $\alpha = \cos{\theta,\ \beta = \sin{\theta,\ \gamma^{2} = \cos{2\theta}}}$
D) $\alpha = \sin{\theta,\ \beta = \cos{\theta,\ \gamma^{2} = \cos{2\theta}}}$
Explanation

(b)
Since, $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 0$
Also, $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = \cos\theta$
Now, $\overrightarrow{\mathbf{c}} = \alpha\overrightarrow{\mathbf{a}} + \beta\overrightarrow{\mathbf{b}} + \gamma\left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right)$
$\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{c}} = \alpha\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{a}} + \beta\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \gamma\overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} \right)$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} \right|\left| \overrightarrow{\mathbf{c}} \right|\cos{\theta = \alpha + 0 + 0 \Longrightarrow \cos{\theta = \alpha\ \ \ \ \ \lbrack\because\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 0\rbrack}}$
$and\ \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = \alpha\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} + \beta\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{b}} + \gamma(\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}) \bullet \overrightarrow{\mathbf{b}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\cos{\theta = \beta \Longrightarrow \cos{\theta = \beta\ \ \ \ \ }}\$

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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