Q. If $|\ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{\ }\left| \mathbf{=} \right|\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}\ |$, then angle between $\ \overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}\$ is $(\ \overrightarrow{\mathbf{a}}\mathbf{\ \neq}\overrightarrow{\mathbf{0}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{\neq}\overrightarrow{\mathbf{0}})$

A) $\frac{\pi}{3}$
B) $\frac{\pi}{6}$
C) $\frac{\pi}{4}$
D) $\frac{\pi}{2}$
Explanation

(d)
$\because\ |\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{\ }\left| \mathbf{=} \right|\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}|$
$\Rightarrow \ |\ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\left| \ ^{2} = \right|\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}\ \left. \ \right|^{2}$
$\Rightarrow a^{2} + b^{2} + 2\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} = a^{2} + b^{2} - 2\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}}$
$\Rightarrow 4\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} = 0$
$\Rightarrow \ \overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} = 0$
$\therefore$ Angle between $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ is $\frac{\pi}{2}$.

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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