Q. Let $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ be non-zero vectors such that $\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \times \overrightarrow{\mathbf{c}} = - \frac{1}{4}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|\overrightarrow{\mathbf{a}}.$ If θ is the acute angle between vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$, then the angle between $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{c}}$ is equal to
Explanation
(d)
$Given,\ \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \times \overrightarrow{\mathbf{c}} = - \frac{1}{4}\left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\overrightarrow{\mathbf{a}}$
$\Longrightarrow \left( \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} \right)\overrightarrow{\mathbf{b}} - \left( \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{b}} \right)\overrightarrow{\mathbf{a}} = - \frac{1}{4}\left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\overrightarrow{\mathbf{a}}$
On comparing both sides, we get
$\left( \overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} \right)\overrightarrow{\mathbf{b}} = 0$
$\left| \overrightarrow{\mathbf{c}} \right|\overrightarrow{\mathbf{a}}|\cos{\theta = 0}$
$\Longrightarrow \ \theta = \frac{\pi}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.