Q. If $\overrightarrow{\mathbf{a}}\$ is perpendicular to $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}\left| \overrightarrow{\mathbf{a}} \right| = 2,\ |\overrightarrow{\mathbf{b}}| = 3,\ |\overrightarrow{\mathbf{c}}| = 4$ and the angle between $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}\ is\frac{2\pi}{3}$, then $\lbrack\overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\rbrack$is equal to
Explanation
(c)
$\therefore\ \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}} \right\rbrack = \overrightarrow{\mathbf{a}} \bullet \left( \left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\sin\frac{2\pi}{3}\widehat{\mathbf{n}} \right)$
$= \left| \overrightarrow{\mathbf{a}} \right||\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|\left( \sin\frac{2\pi}{3} \right)$
$\left\lbrack \because\ \overrightarrow{\mathbf{a}} \bullet \widehat{\mathbf{n}} = \left| \overrightarrow{\mathbf{a}} \right|\widehat{\mathbf{n}} \middle| \cos{0{^\circ}} = \left| \overrightarrow{\mathbf{a}} \right| \right\rbrack$
$= 2 \times 3 \times 4 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.