Q. If the magnitudes of scalar and vector products of two vectors and are 6 and $6\sqrt{3}$ respectively, then the angle between two vectors is
A) 15${^\circ}$
B) 30${^\circ}$
C) 60${^\circ}$
D) 75${^\circ}$
Explanation
(c) $\overrightarrow{A} \bullet \overrightarrow{B} = ABcos\theta = 6$
and $\left| \overrightarrow{A} \times \overrightarrow{B} \right| = ABsin\theta = 6\sqrt{3}$
$\therefore\ \frac{ABsin\theta}{ABcos\theta} = \frac{6\sqrt{3}}{6} = \sqrt{3}$
or tan$\theta = \sqrt{3}$
and $\theta = 60{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.