Question

If $|\overrightarrow{A}\times \overrightarrow{B}|=\frac{(\overrightarrow{A}\cdot \overrightarrow{B})}{\sqrt{3}}$ then the value of angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is:

• A. $0^0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (D)

$|\vec{A}\times \vec{B}|=\frac{(\vec{A}\cdot \vec{B})}{\sqrt{3}}$
$AB\sin \theta =\frac{AB\cos \theta }{\sqrt{3}}$
$\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}$
$\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\theta =\frac{\pi }{6}$