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Q. 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:

Motion in a Plane

Solution:

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