Find the angle between two vectors $\vec a$ and $\vec b$ with magnitudes 1 and 2 respectively and when $\vec a.\vec b=1$.

$\frac{\pi}{3}$

The correct answer is Option (1) → $\frac{\pi}{3}$ ##

Given $\vec a\cdot \vec b=1,|\vec a|=1$ and $|\vec b|=2$.

We have,

$\theta=\cos^{-1}\left(\frac{\vec a\cdot \vec b}{|\vec a||\vec b|}\right)=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$