If $\vec a.\vec b$ and $\sqrt{3}\vec a-\vec b$ are three unit vectors, then the angle between $\vec a$ and $\vec b$ is:

π/6

The correct answer is Option (1) → π/6

$|a|=|b|=1$

Given $|\sqrt{3}\,a-b|=1$

$ (\sqrt{3}\,a-b)\cdot(\sqrt{3}\,a-b)=1$

$3|a|^{2}-2\sqrt{3}\,(a\cdot b)+|b|^{2}=1$

$3-2\sqrt{3}\,(a\cdot b)+1=1$

$4-2\sqrt{3}\,(a\cdot b)=1$

$-2\sqrt{3}\,(a\cdot b)=-3$

$a\cdot b=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$

$\cos\theta=\frac{\sqrt{3}}{2}\;$ so $\;\theta=\frac{\pi}{6}$

The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}\,$.