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:
Answer & explanation
Correct answer: option 1
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 $\frac{\pi}{6}\,$.