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

π/6

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

Given: $\vec{a}$ and $\vec{b}$ are unit vectors.

$\sqrt{3}\,\vec{a}-\vec{b}$ is also a unit vector.

So its magnitude is $1$:

$|\sqrt{3}\,\vec{a}-\vec{b}|^{2}=1$

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

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

Since $\vec{a}$ and $\vec{b}$ are unit vectors:

$= 3 - 2\sqrt{3}\cos\theta + 1$

Thus:

$3 + 1 - 2\sqrt{3}\cos\theta = 1$

$4 - 1 = 2\sqrt{3}\cos\theta$

$3 = 2\sqrt{3}\cos\theta$

$\cos\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$

So:

$\theta = 30^\circ$

The angle between $\vec{a}$ and $\vec{b}$ is $30^\circ$.