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

$\frac{5\pi}{6}$

The correct answer is Option (4) → $\frac{5\pi}{6}$

Let $\vec{a}$ and $\vec{b}$ be unit vectors, and $\sqrt{3}\vec{a} + \vec{b}$ is also a unit vector.

Then: $|\sqrt{3}\vec{a} + \vec{b}| = 1$

Square both sides: $|\sqrt{3}\vec{a} + \vec{b}|^2 = 1^2 = 1$

Compute dot product: $(\sqrt{3}\vec{a} + \vec{b}) \cdot (\sqrt{3}\vec{a} + \vec{b}) = 3|\vec{a}|^2 + 2\sqrt{3} (\vec{a} \cdot \vec{b}) + |\vec{b}|^2$

Since $|\vec{a}| = |\vec{b}| = 1$: $3 + 2\sqrt{3} (\vec{a} \cdot \vec{b}) + 1 = 4 + 2\sqrt{3} (\vec{a} \cdot \vec{b})$

Set equal to 1: $4 + 2\sqrt{3} (\vec{a} \cdot \vec{b}) = 1 \Rightarrow 2\sqrt{3} (\vec{a} \cdot \vec{b}) = -3 \Rightarrow \vec{a} \cdot \vec{b} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$

Thus, angle between $\vec{a}$ and $\vec{b}$: $\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = 150^\circ = \frac{5\pi}{6}$