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

Given that $\vec{a}$ and $\vec{b}$ are unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.

Also given that $\sqrt{3}\vec{a} - \vec{b}$ is a unit vector.

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Since $\sqrt{3}\vec{a} - \vec{b}$ is a unit vector:

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

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

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Expanding the left side using the dot product formula:

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

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

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

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

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Setting this equal to 1:

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

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

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

$= \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$= \frac{3\sqrt{3}}{6}$

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

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Using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$:

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

$= \frac{\sqrt{3}/2}{1 \times 1}$

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

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Since $\cos\theta = \frac{\sqrt{3}}{2}$, the angle $\theta = \frac{\pi}{6}$.

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