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:

Given that $\vec{a}$, $\vec{b}$, and $\sqrt{3}\vec{a} + \vec{b}$ are unit vectors:

$|\vec{a}| = 1$

$|\vec{b}| = 1$

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

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

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

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Expanding the magnitude:

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

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

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Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$:

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

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

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

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Setting equal to 1:

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

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

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

$\vec{a} \cdot \vec{b} = \dfrac{-3}{2\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\vec{a} \cdot \vec{b} = \dfrac{-3\sqrt{3}}{6}$

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

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

$\dfrac{-\sqrt{3}}{2} = (1)(1)\cos\theta$

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

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For angles between vectors in the range $[0, \pi]$:

$\theta = \dfrac{5\pi}{6}$