Find the magnitude of two vectors

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

It is given that $|\vec{a}|=|\vec{b}|, \vec{a} \cdot \vec{b}=\frac{1}{2}$, and $\theta=60^{\circ}$.              $\ldots(1)$

We know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.

$\therefore \frac{1}{2}=|\vec{a}||\vec{a}| \cos 60^{\circ}$                    [Using (1)]

$\Rightarrow \frac{1}{2}=|\vec{a}|^{2} \times \frac{1}{2}$

$\Rightarrow|\vec{a}|^{2}=1$

$\Rightarrow|\vec{a}|=|\vec{b}|=1$