Find the magnitude of two vectors $\vec a$ and $\vec b$ having the same magnitude and such that the angle between them is ${60^\circ }$ and their scalar product is $\cfrac{1}{2}$.

We have, $|\vec a| = |\vec b|,\,\,\,\theta = {60^\circ }$ and $\vec a \cdot \vec b = \cfrac{1}{2}$

Now, $\cos \theta = \cfrac{{\vec a \cdot \vec b}}{{|\vec a||\vec b|}} \Rightarrow \cos {60^\circ } = \cfrac{{\cfrac{1}{2}}}{{|\vec a{|^2}}} \Rightarrow \cfrac{1}{2} = \cfrac{{\cfrac{1}{2}}}{{|\vec a{|^2}}}$
$\Rightarrow |\vec a{|^2} = 1 \Rightarrow |\vec a| = 1$ So, $|\vec b| = 1$