Question:
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 $\frac{1}{2}$.
Solution:
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)]
$\Rightarrow \frac{1}{2}=|\vec{a}|^{2} \times \frac{1}{2}$
$\Rightarrow|\vec{a}|^{2}=1$
$\Rightarrow|\vec{a}|=|\vec{b}|=1$