Question

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

Answer 1

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

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

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

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

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

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

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