Question

If $|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}|=|\overrightarrow{B}|$, then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

• A. ${120}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. $0^o$

Answer 1

The correct option is A ${120}^o$

It is given $|\overrightarrow{A}+\overrightarrow{B}|=\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|=k$

$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{{\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta }$

$k^2=k^2+k^2+{2k}^2\cos \theta$

$k^2={2k}^2+2k^2\cos \theta$

$-k^2=2k^2\cos \theta$

$\cos \theta =\frac{-1}{2}$

$\cos (180-\theta )=\frac{1}{2}$

$\theta ={120}^0$