Question

If $|\overrightarrow{A}-\overrightarrow{B}|=\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|$, the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

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

Answer 1

The correct option is A ${60}^o$

$|\overrightarrow{A}-\overrightarrow{B}|=|\overrightarrow{A}|=|\overrightarrow{B}|$

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

$|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2-2|\overrightarrow{A}||\overrightarrow{B}|\cos \theta =|\overrightarrow{A}{|}^2$

$2|\overrightarrow{A}||\overrightarrow{B}|\cos \theta =|\overrightarrow{B}{|}^2$

$\cos \theta =\frac{|\overrightarrow{B}|}{2|\overrightarrow{A}|}---\left(i\right)$

$|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2-2|\overrightarrow{A}||\overrightarrow{B}|\cos \theta =|\overrightarrow{B}{|}^2$

$\cos \theta =\frac{|\overrightarrow{A}|}{2|\overrightarrow{B}|}---\left(ii\right)$

Multiply both equation

${\cos }^2\theta =\frac{|\overrightarrow{B}|}{2|\overrightarrow{A}|}\times \frac{|\overrightarrow{A}|}{2|\overrightarrow{B}|}$

${\cos }^2\theta =\frac{1}{4}$

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

$\theta ={60}^o$