Question

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

• A. $0^{\circ }$
• B. ${90}^{\circ }$
• C. ${\tan }^{-1}\left(\frac{A}{B}\right)$
• D. ${\tan }^{-1}\left(\frac{A-B}{A+B}\right)$

Answer 1

The correct option is B ${90}^{\circ }$

Let $\overrightarrow{R_1}$ and $\overrightarrow{R_2}$ be the resultant vectors of $(\overrightarrow{A}+\overrightarrow{B})$ and $(\overrightarrow{A}-\overrightarrow{B})$ respectively and $\theta$ be the angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$
Then we have,
$|\overrightarrow{R_1}|=\sqrt{A^2+B^2+2AB\cos \theta }$
and $|\overrightarrow{R_2}|=\sqrt{A^2+B^2+2AB\cos (180-\theta )}$
Since, $|\overrightarrow{R_1}|$ will be equal to $|\overrightarrow{R_2}|$
$\Rightarrow \sqrt{A^2+B^2+2AB\cos \theta }=\sqrt{A^2+B^2+2AB\cos (180-\theta )}$
$\Rightarrow 2AB\cos \theta =-2AB\cos \theta \Rightarrow 4AB\cos \theta =0$
$\Rightarrow \cos \theta =0\Rightarrow \theta ={90}^{\circ }$