Question

If $\left(\overrightarrow{A}+\overrightarrow{B}\right|=\left(\overrightarrow{A}-\overrightarrow{B}\right|$ then it can be said that
vector A and vector B are
to each other.

• A. perpendicular
• B. parallel
• C. $\mathrm{at}{45}^o\mathrm{to}\mathrm{each}\mathrm{other}$

Answer 1

The correct option is A perpendicular

Magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is calculated as

$\sqrt{A^2+B^2+2AB\cos \theta }$
similarly, magnitude of $\overrightarrow{A}-\overrightarrow{B}$ will be
= $\sqrt{A^2+{(-B)}^2+2A(-B)\cos \theta }$
Now $\left(\overrightarrow{A}+\overrightarrow{B}\right|=\left(\overrightarrow{A}-\overrightarrow{B}\right|$

$\Rightarrow \sqrt{A^2+B^2+2AB\cos \theta }=$

$\sqrt{A^2+{(-B)}^2+2A(-B)\cos \theta }$
$\Rightarrow A^2+B^2+2AB\cos \theta =A^2+B^2-2AB\cos \theta$
$\Rightarrow 4AB\cos \theta =0$
since A and B cannot be zero, so
$\cos \theta =0\theta ={90}^o$
hence A and B are perpendicular to each other.