Question

If $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$, then which of the following options is correct?

• A. $\overrightarrow{a}\perp \overrightarrow{b}$
• B. $\overrightarrow{a}||\overrightarrow{b}$
• C. $\overrightarrow{a}\text{and}\overrightarrow{b}\text{are antiparallel}$
• D. $\overrightarrow{a}\text{and}\overrightarrow{b}\text{are inclined at angle of}{60}^{\circ }$

Answer 1

The correct option is A $\overrightarrow{a}\perp \overrightarrow{b}$

We have$|\overrightarrow{a}\times \overrightarrow{b}{|}^2=(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})$
$=\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}$
$=a^2+b^2+2\overrightarrow{a}.\overrightarrow{b}$
Similarly,
$|\overrightarrow{a}-\overrightarrow{b}{|}^2=(\overrightarrow{a}-\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})$
$=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}$
$\text{If}|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$
$a^2+b^2+2\overrightarrow{a}.\overrightarrow{b}=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}$
$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=0\Rightarrow \overrightarrow{a}\perp \overrightarrow{b}$