Question

If the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ and the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ are equal then the angle between $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{2}$

Projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ $=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}$, and projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ $=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|}$
$\therefore \frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|}$
$\therefore |\overrightarrow{b}|=|\overrightarrow{a}|$.
Now, angle between $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ be $\theta$.
Then $\cos \theta =\frac{(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}-\overrightarrow{b}|}$
or, $\cos \theta =\frac{|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2}{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}-\overrightarrow{b}|}$
or, $\cos \theta =0$ [Since, $|\overrightarrow{b}|=|\overrightarrow{a}|$]
or, $\theta =\frac{\pi }{2}$