Question

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors, maximum value of $\frac{{|\overline{a}+\overline{b}|}^2-{|\overline{a}-\overline{b}|}^2}{{|\overline{a}+\overline{b}|}^2+{|\overline{a}-\overline{b}|}^2}$ equal to

• A. $6$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Assum the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$

to be $\theta$.

Since $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vector , $|\overrightarrow{a}|=|\overrightarrow{b}|=1$

$|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

$=1+1+2+\left(1\right)\left(1\right)\cos \theta$

$2(1+\cos \theta )$

$|\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

$=1+1+2-\left(1\right)\left(1\right)\cos \theta$

$=2(1-\cos \theta )$

Then,

$\frac{|\overrightarrow{a}+\overrightarrow{b}{|}^2-|\overrightarrow{a}-\overrightarrow{b}{|}^2}{|\overrightarrow{a}+\overrightarrow{b}{|}^2+|\overrightarrow{a}+\overrightarrow{b}{|}^2}=\frac{2(1+\cos \theta )-2(1-\cos \theta )}{2(1+\cos \theta )+2(1-\cos \theta )}$

$=\frac{4\cos \theta }{4}=\cos \theta$ which is maximum at $\theta =0^0$

$\therefore$ Maximum value of given expression is $1$ at $\theta =0^0$, is $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel.