Question

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

• A. $6$
• B. $4$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

Calculation.....

Here Two unit vectors is a and b.

Find the value of......$\frac{{\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2-{\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2}{{\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2-{\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2}$

Two equation,

$\Rightarrow {\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=a^2+b^2+2ab$

$\Rightarrow {\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2=a^2+b^2-2ab$

Now the vector is,

$\Rightarrow \frac{(a^2+b^2+2ab)-(a^2+b^2-2ab)}{(a^2+b^2+2ab)+(a^2+b^2-2ab)}$

$\Rightarrow \frac{a^2+b^2+2ab-a^2-b^2+2ab}{a^2+b^2+2ab+a^2+b^2-2ab}$

Then we have,

$=\frac{4ab}{2a^2+2b^2}$ , ($\because aandb$ are two unit vectors)

$=\frac{4}{2+2}=1$

So that the value is equal to 1.