Question

Cosine of an angle between the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ if $\left|\overrightarrow{a}\right|=2,\left|\overrightarrow{b}\right|=1$ and $\overline{a}\wedge \overrightarrow{b}={60}^{\circ }$ is:

• A. $\sqrt{3/7}$
• B. $9/\sqrt{21}$
• C. $3/\sqrt{7}$
• D. none

Answer 1

The correct option is A $\sqrt{3/7}$

$cos\theta =\frac{(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}|.|\overrightarrow{a}-\overrightarrow{b}|}=\frac{a^2-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{b}-b^2}{\sqrt{(a^2+b^2+2ab.cos\alpha )}.\sqrt{(a^2+b^2-2ab.cos\alpha }}$

Which gives, $cos\theta =\frac{4-1}{\sqrt{4+1+\mathrm{2.2.1.}cos60^0}.\sqrt{4+1-\mathrm{2.2.1.}cos60^0}}=\frac{3}{\sqrt{7}.\sqrt{3}}=\sqrt{\frac{3}{7}}$