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 $\overrightarrow{a}\wedge \overrightarrow{b}={60}^o$ is

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

Answer 1

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

According to question,

the vector are....$\overrightarrow{a}+\overrightarrow{b}and\overrightarrow{a}-\overrightarrow{b}$

$\begin{array}{c}\left|\overrightarrow{a}\right|=2\\ \left|\overrightarrow{b}\right|=1,and(\overrightarrow{a},\overrightarrow{b})={60}^0\\ (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})=\overrightarrow{a}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}\\ ={\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2{[\cos \theta =\cos 60}^0=\frac{1}{2}\\ =4-1=3\\ if,\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{{\left|a\right|}^2+{\left|b\right|}^2+2\left|a\right|\left|b\right|}\cos \theta \\ =\sqrt{4+1+2.2.\frac{1}{2}}=\sqrt{7}\\ \therefore \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{7},\left|\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{3}\\ \\ \Rightarrow \left|\overrightarrow{a}+\overrightarrow{b}\right|\left|\overrightarrow{a}-\overrightarrow{b}\right|\cos \theta =3\\ \sqrt{7}.\sqrt{3}\cos \theta =3\\ \therefore \cos \theta =\sqrt{\frac{3}{7}}\\ so,thatthecorrectoptionisA.\end{array}$