Question

Cosine of an angle between the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ if $|\overrightarrow{a}|=2,|\overrightarrow{b}|=1$ and $\overrightarrow{a}$ ^ $\overrightarrow{b}={60}^o$ is?

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

Answer 1

The correct option is D None

$\begin{array}{c}Letanglebetween\left(\overrightarrow{a}+\overrightarrow{b}\right)and\left(\overrightarrow{a}-\overrightarrow{b}\right)\\ are\left(\theta \right)\\ \left|a\right|=2and\left|b\right|=1\left(given\right)\\ \Rightarrow \frac{\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|.\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\cos \theta \\ \Rightarrow \frac{\overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}}{{\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2}=\cos \theta \\ \Rightarrow \frac{{\left|\overrightarrow{a}\right|}^2-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2}{{\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2}=\cos \theta \left[\because \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}\right]\\ \Rightarrow \frac{2^2-1^2}{2^2-1^2}=\cos \theta \\ 1=\cos \theta \\ \therefore Anglebetween\left(\overrightarrow{a}+\overrightarrow{b}\right)and\left(\overrightarrow{a}-\overrightarrow{b}\right)is{0}^{\circ }Ans.\\ \therefore \theta =0^{\circ }\end{array}$