Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors perpendicular to each other and $|\overrightarrow{a}|=|\overrightarrow{b}|=1$ . Determine the angle between $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{2a}+\overrightarrow{b}$

• A. $\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{3}{2\sqrt{5}}\right)$
• C. $\theta ={\cos }^{-1}\left(\frac{3}{\sqrt{10}}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{2}{3\sqrt{5}}\right)$

Answer 1

The correct option is C $\theta ={\cos }^{-1}\left(\frac{3}{\sqrt{10}}\right)$

Given $|\overrightarrow{a}|=|\overrightarrow{b}|=1$
$\cos \theta =\frac{(\overrightarrow{a}+\overrightarrow{b}).(2\overrightarrow{a}+\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{2a}+\overrightarrow{b}|}$
$=\frac{2\overrightarrow{a^2}+\overrightarrow{b^2}}{\sqrt{a^2+b^2+2ab\cos {90}^o}\sqrt{(2a{)}^2+b^2+2(2a)b\cos {90}^o}}$
$\frac{2\overrightarrow{a^2}+\overrightarrow{b^2}}{\sqrt{2}\sqrt{5}}=\frac{3}{\sqrt{10}}$
$\theta ={\cos }^{-1}\left(\frac{3}{\sqrt{10}}\right)$