Question

If $\overrightarrow{a}\text{and}\overrightarrow{b}$ are two vectors perpendicular to each other and $|\overrightarrow{a}|=\overrightarrow{b}|=1$, determine the angle between $(\overrightarrow{a}+\overrightarrow{b})\text{and}(2\overrightarrow{a}+\overrightarrow{b})$

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

Answer 1

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

Given $|\overrightarrow{a}|=|\overrightarrow{b}|=1$
$\cos \theta =\frac{(\overrightarrow{a}+\overrightarrow{b}).(2\overrightarrow{a}+\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}||2\overrightarrow{a}+\overrightarrow{b}|}$
Since $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, therefore $\overrightarrow{a}.\overrightarrow{b}=0$
$\Rightarrow \cos \theta =\frac{2a^2+b^2}{\sqrt{a^2+b^2+2ab\cos 90}\sqrt{(2a{)}^2+b^2+2(2a)b\cos 90}}$
$=\frac{2a^2+b^2}{\sqrt{2}\sqrt{5}}=\frac{3}{\sqrt{10}}$
$\theta ={\cos }^{-1}\left(\frac{3}{\sqrt{10}}\right)$