Question

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi }{4}$. If $\theta$ is the angle between the vectors $\hat{a}+\hat{b}$ and $(\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}\left)\right)$, then the value of $164{\cos }^2\theta$ is equal to :

• A. $45+18\sqrt{2}$
• B. $90+27\sqrt{2}$
• C. $54+90\sqrt{2}$
• D. $90+3\sqrt{2}$

Answer 1

The correct option is B $90+27\sqrt{2}$

$\hat{a}\cdot \hat{b}=\frac{1}{\sqrt{2}}$ and $|\overrightarrow{a}\times \overrightarrow{b}|=\frac{1}{\sqrt{2}}$

$\frac{(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}\left)\right)}{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}\left)\right|}=\cos \theta$

$\Rightarrow \cos \theta =\frac{1+3\hat{a}\hat{b}+2}{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}\left)\right|}$

${|\hat{a}+\hat{b}|}^2=2+\sqrt{2}$

${|\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}\left)\right|}^2=1+4+4|\hat{a}\times \hat{b}{|}^2+4\hat{a}\hat{b}$

$=5+4\cdot \frac{1}{2}+\frac{4}{\sqrt{2}}=7+2\sqrt{2}$

$\text{So},{\cos }^2\theta =\frac{{(3+\frac{3}{\sqrt{2}})}^2}{(2+\sqrt{2})(7+2\sqrt{2})}=\frac{9\sqrt{2}(5\sqrt{2}+3)}{164}$

$\Rightarrow 164{\cos }^2\theta =90+27\sqrt{2}$