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 $(\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))$, then the value of $164{\cos }^2\theta$ is equal to :

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

Answer 1

Answer: (A)

Sol. ${\hat{a}}^{\wedge }\hat{b}=\frac{\pi }{4}=\varphi$
$\hat{a}\cdot \hat{b}=|\hat{a}||\hat{b}|\cos \varphi$
$\hat{a}\cdot \hat{b}=\cos \varphi =\frac{1}{\sqrt{2}}$
$\cos \theta =\frac{(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b})|}$
$|\hat{a}+\hat{b}{|}^2=(\hat{a}+\hat{b})\cdot (\hat{a}+\hat{b})$
$|\hat{a}+\hat{b}{|}^2=2+2\hat{a}\cdot \hat{b}$
$=2+\sqrt{2}$
$\hat{a}\times \hat{b}=|\hat{a}||\hat{b}|\sin \varphi \hat{n}$
$\hat{a}\times \hat{b}=\frac{\hat{n}}{\sqrt{2}}$
when $\hat{n}$ is vector $\perp \hat{a}$ and $\hat{b}$
let $\overrightarrow{c}=\hat{a}\times \hat{b}$
We know.
$\vec{c}\cdot \vec{a}=0$
$\vec{c}\cdot \vec{b}=0$
$|\hat{a}+2\hat{b}+2\vec{c}{|}^2$
$=1+4+\frac{(4)}{2}+4\hat{a}\cdot \hat{b}+8\hat{b}\cdot \vec{c}+4\vec{c}\cdot \hat{a}$
$=7+\frac{4}{\sqrt{2}}=7+2\sqrt{2}$
Now
$(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2\vec{c})$
$=|\hat{a}{|}^2+2\hat{a}\cdot \hat{b}+0+\hat{b}\cdot \hat{a}+2|\hat{b}{|}^2+0$
$=1+\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}+2$
$=3+\frac{3}{\sqrt{2}}$
$\cos \theta =\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$
${\cos }^2\theta =\frac{9(\sqrt{2}+1{)}^2}{2(2+\sqrt{2})(7+2\sqrt{2})}$
${\cos }^2\theta =\left(\frac{9}{2\sqrt{2}}\right)\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}$
$164{\cos }^2\theta =\frac{(82)(9)}{\sqrt{2}}\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}\frac{(7-2\sqrt{2})}{(7-2\sqrt{2})}$
$=\frac{(82)}{\sqrt{2}}\frac{(9)[7\sqrt{2}-4+7-2\sqrt{2}]}{(41)}$
$=(9\sqrt{2})[5\sqrt{2}+3]$
$=90+27\sqrt{2}$