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}=\phi$
$\hat{a} \cdot \hat{b}=|\hat{a}||\hat{ b }| \cos \phi$
$\hat{a} \cdot \hat{b}=\cos \phi=\frac{1}{\sqrt{2}}$
$\cos \theta=\frac{(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b})|}$
$|\hat{a}+\hat{b}|^2=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})$
$|\hat{a}+\hat{b}|^2=2+2 \hat{a} \cdot \hat{b}$
$=2+\sqrt{2}$
$\hat{a} \times \hat{b}=|\hat{a}||\hat{b}| \sin \phi \hat{n}$
$\hat{ a } \times \hat{ b }=\frac{\hat{ n }}{\sqrt{2}} \quad$
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}$