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

Answer

Correct option: A.
$90+27 \sqrt{2}$
a
$\hat{a}^{\wedge} \hat{b}=\frac{\pi}{4}=\phi$

$\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \phi$

$\hat{a} \cdot \hat{b}=\cos \phi=\frac{1}{\sqrt{2}}$

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

$|\hat{a}+\hat{b}|^{2}=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})$

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

$=2+\sqrt{2}$

$\hat{a} \times \hat{b}=|\hat{a}||\hat{b}| \sin \phi \hat{n}$

$\hat{ a } \times \hat{ b }=\frac{\hat{ n }}{\sqrt{2}} \quad$ when $\hat{ n }$ is vector $\perp \hat a$ and $\hat{ b }$

let $\vec{c}=\hat{a} \times \hat{b}$

We know.

$\vec{c} \cdot \vec{a}=0$

$\overrightarrow{ c } \cdot \overrightarrow{ b }=0$

$|\hat{a}+2 \hat{b}+2 \vec{c}|^{2}$

$=1+4+\frac{(4)}{2}+4 \hat{ a } \cdot \hat{ b }+8 \hat{ b } \cdot \overrightarrow{ c }+4 \overrightarrow{ c } \cdot \hat{ a }$

$=7+\frac{4}{\sqrt{2}}=7+2 \sqrt{2}$

Now,$(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2 \vec{c})$

$=|\hat{a}|^{2}+2 \hat{a} \cdot \hat{b}+0+\hat{b} \cdot \hat{a}+2|\hat{b}|^{2}+0$

$=1+\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}+2$

$=3+\frac{3}{\sqrt{2}}$

$\cos \theta=\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}} \sqrt{7+2 \sqrt{2}}}$

$\cos ^{2} \theta=\frac{9(\sqrt{2}+1)^{2}}{2(2+\sqrt{2})(7+2 \sqrt{2})}$

$\cos ^{2} \theta=\left(\frac{9}{2 \sqrt{2}}\right) \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})}$

$164 \cos ^{2} \theta=\frac{(82)(9)}{\sqrt{2}} \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})} \frac{(7-2 \sqrt{2})}{(7-2 \sqrt{2})}$

$=\frac{(82)}{\sqrt{2}} \frac{(9)[7 \sqrt{2}-4+7-2 \sqrt{2}]}{(41)}$

$=(9 \sqrt{2})[5 \sqrt{2}+3]$

$=90+27 \sqrt{2}$

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