Mathematics - Vector Question with Solution | TestHub

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

Now let angle between $\hat{a} \& \hat{b}=\frac{\pi }{4}=\varphi$

So, $\hat{a}·\hat{b}=\left|\hat{a}\right|\left|\hat{b}\right|\cos \varphi$

$\Rightarrow \hat{a}·\hat{b}=\cos \varphi =\frac{1}{\sqrt{2}}$

Now finding angle between $\left(\hat{a}+\hat{b}\right) \& \left(\hat{a}+2\hat{b}+2\left(\hat{a}\times \hat{b}\right)\right)$ we get,

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

Now finding the value of ${\left|\hat{a}+\hat{b}\right|}^2=\left(\hat{a}+\hat{b}\right)·\left(\hat{a}+\hat{b}\right)$

$\Rightarrow {\left|\hat{a}+\hat{b}\right|}^2=2+2\hat{a}·\hat{b}$$=2+\sqrt{2}$

Also value of $\hat{a}\times \hat{b}=\left|\hat{a}\right|\left|\hat{b}\right|\sin \varphi \hat{n}$

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

Let $\overrightarrow{c}=\hat{a}\times \hat{b}$

We know that, $\overrightarrow{c}·\overrightarrow{a}=0$ and $\overrightarrow{c}·\overrightarrow{b}=0$ as perpendicular vector dot product is zero.

Now value of ${\left|\hat{a}+2\hat{b}+2\overrightarrow{c}\right|}^2$

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

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

Now finding dot product we get, $\left(\hat{a}+\hat{b}\right)·\left(\hat{a}+2\hat{b}+2\overrightarrow{c}\right)$

$={\left|\hat{a}\right|}^2+2\hat{a}·\hat{b}+0+\hat{b}·\hat{a}+2{\left|\hat{b}\right|}^2+0$

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

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

Now putting all the values in equation $\left(1\right)$ we get,

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

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

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

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

$\Rightarrow 164{\cos }^2\theta =\frac{\left(82\right)}{\sqrt{2}}\frac{\left(9\right)\left[7\sqrt{2}-4+7-2\sqrt{2}\right]}{\left(41\right)}$

$\Rightarrow 164{\cos }^2\theta =\left(9\sqrt{2}\right)\left[5\sqrt{2}+3\right]$

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