Let ^ a and ^ b be two unit vectors such that the angle between...

Let \overset{^}{a} and \overset{^}{b} be two unit vectors such that the angle between them is $\dfrac{\pi}{4}$. If $\theta$ is the angle between the vectors \left(\overset{^}{a} + \overset{^}{b}\right)and \left(\overset{^}{a} + 2\overset{^}{b} + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right)$then the value of$164\cos^{2}\theta$is equal to</p><p>Now let angle between$\overset{^}{a} \text{ and } \overset{^}{b} = \dfrac{\pi}{4} = ϕ$</p><p>So,$\overset{^}{a} \cdot \overset{^}{b} = \left|\overset{^}{a}\right|\left|\overset{^}{b}\right|cosϕ$</p><p>$⇒ \overset{^}{a} \cdot \overset{^}{b} = cosϕ = \dfrac{1}{\sqrt{2}}</p><p>Now finding angle between \left(\overset{^}{a} + \overset{^}{b}\right) \text{ and }\left(\overset{^}{a} + 2\overset{^}{b} + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right)$we get,</p><p>$cos\theta = \dfrac{\left(\overset{^}{a} + \overset{^}{b}\right) \cdot \left(\overset{^}{a} + 2\overset{^}{b} + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right)}{\left|\overset{^}{a} + \overset{^}{b}\right|\left|\overset{^}{a} + 2\overset{^}{b} + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right|} . . . . \left(1\right)$</p><p>Now finding the value of${\left|\overset{^}{a} + \overset{^}{b}\right|}^{2} = \left(\overset{^}{a} + \overset{^}{b}\right) \cdot \left(\overset{^}{a} + \overset{^}{b}\right)$</p><p>$⇒ {\left|\overset{^}{a} + \overset{^}{b}\right|}^{2} = 2 + 2\overset{^}{a} \cdot \overset{^}{b}$$= 2 + \sqrt{2}$

Also value of \overset{^}{a} \times \overset{^}{b} = \left|\overset{^}{a}\right|\left|\overset{^}{b}\right|sinϕ\overset{^}{n}

⇒ \overset{^}{a} \times \overset{^}{b} = \overset{^}{n}\sqrt{2} when \overset{^}{n} is vector $⊥$to \overset{^}{a} and \overset{^}{b}

Let \overset{\to}{c} = \overset{^}{a} \times \overset{^}{b}

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

Now value of {\left|\overset{^}{a} + 2\overset{^}{b} + 2\overset{\to}{c}\right|}^{2}

= 1 + 4 + \dfrac{\left(4\right)}{2} + 4\overset{^}{a} \cdot \overset{^}{b} + 8\overset{^}{b} \cdot \overset{\to}{c} + 4\overset{\to}{c} \cdot \overset{^}{a}

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

Now finding dot product we get, \left(\overset{^}{a} + \overset{^}{b}\right) \cdot \left(\overset{^}{a} + 2\overset{^}{b} + 2\overset{\to}{c}\right)$</p><p>$= {\left|\overset{^}{a}\right|}^{2} + 2\overset{^}{a} \cdot \overset{^}{b} + 0 + \overset{^}{b} \cdot \overset{^}{a} + 2{\left|\overset{^}{b}\right|}^{2} + 0$</p><p>$= 1 + \dfrac{2}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} + 2$</p><p>$= 3 + \dfrac{3}{\sqrt{2}}$</p><p>Now putting all the values in equation \left(1\right)$ we get,

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

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

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

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

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

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

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