Let a and b be two unit vectors such that the angle between them is π 4. If θ is the angle between the vectors ( a+ b) and ( a+2 b+2( a× b)) then the…

Let $\hat{a}$a^ and $\hat{b}$b^ be two unit vectors such that the angle between them is $\frac{\pi }{4}$4π​. If $\theta$θ is the angle between the vectors $(\hat{a}+\hat{b})$(a^+b^) and $(\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))$(a^+2b^+2(a^×b^)) then the value of $164{\mathrm{cos}}^{2}\theta$164cos2θ is equal to

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

So, $\hat{a}\cdot \hat{b}=|\hat{a}||\hat{b}|\mathrm{cos}\phi$a^⋅b^=∣a^∣∣b^∣cosϕ

$$\Rightarrow \hat{a}\cdot \hat{b}=\mathrm{cos}\phi =\frac{1}{\sqrt{2}}$$⇒a^⋅b^=cosϕ=2​1​

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

$$\mathrm{cos}\theta =\frac{(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b})|}....(1)$$cosθ=∣a^+b^∣∣a^+2b^+2(a^×b^)∣(a^+b^)⋅(a^+2b^+2(a^×b^))​....(1)

Now finding the value of ${|\hat{a}+\hat{b}|}^{2}=(\hat{a}+\hat{b})\cdot (\hat{a}+\hat{b})$∣a^+b^∣2=(a^+b^)⋅(a^+b^)

$\Rightarrow {|\hat{a}+\hat{b}|}^{2}=2+2\hat{a}\cdot \hat{b}$⇒∣a^+b^∣2=2+2a^⋅b^$=2+\sqrt{2}$=2+2​

Also value of $\hat{a}\times \hat{b}=|\hat{a}||\hat{b}|\mathrm{sin}\phi \hat{n}$a^×b^=∣a^∣∣b^∣sinϕn^

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

Let $\vec{c}=\hat{a}\times \hat{b}$c=a^×b^

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

Now value of ${|\hat{a}+2\hat{b}+2\vec{c}|}^{2}$∣a^+2b^+2c∣2

$$=1+4+\frac{(4)}{2}+4\hat{a}\cdot \hat{b}+8\hat{b}\cdot \vec{c}+4\vec{c}\cdot \hat{a}$$=1+4+2(4)​+4a^⋅b^+8b^⋅c+4c⋅a^

$$=7+\frac{4}{\sqrt{2}}=7+2\sqrt{2}$$=7+2​4​=7+22​

Now finding dot product we get, $(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2\vec{c})$(a^+b^)⋅(a^+2b^+2c)

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

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

$$=3+\frac{3}{\sqrt{2}}$$=3+2​3​

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

$$\mathrm{cos}\theta =\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$$cosθ=2+2​​7+22​​3+2​3​​

$$\Rightarrow {\mathrm{cos}}^{2}\theta =\frac{9{(\sqrt{2}+1)}^{2}}{2(2+\sqrt{2})(7+2\sqrt{2})}$$⇒cos2θ=2(2+2​)(7+22​)9(2​+1)2​

$$\Rightarrow {\mathrm{cos}}^{2}\theta =(\frac{9}{2\sqrt{2}})\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}$$⇒cos2θ=(22​9​)(7+22​)(2​+1)​

$$\Rightarrow 164{\mathrm{cos}}^{2}\theta =\frac{(82)(9)}{\sqrt{2}}\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}\frac{(7-2\sqrt{2})}{(7-2\sqrt{2})}$$⇒164cos2θ=2​(82)(9)​(7+22​)(2​+1)​(7−22​)(7−22​)​

$$\Rightarrow 164{\mathrm{cos}}^{2}\theta =\frac{(82)}{\sqrt{2}}\frac{(9)[7\sqrt{2}-4+7-2\sqrt{2}]}{(41)}$$⇒164cos2θ=2​(82)​(41)(9)[72​−4+7−22​]​

$$\Rightarrow 164{\mathrm{cos}}^{2}\theta =(9\sqrt{2})[5\sqrt{2}+3]$$⇒164cos2θ=(92​)[52​+3]

$$\Rightarrow 164{\mathrm{cos}}^{2}\theta =90+27\sqrt{2}$$⇒164cos2θ=90+272​