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

We have two unit vectors $$\hat{a}$$ and $$\hat{b}$$ with the angle between them equal to $$\frac{\pi}{4}$$, so $$|\hat{a}| = |\hat{b}| = 1$$ and $$\hat{a} \cdot \hat{b} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$. Also, $$|\hat{a} \times \hat{b}| = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$.

We define $$\vec{p} = \hat{a} + \hat{b}$$ and $$\vec{q} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$$, and we need to find $$164\cos^2\theta$$ where $$\theta$$ is the angle between $$\vec{p}$$ and $$\vec{q}$$.

We compute $$|\vec{p}|^2 = |\hat{a}|^2 + 2(\hat{a}\cdot\hat{b}) + |\hat{b}|^2 = 1 + \frac{2}{\sqrt{2}} + 1 = 2 + \sqrt{2}$$.

Now we find $$|\vec{q}|^2$$. We note that $$\hat{a} \times \hat{b}$$ is perpendicular to both $$\hat{a}$$ and $$\hat{b}$$, so $$\hat{a} \cdot (\hat{a} \times \hat{b}) = 0$$ and $$\hat{b} \cdot (\hat{a} \times \hat{b}) = 0$$.

$$|\vec{q}|^2 = |\hat{a} + 2\hat{b}|^2 + 4|\hat{a}\times\hat{b}|^2 = \left(1 + 4\cdot\frac{1}{\sqrt{2}} + 4\right) + 4\cdot\frac{1}{2} = 5 + 2\sqrt{2} + 2 = 7 + 2\sqrt{2} + 2\sqrt{2} = 5 + 4\cdot\frac{1}{\sqrt{2}} + 4 + 2$$

Let me compute this more carefully. $$|\hat{a}+2\hat{b}|^2 = 1 + 4(\hat{a}\cdot\hat{b}) + 4 = 5 + \frac{4}{\sqrt{2}} = 5 + 2\sqrt{2}$$. And $$|2(\hat{a}\times\hat{b})|^2 = 4\cdot\frac{1}{2} = 2$$. Since $$(\hat{a}+2\hat{b})$$ is perpendicular to $$2(\hat{a}\times\hat{b})$$, we get $$|\vec{q}|^2 = 5 + 2\sqrt{2} + 2 = 7 + 2\sqrt{2}$$.

Next, $$\vec{p}\cdot\vec{q} = (\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})) = (\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}) + 0$$, since $$(\hat{a}+\hat{b})\cdot(\hat{a}\times\hat{b}) = 0$$ (as $$\hat{a}\times\hat{b}$$ is perpendicular to both $$\hat{a}$$ and $$\hat{b}$$).

$$\vec{p}\cdot\vec{q} = \hat{a}\cdot\hat{a} + 2(\hat{a}\cdot\hat{b}) + \hat{b}\cdot\hat{a} + 2(\hat{b}\cdot\hat{b}) = 1 + \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}} = 3 + \frac{3\sqrt{2}}{2}$$

Now $$\cos^2\theta = \frac{(\vec{p}\cdot\vec{q})^2}{|\vec{p}|^2\,|\vec{q}|^2}$$.

We compute the numerator: $$\left(3 + \frac{3\sqrt{2}}{2}\right)^2 = 9 + 2\cdot 3\cdot\frac{3\sqrt{2}}{2} + \frac{9\cdot 2}{4} = 9 + 9\sqrt{2} + \frac{9}{2} = \frac{27}{2} + 9\sqrt{2} = \frac{27 + 18\sqrt{2}}{2}$$.

The denominator: $$(2+\sqrt{2})(7+2\sqrt{2}) = 14 + 4\sqrt{2} + 7\sqrt{2} + 2\cdot 2 = 14 + 11\sqrt{2} + 4 = 18 + 11\sqrt{2}$$.

So $$\cos^2\theta = \frac{27 + 18\sqrt{2}}{2(18+11\sqrt{2})}$$.

We rationalize by multiplying numerator and denominator by $$(18 - 11\sqrt{2})$$:

Denominator factor: $$(18+11\sqrt{2})(18-11\sqrt{2}) = 324 - 242 = 82$$.

Numerator factor: $$(27+18\sqrt{2})(18-11\sqrt{2}) = 486 - 297\sqrt{2} + 324\sqrt{2} - 198\cdot 2 = 486 - 396 + (324-297)\sqrt{2} = 90 + 27\sqrt{2}$$.

So $$\cos^2\theta = \frac{90+27\sqrt{2}}{2\cdot 82} = \frac{90+27\sqrt{2}}{164}$$.

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

Hence, the correct answer is Option A.