Question

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

• A. 90+27$\sqrt2$
• B. 45+18$\sqrt2$
• C. 90+3$\sqrt2$
• D. 54+90$\sqrt2$

Answer 1

Answer: (A)

90+27$\sqrt2$

Answer 2

$\hat a$⋅$\hat b$=$\frac{1}{2}$ and =|$\vec a$×$\vec b$|=$\frac{1}{\sqrt2}$
$\frac{(\hat a+\hat b)⋅(\hat a+2\hat b+2(\hat a×\hat b)}{|\hat a+\hat b||\hat a+2\hat b+2(\hat a×\hat b)|}$=cos⁡θ
|$\hat a$+$\hat b$|2=2+$\sqrt2$
|$\hat a$+2$\hat b$+2($\hat a$×$\hat b$)|2=1+4+4|$\hat a$×$\hat b$|2+4\hat a$$\hat b
=5+4⋅$\frac{1}{2}$+$\frac{4}{\sqrt2}$=7+$2\sqrt2$
Hence, cos2θ⁡=$\frac{(3+\frac{3}{\sqrt2})^2}{(2+√2)(7+2√2)}$=$\frac{9\sqrt2(5\sqrt2+3)}{164}$
⇒164cos2⁡θ=90+27$\sqrt2$
So, the correct option is (A): 90+27$\sqrt2$