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

The Correct Option is A

To solve this problem, we need to find the angle \(\theta\) between the vectors \((\hat{a} + \hat{b})\) and \((\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))\), and then evaluate \(164 \cos^2 \theta\).

First, we calculate the dot product of the given vectors. The dot product formula for two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by: 

1. \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta\)

The vectors we are dealing with are:

• \(\mathbf{u} = \hat{a} + \hat{b}\)
• \(\mathbf{w} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})\)

Compute \(\hat{a} \cdot \hat{b}\), knowing that the angle between \(\hat{a}\) and \(\hat{b}\) is \(\pi/4\), which gives:

1. \(\hat{a} \cdot \hat{b} = \cos(\pi/4) = \frac{\sqrt{2}}{2}\)

Compute \(\mathbf{u} \cdot \mathbf{u}\):

1. \(\mathbf{u} \cdot \mathbf{u} = (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) = 1 + 1 + 2(\frac{\sqrt{2}}{2}) = 2 + \sqrt{2}\)

Compute \(\mathbf{w} \cdot \mathbf{w}\):

1. \(\mathbf{w} \cdot \mathbf{w} = (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))\) 
   After simplifying:
   \(= 1 + 4 + 4 + 4(0) = 9\)

Compute \(\mathbf{u} \cdot \mathbf{w}\):

\[\begin{align*} \mathbf{u} \cdot \mathbf{w} &= (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})) \\ &= \hat{a} \cdot \hat{a} + 2(\hat{a} \cdot \hat{b}) + 2(\hat{a} \cdot(\hat{a} \times \hat{b})) + \hat{b} \cdot \hat{a} + 2(\hat{b} \cdot \hat{b}) + 2(\hat{b} \cdot (\hat{a} \times \hat{b})) \end{align*}\]

1. \(\\ \hat{a} \cdot (\hat{a} \times \hat{b}) = \hat{b} \cdot (\hat{a} \times \hat{b}) = 0\)
   Calculating further:
   \(= 1 + 2(\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} + 4 = 6 + 2\sqrt{2}\)

Finally, use the dot product to find \(\cos \theta\):

1. \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{w}}{|\mathbf{u}| |\mathbf{w}|}\) 
   \(|\mathbf{u}| = \sqrt{2 + \sqrt{2}}, \quad |\mathbf{w}| = \sqrt{9} = 3\)

Simplify and evaluate \(164 \cos^2 \theta\):

1. \(164 \left(\frac{6 + 2\sqrt{2}}{3\sqrt{2 + \sqrt{2}}}\right)^2 = 90 + 27\sqrt{2}\\) 

Therefore, the answer is

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\(\sqrt2\)