Test Index

Vector Algebra Part 2

Section: Mathematics

© examsnet.com

Question : 91 of 100

Marks: +1, -0

Let

\hat{a}

\hat{b}

be two unit vectors such that the angle between them is

\frac{\pi}{4}

\theta

is the angle between the vectors

(\hat{a}+\hat{b})

(\hat{a}+2\hat{b}+2(\hat{a} \times \hat{b}))

, then the value of

164 \cos^2 \theta

[29-Jul-2022-Shift-1]

$90+27\sqrt{2}$
$45+18\sqrt{2}$
$90+3\sqrt{2}$
$54+90\sqrt{2}$

Solution:

\hat{a} \cdot \hat{b} = \frac{1}{\sqrt{2}}

and

|\hat{a} \times \hat{b}| = \frac{1}{\sqrt{2}}

\frac{\widehat{(\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b}))}}{\widehat{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})|}} = \cos \theta

\Rightarrow \cos \theta = \frac{\widehat{1+3 a \hat{b}+2}}{\widehat{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})|}}

|\hat{a}+\hat{b}|^2 = 2+\sqrt{2}

|\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})|^2 = 1+4+4|\hat{a}\times\hat{b}|^2+4\hat{a}\hat{b}

=5+4 \cdot \frac{1}{2}+\frac{4}{\sqrt{2}} = 7+2\sqrt{2}

So,

\cos^2 \theta = \frac{\left(3+\frac{3}{\sqrt{2}}\right)^2}{(2+\sqrt{2})(7+2\sqrt{2})} = \frac{9\sqrt{2}(5\sqrt{2}+3)}{164}

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

© examsnet.com

Go to Question:

Prev Question

Next Question