Question

If $\vec{a}$ and $\vec{b}$ are two vectors of magnitude $2$, each inclined at an angle $60^\circ$, then angle between $\vec{a}$ and $\vec{a} +\vec{b}$ is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $90^\circ$

Answer 1

Answer: (A)

Let $ \theta $ be angle between $\vec{a}$ and $\vec{b}$ then $ \theta = 60^\circ$ (given)
Since , $|\vec{a} + \vec{b}| = |\vec{a}|^2 |+ |\vec{b}|^2 + 2\vec{a} .\vec{b} $
$ = 4 + 4 + (2 \times 2 \times 2 \times \cos 60^\circ)$
$ 8 + 8 \, \cos 60^\circ = 8 + 4 = 12$
$ \Rightarrow \:\:\: |\vec{a} + \vec{b}| = \sqrt{12} = 2 \sqrt{3}$
Now, $\vec{a}( \vec{a} +\vec{b} ) = |\vec{a}||\vec{a} +\vec{b}| \cos \, x$
where x is angle between $\vec{a} $ and $\vec{a} + \vec{b}$
$\Rightarrow \:\:\: \vec{a} . \vec{a} + \vec{a} .\vec{b} = 4 \sqrt{3} \cos \: x$
$ 4 + 2 \times 2 \: \cos 60^\circ = 4 \sqrt{3} \, \cos x$
$\Rightarrow \:\: 6 = 4\sqrt{3} \:\:\: \cos x$
$\Rightarrow \:\:\: \cos x = \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}$
$\Rightarrow \:\:\: x = 30^\circ$