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 $\vec{a}+\vec{b}$
$\Rightarrow \vec{a}.\vec{a}+\vec{a}.\vec{b}=4\sqrt{3}\cos x$
$4+2\times 2\cos 60^{\circ }=4\sqrt{3}\cos x$
$\Rightarrow 6=4\sqrt{3}\cos x$
$\Rightarrow \cos x=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$
$\Rightarrow x=30^{\circ }$