Q 11-03-138JEE MainJEE Main 2021 (26 Aug, Shift 2)Medium
The angle between vector $(\vec{A})$ and $(\vec{A} - \vec{B})$ is:
Answer: (B) $\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)$
In the figure $\vec{B}$ is drawn from the head of $\vec{A}$ with an interior angle of $120^\circ$, so the angle between $\vec{A}$ and $\vec{B}$ (tail to tail) is $60^\circ$.
$\vec{A} - \vec{B}$ has a component $A - B\cos60^\circ = A - \dfrac{B}{2}$ along $\vec{A}$ and $B\sin60^\circ = \dfrac{\sqrt{3}}{2}B$ perpendicular to it.
$$\tan\beta = \frac{\frac{\sqrt{3}}{2}B}{A - \frac{B}{2}} = \frac{\sqrt{3}B}{2A - B}$$
Solution by Sreeraj P, M.Sc Physics