Question

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are defined as $\overrightarrow{A}=a\hat{i}$ and $\overrightarrow{B}=a(\cos \omega t\hat{i}+\sin \omega t\hat{j})$, where $a$ is a constant and $\omega =\frac{\pi }{6}rad{s}^{-1}$. If $|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{3}|\overrightarrow{A}-\overrightarrow{B}|$ at time $t=\tau$ for the first time, then the value of $\tau$, in seconds, is

Answer 1

For the given situation,
L.H.S. = $|\overrightarrow{A}+\overrightarrow{B}|=2a\cos \frac{\omega t}{2}$
R.H.S. = $\sqrt{3}$ $|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{3}\times 2a\sin \frac{\omega t}{2}$

From the standard mathematical equation
L.H.S. = R.H.S.
$\Rightarrow$ $2a\cos \frac{\omega t}{2}=\sqrt{3}\left(2a\sin \frac{\omega t}{2}\right)$

$\Rightarrow$ $\tan \frac{\omega t}{2}=\frac{1}{\sqrt{3}}$

$\Rightarrow$ $\frac{\omega t}{2}=\frac{\pi }{6}\Rightarrow \omega t=\frac{\pi }{3}$

$\Rightarrow$ $\frac{\pi }{6}t=\frac{\pi }{3}$
​​​​​​​ $t=2.00sec$