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Two vectors $\dot{A}$ and $\dot{B}$ are defined as $\dot{A}=a \hat{i}$ and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega t \hat{\mathrm{i}}+\sin \omega t \hat{j}$ ), where a is a constant and $\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1}$. If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $t=\tau$ for the first time, the value of $\tau$, in, seconds, is. . . . . .
$1$
$2$
$5$
$6$
Solution
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=2 \mathrm{a} \cos \frac{\omega \mathrm{t}}{2}$
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=2 \mathrm{a} \sin \frac{\omega \mathrm{t}}{2}$
So, $2 \mathrm{a} \cos \frac{\omega \mathrm{t}}{2}=\sqrt{3}\left(2 \mathrm{a} \sin \frac{\omega \mathrm{t}}{2}\right)$
$\tan \frac{\omega \mathrm{t}}{2}=\frac{1}{\sqrt{3}}$
$\frac{\omega \mathrm{t}}{2}=\frac{\pi}{6}$
$\omega \mathrm{t}=\frac{\pi}{3}$
Now, $\omega=\frac{\pi}{6} \mathrm{rad} \mathrm{s}^{-1}$
$\frac{\pi}{6} t=\frac{\pi}{3}$
$t=2 s$
