Discussion Forum

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

Answer:

Explanation:

$A= a\hat{i}$  and  $B= a\cos\omega \hat{i}+a \sin \omega t\hat{j}$

  $A+B= (a+a\cos\omega t )\hat{i}+a \sin \omega t\hat{j}$

  $A-B= (a-a\cos\omega t )\hat{i}+a \sin \omega t\hat{j}$

    $\mid A+B\mid= \sqrt{3}\mid A-B\mid$

      $\sqrt{(a+a\cos\omega t)^{2}+(a\sin\omega t)^{2}}=\sqrt{3}\sqrt{(a-a\cos \omega t)^{2}+(a \sin \omega t)^{2}}$

  $\Rightarrow$      $2 \cos\frac{\omega t}{2}=\pm \sqrt{3}\times 2\sin \frac{\omega t}{2}$

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

   $ \frac{\omega t}{2}=n\pi\pm \frac{\pi}{6}$

   $ \frac{\pi}{12}t=n\pi\pm \frac{\pi}{6}$

     $t=(12n\pm 2)s$

         = 2 s