Question

Two vectors having equal magnitudes A make an angle $\theta$ against each other. Find the magnitude and direction of the resultant.

Answer 1

We have,

Two vectors having equal magnitude makes an $\angle \theta$

Let the magnitude of each vector be A is:

$R=\sqrt{(A^2+A^2+2A\times Acos\theta )}$

$=\sqrt{\left[2A^2\right(1+cos\theta \left)\right]}$

$\sqrt{2A^2\times 2cos\frac{\theta }{2}}$

$=2Acos\frac{\theta }{2}$

Now,

Direction :

Let angle between any one of vectors and thne resulatant be $\theta$

$tan\theta =\frac{Asin\theta }{Acos\theta +A}$

$=\frac{Sin\theta }{1+cos\theta }$

$=\frac{2sin\frac{\theta }{2}.cos\frac{\theta }{2}}{2co{s}^2\frac{\theta }{2}}$

By using, $sin\theta =2sin\frac{\theta }{2}cos\frac{\theta }{2}$ and $1+cos\theta =2co{s}^2\frac{\theta }{2}$

$=tan\frac{\theta }{2}$

It means that, $tan{\theta }^{\prime }=tan\frac{\theta }{2}$

$\theta =\frac{\theta }{2}$