Question

Two vectors having equal magnitudes of A make an angle θ with each other. Find the magnitude B of the resultant vector and the angle it makes with any one of the vectors?

• A. B = A $cos\frac{\theta }{2}$ , $\alpha =0$
• B. B = 2 A cos $\frac{\theta }{2}$ , $\alpha =\frac{\theta }{2}$
• C. B = A cos $\frac{\theta }{2}$ , $\alpha =\frac{\theta }{2}$
• D. B = 2 A cos $\frac{\theta }{2}$ , $\alpha =\theta$

Answer 1

The correct option is B

B = 2 A cos $\frac{\theta }{2}$ , $\alpha =\frac{\theta }{2}$

The magnitude of the resultant will be

$B=\sqrt{A^2+A^2+2AAcos\theta }$

$=\sqrt{2A^2+(1+cos\theta )}$

$=\sqrt{4A^{2}co{s}^2\frac{\theta }{2}}$

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

The resultant will make an angle $\alpha$ with the first vector where,

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

=$\frac{2Asin\frac{\theta }{2}cos\frac{\theta }{2}}{2Aco{s}^2\frac{\theta }{2}}$

( sin 2 $\theta$ = 2 sin $\theta$ cos $\theta$ , cos 2 $\theta$ = 2 $co{s}^2$ $\theta$ - 1 )

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

Therefore , $\alpha =\frac{\theta }{2}$