Two vectors of equal magnitude P are inclined at some angle such that the difference in magnitude of resultant and magnitude of either of the vectors is 0.732 times either of the magnitude of vectors. If the angle between them is increased by half of its initial value then find the magnitude of difference of the vectors            

$\text{R}=\sqrt{{\text{P}}^{\text{2}}+{\text{Q}}^{\text{2}}+2\text{PQ}\cos \theta }$

$=\sqrt{{\text{P}}^{\text{2}}+{\text{P}}^2+2{\text{P}}^{\text{2}}\cos \theta }$

$=\sqrt{2{\text{P}}^{\text{2}}+2{\text{P}}^{\text{2}}\cos \theta }$

$=\sqrt{2{\text{P}}^2\left(1+\cos \theta \right)}$           $1+\cos \theta =1+{\cos }^2\left(\text{θ/}2\right)\left[\because \cos 2\theta =2{\cos }^2\theta -1\right]$

$=\sqrt{2{\text{P}}^2.2{\cos }^2\frac{\theta }{2}}$                                  $=1+2{\cos }^2\frac{\theta }{2}-1$

$\text{R}=\sqrt{4{\text{P}}^2{\cos }^2\frac{\theta }{2}}$                                  $=2{\cos }^2\frac{\theta }{2}$

$\text{R}=2\text{P}\cos \frac{\theta }{2}$

$\text{R}-\text{P}=2\text{P}\cos \frac{\theta }{2}-\text{P}=0.732\text{P}$

$\text{2Pcos}\frac{\theta }{2}-\text{P}=0.732\text{P}$

$\text{P}\left(2\cos \frac{\theta }{2}-1\right)=0.732\text{P}$

$\text{2cos}\frac{\theta }{2}=1.732$

$\text{2cos}\frac{\theta }{2}=\sqrt{3}$

$\text{cos}\frac{\theta }{2}=\frac{\sqrt{3}}{2}$

$\text{cos}\frac{\theta }{2}=\cos 30$

$\frac{\theta }{2}=30°$

$\theta =60°$ , if angle is increased by half of its initial value

${\theta }_1=\theta +\frac{\theta }{2}=60+30=90°$

$\text{R}=\sqrt{{\text{P}}^2+{\text{Q}}^2-2\text{PQ}\cos 90°}$

$=\sqrt{2{\text{P}}^{\text{2}}-2{\text{P}}^{\text{2}}\times 0}=\sqrt{{\text{2P}}^{\text{2}}}$

$\text{R}=\sqrt{2}\text{P}$