The resultant of two vectors $\overrightarrow{\mathrm{P}}\mathrm{and}\overrightarrow{\mathrm{Q}}\mathrm{is} \overrightarrow{\mathrm{R}}$. If the magnitude of $\overrightarrow{\mathrm{Q}}$ is doubled, the new resultant becomes perpendicular to $\overrightarrow{\mathrm{P}}$, then the magnitude of $\overrightarrow{R}$ is  

${\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ}\mathrm{cos\theta }$ ..........(1)

when Q is doubled

$\overrightarrow{{\mathrm{R}}_1}=\overrightarrow{\mathrm{P}}+2\overrightarrow{\mathrm{Q}}\mathrm{and}\overrightarrow{{\mathrm{R}}_1}\perp \overrightarrow{\mathrm{P}}$

If $\alpha$ is the angle between $\overrightarrow{{\mathrm{R}}_1}$ and $\overrightarrow{\mathrm{P}}$

$\tan \mathrm{\alpha }=\frac{2\mathrm{Qsin\theta }}{\mathrm{P}+2\mathrm{Qcos\theta }}$

$\Rightarrow \mathrm{P}+2\mathrm{Qcos\theta }=0$

From (1)

${\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2+\mathrm{P}(-\mathrm{P})={\mathrm{Q}}^2$

R = Q