Two vectors P and Q have equal magnitudes. If the magnitude of P + Q is n times the magnitude of P − Q, then angle between P and Q is

According to question,

$\left|\mathrm{P}\right|=\left|\mathrm{Q}\right|$               ...(i)

Also,     $|\mathrm{P}+\mathrm{Q}|=\mathrm{n}|\mathrm{P}-\mathrm{Q}|$         ...(ii)

On squaring Eq. (ii) both sides, we get

$|\mathrm{P}+\mathrm{Q}{|}^2={\mathrm{n}}^2|\mathrm{P}-\mathrm{Q}{|}^2$

$\begin{array}{l}\Rightarrow {\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQcos}\mathrm{\theta }={\mathrm{n}}^2\left({\mathrm{P}}^2+{\mathrm{Q}}^2-2\mathrm{PQcos}\mathrm{\theta }\right)\\ \Rightarrow 1+1+2\cos \mathrm{\theta }={\mathrm{n}}^2(1+1-2\cos \mathrm{\theta })\\ \Rightarrow 2+2\cos \mathrm{\theta }={\mathrm{n}}^2(2-2\cos \mathrm{\theta })\\ \Rightarrow 1+\cos \mathrm{\theta }={\mathrm{n}}^2(1-\cos \mathrm{\theta })\end{array}$

$\begin{array}{l}\Rightarrow \frac{1+\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}=\frac{{\mathrm{n}}^2}{1}\Rightarrow \frac{1}{\cos \mathrm{\theta }}=\frac{{\mathrm{n}}^2+1}{{\mathrm{n}}^2-1}\\ \Rightarrow \cos \mathrm{\theta }=\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2+1}\Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2+1}\right)\end{array}$