The magnitudes of two vectors $\overrightarrow{\mathrm{P}} \mathrm{and} \overrightarrow{\mathrm{Q}}$ differ by 1. The magnitude of their resultant makes an angle of tan^–1(3/4) with P. The angle between P and Q is

(i) P – Q = 1
$\text{ (ii) t}\mathrm{an}\mathrm{\alpha }=\frac{3}{4}=\frac{\mathrm{Qsin}\mathrm{\theta }}{\mathrm{P}+\mathrm{Qcos}\mathrm{\theta }}$
where $\mathrm{\theta }$ is the angle between P and Q, $\mathrm{\alpha }$ is the angle between P and R
$\therefore$ When  $\mathrm{\theta }$ = 90^o can only satisfy both the conditions (i) & (ii)