Physics - Vectors Question with Solution | TestHub

$\tan {90}^{\mathrm{o}}=\frac{Q\mathrm{s}\mathrm{i}\mathrm{n}\theta }{P+Q\mathrm{c}\mathrm{o}\mathrm{s}\theta }\Rightarrow P+Q\mathrm{c}\mathrm{o}\mathrm{s}\theta =0$
$\Rightarrow \mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{-P}{Q}\hspace{0.17em}$…(1)
$\because R^2=P^2+Q^2+2PQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\cos \theta$
$\because P=R$
$\therefore \frac{-P}{Q}=\frac{1}{2\mathrm{c}\mathrm{o}\mathrm{s}\theta }$…(2)
From (1) & (2)
$\mathrm{c}\mathrm{o}\mathrm{s}\theta =\pm \frac{1}{\sqrt{2}}$
$\therefore \theta =\frac{3\pi }{4}$