Physics - Vectors Question with Solution | TestHub

So angle between $\left(2\overrightarrow{P}+\overrightarrow{Q}\right)$ and $\overrightarrow{Q}$ is ${90}^{\mathrm{o}}$

Alternate solution

$\left|\overrightarrow{\text{P}}+\overrightarrow{\text{Q}}\right|=\left|\overrightarrow{\text{P}}\right|$
${\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{P}\mathrm{Q}\cos \mathrm{\theta }={\mathrm{P}}^2$
$\Rightarrow \mathrm{Q}+2\mathrm{P}\mathrm{Q}\cos \mathrm{\theta }=\mathrm{O}\mathrm{}$
$\Rightarrow \cos \mathrm{\theta }=-\frac{\mathrm{Q}}{2\mathrm{P}}$

$\tan \mathrm{\alpha }=\frac{2\mathrm{P}\sin \mathrm{\theta }}{2\mathrm{P}\cos \mathrm{\theta }+\mathrm{Q}}=\mathrm{\infty }$
$\because \left[2\mathrm{P}\cos \mathrm{\theta }+\mathrm{Q}=0\right]$
$\mathrm{\alpha }={90}^{\mathrm{o}}$