Question

Magnitude of resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is equal to the magnitude of $\overrightarrow{P}$. Find the angle between $\overrightarrow{Q}$ and the resultant of $2\overrightarrow{P}$ and $\overrightarrow{Q}$.

• A. ${90}^{\circ }$.
• B. ${60}^{\circ }$.
• C. ${75}^{\circ }$.
• D. ${90}^{\circ }$.

Answer 1

The correct option is D ${90}^{\circ }$.

Given $|\overrightarrow{P}+\overrightarrow{Q}|=\left|\overrightarrow{P}\right|$
Let $\theta$ be the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$.
$\Rightarrow P^2+Q^2+2PQ\cos \theta =P^2$
$\Rightarrow Q+2P\cos \theta =0$
$\Rightarrow \cos \theta =-\frac{Q}{2P}$


$\tan \alpha =\frac{2P\sin \theta }{2P\cos \theta +Q}=\infty$
$[\because 2P\cos \theta +Q=0]$
or $\alpha ={90}^{\circ }$

Alternate solution:
From the figure below, making use of the fact that $|\overrightarrow{P}|=|\overrightarrow{P}+\overrightarrow{Q}|$


Angle between $(2\overrightarrow{P}+\overrightarrow{Q})$ and $\overrightarrow{Q}$ is ${90}^{\circ }$.