Question

The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$. If magnitude of $\overrightarrow{Q}$ is doubled, the new resultant is perpendicular to $\overrightarrow{P}.$ Then $|\overrightarrow{R}|$ equals

• A. $|\overrightarrow{P}|$
• B. $|(\overrightarrow{P}+\overrightarrow{Q})|$
• C. $|\overrightarrow{Q}|$
• D. $|\overrightarrow{P}-\overrightarrow{Q}|$

Answer 1

The correct option is C $|\overrightarrow{Q}|$

When $Q$ is doubled, angle between new resultant and $P$, can be written as
$\text{tan}\alpha =\frac{2\text{Q sin}\theta }{P+2\text{Q cos}\theta }$
Given that $\alpha ={90}^{\circ }$
$\therefore P+2\text{Q cos}\theta =0$
$\Rightarrow \text{cos}\theta =-\frac{P}{2Q}$
$\therefore |\overrightarrow{R}|=\sqrt{P^2+Q^2+2\text{PQ cos}\theta }$
$=\sqrt{P^2+Q^2-2.\text{PQ}.\frac{P}{2Q}}$
$|\overrightarrow{R}|=|\overrightarrow{Q}|$, hence option (c) is correct.