Question

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

• A. $P+Q$
• B. $P$
• C. $P-Q$
• D. $Q$

Answer 1

The correct option is D $Q$

$\text{Step 1: Initial resultant between two vectors [Ref. Fig. 1]}$

Resultant of $\overrightarrow{P}$ and $\overrightarrow{Q},$

${\left|\overrightarrow{R}\right|}^2={\left|\overrightarrow{P}\right|}^2+{\left|\overrightarrow{Q}\right|}^2+2\left|\overrightarrow{P}\right|\left|\overrightarrow{Q}\right|\cos \theta$ $....\left(1\right)$

$\text{Step 2: Dot product between new resultant and}\overrightarrow{P}$ $\text{[Ref. Fig. 2]}$

When $\overrightarrow{Q}$ is doubled, new resultant, ${\overrightarrow{R}}_1$ $=\overrightarrow{P}+2\overrightarrow{Q}$ becomes perpendicular to $\overrightarrow{P}$

$\therefore {\overrightarrow{R}}_1.\overrightarrow{P}=0$

$\Rightarrow$ $(\overrightarrow{P}+2\overrightarrow{Q}).\overrightarrow{P}=0$

$\Rightarrow$ $|\overrightarrow{P}{|}^2+2|\overrightarrow{P}||\overrightarrow{Q}|\cos \theta =0$

$\Rightarrow$ $\cos \theta =\frac{-|\overrightarrow{P}|}{2|\overrightarrow{Q}|}$

$\text{Step 3: Solving equation}$

Putting value of $\cos \theta$ in equation $\left(1\right)$

$|\overrightarrow{R}{|}^2=|\overrightarrow{P}{|}^2+|\overrightarrow{Q}{|}^2+2|\overrightarrow{P}||\overrightarrow{Q}|\left[\frac{-|\overrightarrow{P}|}{2|\overrightarrow{Q}|}\right]$

$\Rightarrow$ $|\overrightarrow{R}{|}^2=|\overrightarrow{Q}{|}^2$

$\Rightarrow$ $|\overrightarrow{R}|=|\overrightarrow{Q}|$

Hence, Option(D) is correct.