Question

The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$ . If $\overrightarrow{Q}$ is doubled then the new resultant vector is perpendicular to $\overrightarrow{P}$ . The magnitude of $\overrightarrow{R}$ is :-

• A. $\frac{P^2-Q^2}{2PQ}$
• B. $Q$
• C. $\frac{P}{Q}$
• D. $\frac{P+Q}{P-Q}$

Answer 1

Answer: (B)

$Q$

After the vector $Q$ is doubled ,
The resultant is ${90}^o$ with $P$ hence angle between $P$ and $2Q$ is obtuse
Let the angle be $90+\theta$ and $P$ be along X-axis then resultant would be along Y-axis
Resolving $2Q$ along X-axis and Y-axis
$Qcos\theta =$resultant
$2Qsin\theta =P$
Hence, $sin\theta =\frac{P}{2Q}$
Initially the magnitude of resultant $R$ is

$=\sqrt{P^2+Q^2+2PQcos(90+\theta )}$

$=\sqrt{P^2+Q^2-2PQsin\theta }$

$=\sqrt{P^2+Q^2-2PQ\times \frac{P}{2Q}}$

$=\sqrt{Q^2}$

$=Q$