Question

The resultant of $\overrightarrow{P}$ and $\overrightarrow{Q}$ is perpendicular to $\overrightarrow{P}$. What is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$

• A. ${\cos }^{-1}\left(\frac{P}{Q}\right)$
• B. ${\cos }^{-1}(-\frac{P}{Q})$
• C. ${\sin }^{-1}\left(\frac{P}{Q}\right)$
• D. ${\sin }^{-1}(-\frac{P}{Q})$

Answer 1

The correct option is B ${\cos }^{-1}(-\frac{P}{Q})$

Let $\overrightarrow{R}$ be the resultant vector and $\alpha$ be the angle between $\overrightarrow{R}$ and $\overrightarrow{P}$
Using formula,
$\tan \alpha =\frac{b\sin \theta }{a+b\cos \theta }$
$\tan 90=\frac{Q\sin \theta }{P+Q\cos \theta }$
$\Rightarrow P+Q\cos \theta =0$
$\cos \theta =\frac{-P}{Q}$
$\theta ={\cos }^{-1}\left(\frac{-P}{Q}\right)$