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

Answer 1

Answer: (B)

${\cos }^{-1}(-P/Q)$

$\Rightarrow \tan {90}^o=\frac{Q\sin \theta }{P+Q\cos \theta }\Rightarrow P+Q\cos \theta =0$
$\cos \theta =\frac{-P}{Q}\therefore \theta ={\cos }^{-1}\left(\frac{-P}{Q}\right)$