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. $co{s}^{-1}\left(\frac{P}{Q}\right)$
  
• B. $co{s}^{-1}\left(\frac{-P}{Q}\right)$
  
• C. $si{n}^{-1}\left(\frac{P}{Q}\right)$
  
• D. $si{n}^{-1}\left(\frac{-P}{Q}\right)$

Answer 1

The correct option is B

$co{s}^{-1}\left(\frac{-P}{Q}\right)$

$\Rightarrow {90}^{\circ }=\frac{Qsin\theta }{P+Qcos\theta }\Rightarrow P+Qcos\theta =0cos\theta =\frac{-P}{Q}\therefore \theta =co{s}^{-1}\left(\frac{-P}{Q}\right)$