Question

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

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

Answer 1

Answer: (B)

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

Answer 2

⇒ $\tan 90{^\circ} = \frac{Q\sin\theta}{P + Q\cos\theta}$⇒ $P + Q\cos\theta = 0$

$\cos\theta = \frac{- P}{Q}$∴ $\theta = \cos^{- 1}\left( \frac{- P}{Q} \right)$