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

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} \Rightarrow \tan 90^{0} = \frac{Q \sin θ}{P + Q \cos θ} \\ \Rightarrow P + Q \cos θ = 0 \\ \cos θ = (\frac{- P}{Q}) \\ θ = \cos^{- 1} (\frac{- P}{Q}) \end{array}$

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