question_answer

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

A)                 \[{{\cos }^{-1}}(P/Q)\]

B)                             \[{{\cos }^{-1}}(-P/Q)\]

C)                 \[{{\sin }^{-1}}\,(P/Q)\]

D)                 \[{{\sin }^{-1}}\,(-P/Q)\]

Correct Answer: B

Solution :

                                    Þ \[\tan 90{}^\circ =\frac{Q\sin \theta }{P+Q\cos \theta }\]Þ \[P+Q\cos \theta =0\]                 \[\cos \theta =\frac{-P}{Q}\]\ \[\theta ={{\cos }^{-1}}\left( \frac{-P}{Q} \right)\]