Resultant of overrightarrow P and overrightarrow Q is perpendicular to overrightarrow P . angle betw

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3-1.Vectors

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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

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

Solution

(b) $\tan 90^\circ = \frac{{Q\sin \theta }}{{P + Q\cos \theta }}$

$⇒$ $P + Q\cos \theta = 0$

$\cos \theta = \frac{{ – P}}{Q}$

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

Standard 11

Physics

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