The resultant of overrightarrow P and overri | vedclass

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Question

(b) $	an 90^\circ = \frac{{Q\sin 	heta }}{{P + Q\cos 	heta }}$

$&rArr;$ $P + Q\cos 	heta = 0$

$\cos 	heta = \frac{{ - P}}{Q}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $	an 90^\circ = \frac{{Q\sin 	heta }}{{P + Q\cos 	heta }}$

$&rArr;$ $P + Q\cos 	heta = 0$

$\cos 	heta = \frac{{ - P}}{Q}$ 

$&rArr;$ $	heta = {\cos ^{ - 1}}\left( {\frac{{ - P}}{Q}} ight)$

The resultant of $\overrightarrow P $ and $\overrightarrow Q $ is perpendicular to $\overrightarrow P $. What is the angle between $\overrightarrow P $ and $\overrightarrow Q $

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