The resultant of overrightarrow P and overrig | vedclass

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Question

(B) Let $	heta$ be the angle between vectors $\overrightarrow P$ and $\overrightarrow Q$.The resultant vector $\overrightarrow R = \overrightarrow P

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $	heta$ be the angle between vectors $\overrightarrow P$ and $\overrightarrow Q$.The resultant vector $\overrightarrow R = \overrightarrow P + \overrightarrow Q$ makes an angle $\alpha$ with $\overrightarrow P$,given by $	an \alpha = \frac{Q \sin 	heta}{P + Q \cos 	heta}$.Since the resultant is perpendicular to $\overrightarrow P$,the angle $\alpha = 90^\circ$.Therefore,$	an 90^\circ = \frac{Q \sin 	heta}{P + Q \cos 	heta}$.Since $	an 90^\circ$ is undefined,the denominator must be zero:$P + Q \cos 	heta = 0$$Q \cos 	heta = -P$$\cos 	heta = -\frac{P}{Q}$$	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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