The resultant of `vecP` and `vecQ` is perpendicular to `vecP`. What is the angle between `vecP` and `vecQ`

Options

A. `((P^(2)-Q^(2))/(2PQ))`
B. Q
C. `(P)/(Q)`
D. `(P+Q)/(P-Q)`

Answer

B

Explanation

`R^(2)=P^(2)+Q^(2)+2PQ cos theta`

`tan alpha=(2Q sin theta)/({+2Q cos theta)=oo`
`cos theta=(-P)/(2Q)`
`R^(2)=P^(2)+Q^(2)+2PQ((-P))/(2Q)rArrR=Q`