The resultant of two vectors `vecP andvecQ is vecR`. If the magnitude of `vecQ` is doudled, the new resultant becomes perpendicuar to `vecP`. Then the magnitude of `vecR` is :

Options

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

Answer

B

Explanation

When Q is doubled then `vecQ` becomes `2vecQ`. Suppose that the new resultant becomes `vecR`'
`therefore vecR'=vecP+2vecQ & vecR' bot vecP`
From `tan alpha=(B sin theta)/(A+B cos theta)`
`rArr tan90^(@)=(2Q sin theta)/(P+2Q cos theta) rArr (1)/(0)= (2Q sin theta)/(P+2 Q cos theta)`
` rArr P +2Q cos theta=0 rArr 2Q cos theta= -P`
`therefore cos theta=(-P)/(2Q)" "` ......(i)
Now, according to question
`vecR=vecP+vecQ`
`therefore R=sqrt(P^(2)+Q^(2)+2PQ cos theta)`
Putting value of `costheta` from equation number (i)
`R=sqrt(P^(2)+Q^(2)+2PQ(-(P)/(2Q))`
`rArr R=sqrt(P^(2)+Q^(2)-P^(2))rArr R=sqrt(Q^(2)) `
`therefore R=Q`