The resultant of two vectors `vec(P)` and `vec(Q)` is `vec(R)`. If `vec(Q)` is doubled then the new resultant vector is perpendicular to `vec(P)`. Then magnitude of `vec(R)` is :-

Options

A. `P+Q`
B. P
C. `P-Q`
D. Q

Answer

D

Explanation

`tan 90^(@)=(2Qsin theta)/(P+2Qcos theta)rArroo=(2Qsin theta)/(P+2Q cos theta)`
`rArr P+2Qcos theta=0`
Now `R^(2)=P^(2)+Q^(2)+2PQ cos theta`
`Q^(2)+P[P+2Q cos theta]`
`R^(2)=Q^(2) rArr R=Q`
Alternative method:
`vec(R )=vec(P)+vec(Q)rArr vec(P)=vec(R)-vec(Q)`
and `vec(S)=vec(P)+2vec(Q)=vec(R )-vec(Q)+2vec(Q)=vec(R )+vec(Q)`
Now, `vec(S)` and `vec(P)` are perpendicular(figure), So
`vec(S).vec(P)=0 rArr (vec(R )+vec(Q)).(vec(R )-vec(Q))=0`
`rArr vec(R )=Q^(2) rArr R=Q`