Two forces P and Q acting at a point are such that if P is reveserd, the direction of the resultant is turned through `90^(@)`. Prove that the magnitudes of the forces are equal.

Explanation

`vec(R )=vec(P)+vec(Q),vec(S)= -vec(P)+vec(Q)`.Given that `vec(S)` is `_|_ vec(R )`
So, `vec(R ).vec(S)=orArr(vec(P)+vec(Q)).(-vec(P)+vec(Q))=0`
or `-P^(2)+Q^(2)=0` or `P^(2)=Q^(2) rArr P=Q`
Hence proved.
Alternatively: `beta= 90^(@)-alpha, tan alpha=(Qsin theta)/(P+Q cos theta)`
`tan beta=(Q sin(180-theta))/(P+Q cos theta(180-theta))`
`rArr tan (90^(@)-alpha)=(Q sin theta)/(P-Q cos theta)`
`rArr cot alpha =(Q sin theta)/(P-Q cos theta)`..(ii)
Multiply Eqs.(i) and(ii):`1= (Q^(2) sin^(2) theta)/(P^(2)-Q^(2) cos theta)`
simplify to get `P=Q`