JEE Main · 2005 · 08 May 2005 · Offline · Vector Algebra

Take $F$ along a reference direction and let the larger force $F^{\prime }$ make an angle $\theta$ with it. The resultant is $(F+F^{\prime }\cos \theta ,F^{\prime }\sin \theta )$; for it to be perpendicular to $F$ its component along $F$ must vanish, so $F+F^{\prime }\cos \theta =0$. The resultant then has magnitude $F^{\prime }\sin \theta$, and this equals $\frac{F^{\prime }}{3}$, so $\sin \theta =\frac{1}{3}$. From $\cos \theta =-\frac{F}{F^{\prime }}$ and ${\sin }^2\theta +{\cos }^2\theta =1$ we get $\frac{F^2}{F^{\prime 2}}=1-\frac{1}{9}=\frac{8}{9}$, hence $\frac{F^{\prime }}{F}=\frac{3}{2\sqrt{2}}$. The larger force to the smaller is therefore $3:2\sqrt{2}$.