Let O be an interior point of `DeltaABC` such that `bar(OA)+2bar(OB) + 3bar(OC) = 0`. Then the ratio of a `DeltaABC` to area of `DeltaAOC` is

Answer

3

Explanation

`("Area of" triangleABC)/("Area of" triangleAOC)= (1/2|vecaxxvecb+vecbxxvecc+veccxxveca|)/(1/2 |veca xxvecb|)`
now `veca + 2vecb + 3vecc = vec0`
Cross multiply with `vecb, vecaxxvecb +3vecc xxvecb=vec0`
`Rightarrow vecaxxvecb= (vecb xxvecc)`
`vecaxxvecb = 3/2(veccxxveca) = 3(vecb xxvecc)`
`Let (vecc xxveca)= vecp`
`veca xxvecb = (3vecp)/2, vecbxxvecc= vecp/2`
`Ratio = (|vecaxxvecb=vecb xxvecc +vecc xxveca|)/(|veccxxveca|)`
`(|(3vecp)/2 + vecP/2 +vecp|)/(|vecp|)`
`(3|vecp|)/|vecp|=3`