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

Options

A. 2
B. `3/2`
C. 3
D. `5/2`

Answer

C

Explanation

`vec(OD) = (veca+vec(2b))/(3) = -veca`, `(therefore veca+vec2b+vec3c=vec0)`
`rArr |OD|=|OC|`
Now `(ar(triangleACD))(ar(triangleBCD))=2`
`rArr ar(triangleACD)=(2Delta)/(3)`, where `Delta` is area of `triangleABC`
`ar(triangleAOC)=1/2ar(triangleACD)`
`=1/2 xx (2Delta)/(3) = Delta/3`
`rArr (ar(triangleABC))/(ar(triangleAOC))=3`