Let O be an interior point of `triangleABC` such that `vec(OA) + 2vec(OB)+ 3vec(OC)=veco` . Then find the ratio of the area of `triangleABC` to the area of `triangleBRC` is 1 unit.

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`