Let `G_(1),G_(2) and G_(3)` be the centroids of the trianglular faces OBC,OCA and OAB, respectively, of a tetrahedron OABC. If `V_(1)` denotes the volume of the tetrahedron OABC and `V_(2)` that of the parallelepiped with `OG_(1),OG_(2) and OG_(3)` as three concurrent edges, then prove that `4V_(1)=9V_(2)` .

Options

A. `4V_(1)=9V_(2)`
B. `9V_(1)=4V_(2)`
C. `3V_(1)=2V_(2)`
D. `3V_(2)=2V_(1)`

Answer

A

Explanation

Taking `O` as the origin let the position vectors of `A,B` and `C` be `veca,vecb` and `vecc` respectively. Thenthe position vectors of `G_(1),G_(2)` and `G_(3)` are `(vecb+vecc)/3,(vecc+veca)/3` and `(veca+vecb)/3` respectively.
`:.V_(1)=1/6[(veca,vecb, vecc)]` and `V_(2)=[(vec(OG_(1)),vec(OG_(2)),vec(OG_(3)))]`
`V_(2)=[(vec(OG_(1)),vec(OG_(2)),vec(OG_(3)))]`
`implies V_(2)=[((vecb+vecc)/3,(vecc+veca)/3,(veca+vecb)/3)]`
`impliesV_(2)=1/27[(vecb+vecc,vecc+veca,veca+vecb)]`
`impliesV_(2)=2/27[(veca,vecb,vecc)]impliesV_(2)=2/27xx6V_(1)=9V_(2)=4V_(1)`