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)` .

Explanation

Taking O as the origin , let the position vectors of A,B and C be `veca , vecb and vecc`. Respectively, then the position vectors `G_(1), G_(2) and G-(3) are (vecb +vecc)/3,(vecc + veca)/(3) and (veca+vecb)/3` , respectively. Therefore,
`V_(1)=1/6[vecavecbvecc] and V_(2)=[vec(OG_(1))" "vec(OG_(2))" "vec(OG_(3))]`
`now,V_(2)=[vec(OG_(1))" "vec(OG_(2))" "vec(OG_(3))]`
`= 1/27 [ vecb + vecc vecc+veca veca +vecb]`
`=2/27 [ veca vecb vecc]`
`=2/27 xx6V_(1)or 9V_(2)=4V_(1)`