Let `G_1, G_2a n dG_3` be the centroids of the triangular faces `O B C ,O C Aa n dO A B ,` respectively, of a tetrahedron `O A B Cdot` If `V_1` denotes the volumes of the tetrahedron `O A B Ca n dV_2` that of the parallelepiped with `O G_1,O G_2a n dO G_3` as three concurrent edges, then prove that `4V_1=9V_1dot`

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)`