Let G(1), G(2) and G(3) be the centroid ...

Let `G_(1), G(2) and G_(3)` be the centroid of the triangular faces OBC, OCA and OAB of a tetrahedron OABC. If `V_(1)` denotes the volume of tetrahedron OABC and `V_(2)` that of the parallelepiped with `OG_(1), OG_(2) and OG_(3)` as three concurrent edges, then the value of `(4V_(1))/(V_2)` is (where O is the origin

`4V_(1)=9V_(2)`

`9V_(1)=4V_(2)`

`3V_(1)=2V_(2)`

`3V_(2)=2V_(1)`

Text Solution

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The correct Answer is:

Taking `O` as the origin,let the position vectors of `A,B` and `C` be `veca, vecb` and `vecc` respectively. Then, the 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))]`
Now,
`V_(2)=[(vec(OG)_(1), vec(OG)_(2),vec(OG)_(3))]`
`impliesV_(2)=[((vecb+vecc)/3, (vecc+veca)/3, (veca+vecb)/3)]`
`implies V_(2)=1/27[(vecb+vecc, vecc+veca, veca+vecb)]=2/27[(veca, vecb, vecc)]=2/27xx6V_(1)`
`implies 9V_(2)=4V_(1)`

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