A(1,1,1), B(1,-4,3), C(2,-2,0) and D(8,1,4) a | vedclass

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(D) The centroid of a tetrahedron with vertices $A, B, C, D$ is given by $G = \frac{A+B+C+D}{4}$.Given $A(1,1,1), B(1,-4,3), C(2,-2,0), D(8,1,4)$.The

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$(3,-1,2)$

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(D) The centroid of a tetrahedron with vertices $A, B, C, D$ is given by $G = \frac{A+B+C+D}{4}$.Given $A(1,1,1), B(1,-4,3), C(2,-2,0), D(8,1,4)$.The centroid of the tetrahedron $ABCD$ is $G = \left(\frac{1+1+2+8}{4}, \frac{1-4-2+1}{4}, \frac{1+3+0+4}{4}\right) = \left(\frac{12}{4}, \frac{-4}{4}, \frac{8}{4}\right) = (3,-1,2)$.It is a known property that the centroid of the tetrahedron formed by the centroids of the faces of a tetrahedron is the same as the centroid of the original tetrahedron.Therefore,the centroid of the tetrahedron with vertices $G_1, G_2, G_3, G_4$ is the same as the centroid of $ABCD$,which is $(3,-1,2)$.Thus,option $D$ is correct.

$A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron. $G_1, G_2, G_3$ and $G_4$ are the centroids of the faces $ABC, BCD, CDA$ and $DAB$. Then the centroid of the tetrahedron having $G_1, G_2, G_3, G_4$ as its vertices is

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