A(3,2,-1), B(4,1,1), C(6,2,5) and D(3,3,3) ar | vedclass

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(A) The centroid $G$ of a tetrahedron with vertices $A, B, C, D$ is given by $G = \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_

Vedclass · Accepted Answer

$(4, 2, 2)$

Vedclass · Answer

(A) The centroid $G$ of a tetrahedron with vertices $A, B, C, D$ is given by $G = \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.For the given points $A(3,2,-1), B(4,1,1), C(6,2,5), D(3,3,3)$,the centroid $G$ is:$G = \left(\frac{3+4+6+3}{4}, \frac{2+1+2+3}{4}, \frac{-1+1+5+3}{4}\right) = \left(\frac{16}{4}, \frac{8}{4}, \frac{8}{4}\right) = (4, 2, 2)$.The lines joining a vertex of a tetrahedron to the centroid of the opposite face are concurrent at the centroid of the tetrahedron.Thus,the lines $AG_1, BG_2, CG_3$ and $DG_4$ are concurrent at the point $(4, 2, 2)$.

$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$ are four points. $G_1, G_2, G_3$ and $G_4$ are the centroids of the triangles $\triangle BCD, \triangle CDA, \triangle DAB$ and $\triangle ABC$ respectively. The point of concurrence of the lines $AG_1, BG_2, CG_3$ and $DG_4$ is

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