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

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(C) Point $E$ is the centroid of face $ABC$ (which is a triangle).$	ext{Centroid} = \left[\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2

Vedclass · Accepted Answer

$\left(\frac{4}{9}, \frac{11}{3}, \frac{-10}{9}\right)$

Vedclass · Answer

(C) Point $E$ is the centroid of face $ABC$ (which is a triangle).$	ext{Centroid} = \left[\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight]$$E = \left[\frac{2+(-3)+(-1)}{3}, \frac{3+3+4}{3}, \frac{-4+(-2)+2}{3}ight] = \left[\frac{-2}{3}, \frac{10}{3}, \frac{-4}{3}ight]$Point $F$ is the centroid of face $ACD$.$F = \left[\frac{2+(-1)+3}{3}, \frac{3+4+5}{3}, \frac{-4+2+1}{3}ight] = \left[\frac{4}{3}, \frac{12}{3}, \frac{-1}{3}ight] = \left[\frac{4}{3}, 4, \frac{-1}{3}ight]$Point $G$ is the centroid of face $ABD$.$G = \left[\frac{2+(-3)+3}{3}, \frac{3+3+5}{3}, \frac{-4+(-2)+1}{3}ight] = \left[\frac{2}{3}, \frac{11}{3}, \frac{-5}{3}ight]$Now, $E, F, G$ form a triangle. The centroid of $	riangle EFG$ is given by the average of the coordinates of $E, F,$ and $G$.$	ext{Centroid of } 	riangle EFG = \left[\frac{\frac{-2}{3}+\frac{4}{3}+\frac{2}{3}}{3}, \frac{\frac{10}{3}+4+\frac{11}{3}}{3}, \frac{\frac{-4}{3}+\left(\frac{-1}{3}ight)+\left(\frac{-5}{3}ight)}{3}ight]$$= \left[\frac{\frac{4}{3}}{3}, \frac{\frac{10+12+11}{3}}{3}, \frac{\frac{-10}{3}}{3}ight] = \left[\frac{4}{9}, \frac{33}{9}, \frac{-10}{9}ight] = \left[\frac{4}{9}, \frac{11}{3}, \frac{-10}{9}ight]$

$A(2,3,-4), B(-3,3,-2), C(-1,4,2)$ and $D(3,5,1)$ are the vertices of a tetrahedron. If $E, F, G$ are the centroids of its faces containing the point $A$, then the centroid of the triangle $EFG$ is

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