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

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(D) Let $A=(2,3,-4), B=(-3,3,-2), C=(-1,4,2), D=(3,5,1)$.$G_1$ is the centroid of face $ABD$: $G_1 = \left(\frac{2-3+3}{3}, \frac{3+3+5}{3}, \frac{-4-

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$\left(\frac{5}{9}, \frac{35}{9}, \frac{-5}{9}\right)$

Vedclass · Answer

(D) Let $A=(2,3,-4), B=(-3,3,-2), C=(-1,4,2), D=(3,5,1)$.$G_1$ is the centroid of face $ABD$: $G_1 = \left(\frac{2-3+3}{3}, \frac{3+3+5}{3}, \frac{-4-2+1}{3}\right) = \left(\frac{2}{3}, \frac{11}{3}, -\frac{5}{3}\right)$.$G_2$ is the centroid of face $BCD$: $G_2 = \left(\frac{-3-1+3}{3}, \frac{3+4+5}{3}, \frac{-2+2+1}{3}\right) = \left(-\frac{1}{3}, 4, \frac{1}{3}\right)$.$G_3$ is the centroid of face $ACD$: $G_3 = \left(\frac{2-1+3}{3}, \frac{3+4+5}{3}, \frac{-4+2+1}{3}\right) = \left(\frac{4}{3}, 4, -\frac{1}{3}\right)$.Let $G$ be the centroid of $\Delta G_1 G_2 G_3$:$G = \left(\frac{\frac{2}{3} - \frac{1}{3} + \frac{4}{3}}{3}, \frac{\frac{11}{3} + 4 + 4}{3}, \frac{-\frac{5}{3} + \frac{1}{3} - \frac{1}{3}}{3}\right)$$G = \left(\frac{\frac{5}{3}}{3}, \frac{\frac{11+24}{3}}{3}, \frac{-\frac{5}{3}}{3}\right) = \left(\frac{5}{9}, \frac{35}{9}, -\frac{5}{9}\right)$.

$A(2,3,-4), B(-3,3,-2), C(-1,4,2), D(3,5,1)$ are the vertices of a tetrahedron. If $G_1, G_2$ and $G_3$ are the centroids of the three faces having the vertex $D$ in common,then the centroid of the $\Delta G_1 G_2 G_3$ is

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