If G(2, -1, 2) is the centroid of tetrahedron | vedclass

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(C) Let the coordinates of the vertices be $O(0, 0, 0)$,$A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.The centroid $G$ of the tetrahedr

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$4$

Vedclass · Answer

(C) Let the coordinates of the vertices be $O(0, 0, 0)$,$A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.The centroid $G$ of the tetrahedron $OABC$ is given by $G = \left(\frac{0+x_1+x_2+x_3}{4}, \frac{0+y_1+y_2+y_3}{4}, \frac{0+z_1+z_2+z_3}{4}ight) = (2, -1, 2)$.This implies $\frac{x_1+x_2+x_3}{4} = 2 \Rightarrow x_1+x_2+x_3 = 8$,$\frac{y_1+y_2+y_3}{4} = -1 \Rightarrow y_1+y_2+y_3 = -4$,and $\frac{z_1+z_2+z_3}{4} = 2 \Rightarrow z_1+z_2+z_3 = 8$.The centroid $G_1$ of $	riangle ABC$ is given by $G_1 = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$.Substituting the values,we get $G_1 = \left(\frac{8}{3}, \frac{-4}{3}, \frac{8}{3}ight)$.The distance $\left|\overline{O G_1}ight| = \sqrt{\left(\frac{8}{3}ight)^2 + \left(-\frac{4}{3}ight)^2 + \left(\frac{8}{3}ight)^2} = \sqrt{\frac{64}{9} + \frac{16}{9} + \frac{64}{9}} = \sqrt{\frac{144}{9}} = \sqrt{16} = 4$.

If $G(2, -1, 2)$ is the centroid of tetrahedron $OABC$ where $O=(0, 0, 0)$ and $G_1$ is the centroid of $\triangle ABC$,then $\left|\overline{O G_1}\right|=$

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