O(0,0,0), A(3,1,4), B(1,3,2) and C(0,4,-2) ar | vedclass

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(C) The vertices of the tetrahedron are $O(0,0,0), A(3,1,4), B(1,3,2)$,and $C(0,4,-2)$.The centroid $G$ of the tetrahedron is given by the average of

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$\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right)$

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(C) The vertices of the tetrahedron are $O(0,0,0), A(3,1,4), B(1,3,2)$,and $C(0,4,-2)$.The centroid $G$ of the tetrahedron is given by the average of its vertices: $G = \left(\frac{0+3+1+0}{4}, \frac{0+1+3+4}{4}, \frac{0+4+2-2}{4}\right) = \left(1, 2, 1\right)$.The centroid $G_1$ of the face $ABC$ is given by the average of vertices $A, B$,and $C$: $G_1 = \left(\frac{3+1+0}{3}, \frac{1+3+4}{3}, \frac{4+2-2}{3}\right) = \left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right)$.We need to find the point $P$ that divides the line segment $GG_1$ in the ratio $1:2$. Using the section formula,$P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$ with $m=1, n=2$,$G(1,2,1)$ and $G_1(\frac{4}{3}, \frac{8}{3}, \frac{4}{3})$:$x = \frac{1(\frac{4}{3}) + 2(1)}{1+2} = \frac{\frac{4}{3} + 2}{3} = \frac{10}{9}$.$y = \frac{1(\frac{8}{3}) + 2(2)}{1+2} = \frac{\frac{8}{3} + 4}{3} = \frac{20}{9}$.$z = \frac{1(\frac{4}{3}) + 2(1)}{1+2} = \frac{\frac{4}{3} + 2}{3} = \frac{10}{9}$.Thus,the point is $\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right)$.

$O(0,0,0), A(3,1,4), B(1,3,2)$ and $C(0,4,-2)$ are the vertices of a tetrahedron. If $G$ is the centroid of the tetrahedron and $G_1$ is the centroid of its face $ABC$,then the point which divides $GG_1$ in the ratio $1:2$ is

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