Let A(1, 2, 3), B(-1, 4, 6), C(0, -6, 4), and | vedclass

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(B) The centroid $G$ of a tetrahedron with vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$, and $D(x_4, y_4, z_4)$ is given by $\l

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$\frac{4}{3}$

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(B) The centroid $G$ of a tetrahedron with vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$, and $D(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.For the given vertices $A(1, 2, 3)$, $B(-1, 4, 6)$, $C(0, -6, 4)$, and $D(1, 1, 1)$, the centroid $G$ is $\left(\frac{1-1+0+1}{4}, \frac{2+4-6+1}{4}, \frac{3+6+4+1}{4}\right) = \left(\frac{1}{4}, \frac{1}{4}, \frac{14}{4}\right)$.The centroid $G_1$ of the face $BCD$ is given by $\left(\frac{x_2+x_3+x_4}{3}, \frac{y_2+y_3+y_4}{3}, \frac{z_2+z_3+z_4}{3}\right) = \left(\frac{-1+0+1}{3}, \frac{4-6+1}{3}, \frac{6+4+1}{3}\right) = \left(0, -\frac{1}{3}, \frac{11}{3}\right)$.In a tetrahedron, the centroid $G$ divides the line segment joining a vertex to the centroid of the opposite face in the ratio $3:1$. Specifically, $G$ lies on the median $AG_1$ such that $AG : GG_1 = 3 : 1$.This implies $AG = \frac{3}{4} AG_1$, or $\frac{AG_1}{AG} = \frac{4}{3}$.

Let $A(1, 2, 3)$, $B(-1, 4, 6)$, $C(0, -6, 4)$, and $D(1, 1, 1)$ be the vertices of a tetrahedron. Let $G$ be its centroid and $G_1$ be the centroid of its face $BCD$. Then $\frac{AG_1}{AG} =$

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