Let A=(1,2,0), B=(2,0,-1), C=(0,-2,3) and D=( | vedclass

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(A) The centroid $G_1$ of triangle $ABC$ is given by $\left(\frac{1+2+0}{3}, \frac{2+0-2}{3}, \frac{0-1+3}{3}\right) = \left(1, 0, \frac{2}{3}\right)$

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$\left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right)$

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(A) The centroid $G_1$ of triangle $ABC$ is given by $\left(\frac{1+2+0}{3}, \frac{2+0-2}{3}, \frac{0-1+3}{3}\right) = \left(1, 0, \frac{2}{3}\right)$.The centroid $G_2$ of tetrahedron $ABCD$ is given by $\left(\frac{1+2+0-1}{4}, \frac{2+0-2+2}{4}, \frac{0-1+3-3}{4}\right) = \left(\frac{2}{4}, \frac{2}{4}, \frac{-1}{4}\right) = \left(\frac{1}{2}, \frac{1}{2}, -\frac{1}{4}\right)$.Point $P$ divides $G_1G_2$ in the ratio $4:3$ internally. Using the section formula $\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n}\right)$ with $m=4, n=3$:$x = \frac{4(1/2) + 3(1)}{4+3} = \frac{2+3}{7} = \frac{5}{7}$.$y = \frac{4(1/2) + 3(0)}{4+3} = \frac{2+0}{7} = \frac{2}{7}$.$z = \frac{4(-1/4) + 3(2/3)}{4+3} = \frac{-1+2}{7} = \frac{1}{7}$.Thus, $P = \left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right)$.

Let $A=(1,2,0)$, $B=(2,0,-1)$, $C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in space. Let $G_1$ be the centroid of triangle $ABC$ and $G_2$ be the centroid of tetrahedron $ABCD$. If $P$ divides $G_1G_2$ in the ratio $4:3$ internally, then $P=$

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