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$$
A(2,3,-4), B(-3,3,-2), C(-1,4,2), D(3,5,1)
$$
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
Options:
A(2,3,-4), B(-3,3,-2), C(-1,4,2), D(3,5,1)
$$
are the vertices of a tetrahedron. If $G_1, G_2$ and $G_3$ are the centroids of the three faces having the vertex $D$ in common, then the centroid of the $\Delta G_1 G_2 G_3$ is
Solution:
2830 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{5}{9}, \frac{35}{9}, \frac{-5}{9}\right)$
Let $G_1$ be the centroid of face $A B D$. $G_2$ be the centroid of face $B C D$. $G_3$ be the centroid of face $A C D$.
$$
\begin{aligned}
\therefore G_1 & =\left(\frac{2}{3}, \frac{11}{3},-\frac{5}{3}\right) ; G_2=\left(-\frac{1}{3}, \frac{12}{3}, \frac{1}{3}\right) \\
G_2 & =\left(\frac{4}{3}, \frac{12}{3},-\frac{1}{3}\right)
\end{aligned}
$$
Let $G$ be the centroid the $\Delta G_1 G_2 G_3$
$$
\begin{aligned}
& \therefore G=\left(\frac{\frac{2}{3}-\frac{1}{3}+\frac{4}{3}}{3}, \frac{\frac{11}{3}+\frac{12}{3}+\frac{12}{3}}{3}, \frac{-\frac{5}{3}+\frac{1}{3}-\frac{1}{3}}{3}\right) \\
& \Rightarrow G=\left(\frac{5}{9}, \frac{35}{9},-\frac{5}{9}\right)
\end{aligned}
$$
$$
\begin{aligned}
\therefore G_1 & =\left(\frac{2}{3}, \frac{11}{3},-\frac{5}{3}\right) ; G_2=\left(-\frac{1}{3}, \frac{12}{3}, \frac{1}{3}\right) \\
G_2 & =\left(\frac{4}{3}, \frac{12}{3},-\frac{1}{3}\right)
\end{aligned}
$$
Let $G$ be the centroid the $\Delta G_1 G_2 G_3$
$$
\begin{aligned}
& \therefore G=\left(\frac{\frac{2}{3}-\frac{1}{3}+\frac{4}{3}}{3}, \frac{\frac{11}{3}+\frac{12}{3}+\frac{12}{3}}{3}, \frac{-\frac{5}{3}+\frac{1}{3}-\frac{1}{3}}{3}\right) \\
& \Rightarrow G=\left(\frac{5}{9}, \frac{35}{9},-\frac{5}{9}\right)
\end{aligned}
$$
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