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$\mathrm{A}(1,1,1), \mathrm{B}(1,-4,3), \mathrm{C}(2,-2,0)$ and $\mathrm{D}(8,1,4)$ are the vertices of a tetrahedron. $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3$ and $\mathrm{G}_4$ are the centroids of the faces $\mathrm{ABC}, \mathrm{BCD}, \mathrm{CDA}$ and $\mathrm{DAB}$. Then the centroid of the tetrahedron having $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3, \mathrm{G}_4$ as its vertices is
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2798 Upvotes
Verified Answer
The correct answer is:
$(3,-1,2)$
Given vertices $\mathrm{A}(1,1,1), \mathrm{B}(1,-4,3), \mathrm{C}(2,-2,0)$ and $\mathrm{D}(8,1,4)$.
Centroid of $\triangle \mathrm{ABC}$ is $\mathrm{G}_1=\left(\frac{1+1+2}{3}, \frac{1-4-2}{3}, \frac{1+3+0}{3}\right)$
$$
\mathrm{G}_1=\left(\frac{4}{3}, \frac{-5}{3}, \frac{4}{3}\right)
$$
Similarly, $\mathrm{G}_2=\left(\frac{11}{3}, \frac{+5}{3}, \frac{7}{3}\right), \mathrm{G}_3=\left(\frac{11}{3}, 0, \frac{5}{3}\right)$ and $\mathrm{G}_4=\left(\frac{10}{3}, \frac{-2}{3}, \frac{8}{3}\right)$
Centroid of $G_1, G_2, G_3$ and $G_4$ is
$$
\begin{aligned}
& \frac{1}{4}\left\{\frac{4}{3}+\frac{11}{3}+\frac{11}{3}+\frac{10}{3}\right\},\left\{\frac{-5}{3}-\frac{5}{3}+0-\frac{2}{8}\right\} \\
& \frac{1}{4}\left\{\frac{4}{3}+\frac{7}{3}+\frac{5}{3}+\frac{8}{3}\right\} \\
& \Rightarrow\left(\frac{1}{4} \times \frac{36}{3}, \frac{1}{4} \times \frac{-12}{3}, \frac{1}{4} \times \frac{24}{3}\right) \\
& \Rightarrow(3,-1,2)
\end{aligned}
$$
So, option (d) is correct.
Centroid of $\triangle \mathrm{ABC}$ is $\mathrm{G}_1=\left(\frac{1+1+2}{3}, \frac{1-4-2}{3}, \frac{1+3+0}{3}\right)$
$$
\mathrm{G}_1=\left(\frac{4}{3}, \frac{-5}{3}, \frac{4}{3}\right)
$$
Similarly, $\mathrm{G}_2=\left(\frac{11}{3}, \frac{+5}{3}, \frac{7}{3}\right), \mathrm{G}_3=\left(\frac{11}{3}, 0, \frac{5}{3}\right)$ and $\mathrm{G}_4=\left(\frac{10}{3}, \frac{-2}{3}, \frac{8}{3}\right)$
Centroid of $G_1, G_2, G_3$ and $G_4$ is
$$
\begin{aligned}
& \frac{1}{4}\left\{\frac{4}{3}+\frac{11}{3}+\frac{11}{3}+\frac{10}{3}\right\},\left\{\frac{-5}{3}-\frac{5}{3}+0-\frac{2}{8}\right\} \\
& \frac{1}{4}\left\{\frac{4}{3}+\frac{7}{3}+\frac{5}{3}+\frac{8}{3}\right\} \\
& \Rightarrow\left(\frac{1}{4} \times \frac{36}{3}, \frac{1}{4} \times \frac{-12}{3}, \frac{1}{4} \times \frac{24}{3}\right) \\
& \Rightarrow(3,-1,2)
\end{aligned}
$$
So, option (d) is correct.
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