Question

A $(2,3,-4)$, B $(-3,3,-2)$, C $(-1,4,2)$ and D $(3,5,1)$ are the vertices of a tetrahedron. If $E,F,G$ are the centroids of its faces containing the point $A$, then the centroid of the triangle $EFG$ is

• A. $\left(\frac{1}{9},\frac{15}{9},\frac{-3}{9}\right)$
• B. $\left(\frac{1}{4},\frac{15}{4},\frac{-3}{4}\right)$
• C. $\left(\frac{4}{9},\frac{11}{3},\frac{-10}{9}\right)$
• D. $\left(\frac{-1}{9},\frac{12}{9},\frac{1}{9}\right)$

Answer 1

Answer: (C)

We have,
$A(2,3,-4),B(-3,3,-2),C(-1,4,2),D(3,5,1)$
Now, $E=$ centroid of $△ABC$
$=\left(\frac{2-3-1}{3},\frac{3+3+4}{3},\frac{-4-2+2}{3}\right)$
$=\left(\frac{-2}{3},\frac{10}{3},\frac{-4}{3}\right)$
$F=$ centroid of $△ABD$
$=\left(\frac{2-3+3}{3},\frac{3+3+5}{3},\frac{-4-2+1}{3}\right)$
$=\left(\frac{2}{3},\frac{11}{3},\frac{-5}{3}\right)$
$G=$centroid of $△ACD$
$=\left(\frac{2-1+3}{3},\frac{3+4+5}{3},\frac{-4+2+1}{3}\right)$
$=\left(\frac{4}{3},4,\frac{-1}{3}\right)$
Let centroid of $△EFG$ is $H$
$\therefore H=\left(\frac{\frac{-2}{3}+\frac{2}{3}+\frac{4}{3}}{3},\frac{\frac{10}{3}+\frac{11}{3}+4}{3},\frac{\frac{-4}{3}\frac{-5}{3}\frac{-1}{3}}{3}\right)$
$=\left(\frac{4}{9},\frac{11}{3},\frac{-10}{9}\right)$