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 $\triangle A B C$
$=\left(\frac{2-3-1}{3}, \frac{3+3+4}{3}, \frac{-4-2+2}{3}\right)$
$=\left(\frac{-2}{3}, \frac{10}{3}, \frac{-4}{3}\right)$
$F =$ centroid of $\triangle A B D $
$=\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 $ \triangle A C D $
$=\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 $\triangle EFG$ 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) $
$=\left(\frac{4}{9}, \frac{11}{3}, \frac{-10}{9}\right)$