$A(3,2,-1), B(4,1,1), C(6,2,5)$ are three points. If $D, E, F$ are three points which divide $B C$, $C A, A…

$A(3,2,-1), B(4,1,1), C(6,2,5)$ are three points. If $D, E, F$ are three points which divide $B C$, $C A, A B$ respectively in the same ratio $2: 1$, then the centroid of $\triangle D E F$ is
  1. $\left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$
  2. $(13,5,5)$
  3. $(4,2,1)$
  4. $\left(\frac{11}{3}, \frac{4}{3}, \frac{1}{3}\right)$

Solution


Here, $D=\left(\frac{2 \times 6+4 \times 1}{3}, \frac{2 \times 2+1 \times 1}{3}, \frac{2 \times 5+1 \times 1}{3}\right)$ $=\left(\frac{16}{3}, \frac{5}{3}, \frac{11}{3}\right)$ Similarly, $E=\left(\frac{12}{3}, \frac{6}{3}, \frac{3}{3}\right) \Rightarrow F=\left(\frac{11}{3}, \frac{4}{3}, \frac{1}{3}\right)$ Let centroid of $\triangle D E F$ is $O(x, y, z)$. $ \begin{aligned} & \therefore x=\frac{\frac{16}{3}+\frac{12}{3}+\frac{11}{3}}{3}=\frac{13}{3} \\ & y=\frac{\frac{5}{3}+\frac{6}{3}+\frac{4}{3}}{3}=\frac{5}{3} \text { and } z=\frac{\frac{11}{3}+1+\frac{1}{3}}{3}=\frac{5}{3} \end{aligned} $ Hence, coordinates are $\left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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