Q102 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

Let $A=(1,2,0), B=(2,0,-1), C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in the space. Let $\mathrm{G}_1$ be the centroid of triangle $A B C$ and $G_2$ be the centroid of tetrahedron ABCD. If $P$ divides $G_1 G_2$ in the ratio $4: 3$ internally then $\mathrm{P}=$

  • Correct answer
    A
    $\left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right)$
  • B
    $\left(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}\right)$
  • C
    $\left(\frac{4}{7}, \frac{-2}{7}, \frac{1}{7}\right)$
  • D
    $\left(\frac{1}{7}, \frac{-3}{7}, \frac{5}{7}\right)$

Solution

Given points are $\mathrm{A}=(1,2,0), \mathrm{B}=(2,0,-1)$, $\mathrm{C}=(0,-2,3)$ and $\mathrm{D}=(-1,2,-3)$ Here $\mathrm{G}_1=(1,0,2 / 3)$ and $\mathrm{G}_2=(1 / 2,1 / 2,-1 / 4)$ Now $P$ divides $G_1 G_2$ in the ratio $4: 3$ internally So, $\mathrm{P}=(5 / 7,2 / 7,1 / 7)$

Hence, the correct answer is option: 'A'.

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