TS EAMCET · Maths · Three Dimensional Geometry
\(\mathrm{O}(0,0,0), \mathrm{A}(3,1,4), \mathrm{B}(1,3,2)\) and \(\mathrm{C}(0,4,-2)\) are the vertices of a tetrahedron. If G is the centroid of the tetrahedron and \(\mathrm{G}_1\) is the centroid of its face ABC, then the point which divides \(\mathrm{GG}_1\) in the ratio \(1: 2\) is
- A \(\left(\frac{10}{3}, \frac{20}{3}, \frac{10}{3}\right)\)
- B \(\left(\frac{20}{9}, \frac{10}{9}, \frac{10}{9}\right)\)
- C \(\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right)\)
- D \(\left(\frac{20}{3}, \frac{10}{3}, \frac{10}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right)\)
Step-by-step Solution
Detailed explanation
\( \mathrm{G}_1 = \left(\frac{3+1+0}{3}, \frac{1+3+4}{3}, \frac{4+2-2}{3}\right) = \left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right) \) \( \mathrm{G} = \left(\frac{0+3+1+0}{4}, \frac{0+1+3+4}{4}, \frac{0+4+2-2}{4}\right) = \left(1, 2, 1\right) \)…
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