Let $G_{1}, G_{2} a n d G_{3}$ be the centroids of the triangular faces OBC, OCA and OAB respectively of a tetrahedron OABC (where ‘O’ is the origin). If $V_{1}$ denotes the volume of the tetrahedron OABC and $V_{2}$ that of the parallelepiped with $O G_{1}, O G_{2}$ and $O G_{3}$ as three concurrent edges, then the value of $\frac{198 V_{2}}{V_{1}}$ is

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answer is 88.

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Detailed Solution

Taking ‘O’ as the origin, let the p.v’s of A, B and C be $\bar{a}, \bar{b}, \bar{c}$ respectively. Then the position vectors of $G_{1}, G_{2}, G_{3}$ are $\frac{\bar{b} + \bar{c}}{3}, \frac{\bar{c} + \bar{a}}{3}, \frac{\bar{a} + \bar{b}}{3}$

$V_{1} = \frac{1}{6} [\bar{a} \bar{b} \bar{c}] V_{2} [\vec{O G_{1}} \vec{O G_{2}} \vec{O G_{3}}] V_{2} = \frac{1}{27} [\bar{b} + \bar{c} \bar{c} + \bar{a} \bar{a} + \bar{b}] = \frac{2}{27} [\bar{a} \bar{b} \bar{c}] V_{2} = \frac{2}{27} 6 V_{1} \Rightarrow \frac{198 V_{2}}{V_{1}} = \frac{198 \times 2 \times 6}{27} = 88$