A) 9
B) 12
C) 27
D) 81
Correct Answer: C
Solution :
[c] Let, consider a tetrahedron with vertices 0(0, 0, 0) ; A(a, 0, 0); B(0, b, 0) & C(0, 0, c) Its volume \[(V)=\frac{1}{6}[\vec{a}\,\vec{b}\,\vec{c}]=\frac{1}{6}[\overrightarrow{OA}\,\overrightarrow{OB\,}\,\overrightarrow{OC}]\] Now, centroids of the faces OAB, OAC; OBC & ABC are \[{{G}_{1}}\left( \frac{a}{3},\frac{b}{3},0 \right);{{G}_{2}}\left( \frac{a}{3},0,\frac{c}{3} \right);{{G}_{3}}\left( 0,\frac{b}{3},\frac{c}{3} \right)\] \[\And \,{{G}_{4}}\left( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right)\]respectively Then, \[\overrightarrow{{{G}_{4}}{{G}_{1}}}=\frac{{\vec{a}}}{3};\,\overrightarrow{{{G}_{4}}{{G}_{2}}}=\frac{{\vec{b}}}{3}\And \,\overrightarrow{{{G}_{4}}{{G}_{3}}}=\frac{{\vec{a}}}{3}\] Volume of tetrahedron by centroids \[V'=\frac{1}{6}[\overrightarrow{{{G}_{4}}{{G}_{1}}}\,\overrightarrow{{{G}_{4}}{{G}_{2}}}\,\overrightarrow{{{G}_{4}}{{G}_{3}}}]=\frac{1}{6}\left[ \frac{{\vec{a}}}{3}\frac{{\vec{b}}}{3}\frac{{\vec{c}}}{3} \right]\]
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