Question

O(0,0,0), A(3,1,4), B(1,3,2) and C(0,4,-2) are the vertices of a tetrahedron. If G is the centroid of the tetrahedron and \(G_1\) is the centroid of its face ABC, then the point which divides \(GG_1\) 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. \(\left( \frac{20}{3}, \frac{10}{3}, \frac{10}{3} \right)\)

Hint:

The centroid of a tetrahedron divides the line segment joining a vertex to the centroid of the opposite face in the ratio 3:1. This geometric property can verify calculations.

Answer 1

The Correct Option is C

Step 1: Find Centroid G of Tetrahedron OABC:
\[ G = \frac{O + A + B + C}{4} \] \(x_G = \frac{0+3+1+0}{4} = 1\) \(y_G = \frac{0+1+3+4}{4} = 2\) \(z_G = \frac{0+4+2-2}{4} = 1\) \(G = (1, 2, 1)\).
Step 2: Find Centroid \(G_1\) of Face ABC:
\[ G_1 = \frac{A + B + C}{3} \] \(x_{G1} = \frac{3+1+0}{3} = \frac{4}{3}\) \(y_{G1} = \frac{1+3+4}{3} = \frac{8}{3}\) \(z_{G1} = \frac{4+2-2}{3} = \frac{4}{3}\) \(G_1 = (\frac{4}{3}, \frac{8}{3}, \frac{4}{3})\).
Step 3: Find Point P dividing \(GG_1\) in ratio 1:2:
Using section formula \(P = \frac{1 \cdot G_1 + 2 \cdot G}{1 + 2}\): \(x_P = \frac{1(4/3) + 2(1)}{3} = \frac{4/3 + 6/3}{3} = \frac{10/3}{3} = \frac{10}{9}\). \(y_P = \frac{1(8/3) + 2(2)}{3} = \frac{8/3 + 12/3}{3} = \frac{20/3}{3} = \frac{20}{9}\). \(z_P = \frac{1(4/3) + 2(1)}{3} = \frac{4/3 + 6/3}{3} = \frac{10/3}{3} = \frac{10}{9}\). \(P = (\frac{10}{9}, \frac{20}{9}, \frac{10}{9})\).