# Centroid of a Tetrahedron

## Centroid of a Tetrahedron

Tetrahedron: Let ABC be a triangle and D is a point in the space which is not in the plane of the triangle ABC. Then ABCD is called a tetrahedron.

Note:

• The tetrahedron ABCD has four faces, namely ∆ABC, ∆ACD, ∆ABD, ∆BCD. It has four vertices, namely A, B, C, D and it has six edges, namely AB, AC, AD, BC, BD, CD.
• The Centroid G of the tetrahedron ABCD divides the line joining any vertex to the Centroid of its opposite face in the ratio 3:1

Centroid of a Tetrahedron: The centroid of the tetrahedron formed by the points A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃), D(x₄, y₄, z₄) is $$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right)$$.

Proof: The centroid of ∆BCD is $${{G}_{1}}=\left( \frac{{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{3},\frac{{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{3},\frac{{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{3} \right)$$.

Since the centroid of tetrahedron ABCD divides AG₁ in the ratio 3:1, it follows that

$$G=\left( \frac{3\left( \frac{{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{3} \right)+1({{x}_{1}})}{4},\frac{3\left( \frac{{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{3} \right)+1({{y}_{1}})}{4},\frac{3\left( \frac{{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{3} \right)+1({{z}_{1}})}{4} \right)$$.

$$=\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right)$$.

Example: Find the centroid of the tetrahedron formed by the point (2, 3, 4), (-3, -2, 1), (-1, -4, 1) and (3, 5, 1).

Solution: Given that (2, 3, 4), (-3, -2, 1), (-1, -4, 1) and (3, 5, 1).

The centroid of the tetrahedron formed by the points A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃), D(x₄, y₄, z₄) is $$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right)$$.

$$G=\left( \frac{2-3-1+3}{4},\frac{3-2-4+5}{4},\frac{4+1+1+1}{4} \right)$$.

$$=\left( \frac{1}{4},\frac{1}{2},\frac{7}{4} \right)$$.