Finding center of mass for tetrahedron

Question

I am given a tetrahedron with the following points:

$$\begin{align} P_1 &= (2,0,1)\\ P_2 &= (-1,1,1)\\ P_3 &= (1,0,2)\\ P_4 &= (3,1,4) \end{align}$$

and I am tasked with finding its center of mass, M.

My attempt: $$\vec{OM} = \frac{1}{3}(\vec{OP_1} + \vec{OP_2} + \vec{OP_3})$$

$\vec{OM_1}$ is the coordinate for the center of mass of the base and $\vec{OM_2}$ is the coordinate of the center of mass of the tetrahedron. To find the center of mass of the tetrahedron, I do the following:

$$\begin{align} \vec{OM_2} &= \frac{\vec{OM} - \vec{OP_4}}{2} \\ \vec{OM_2} &= \frac{\frac{1}{3}(2,1,4) - (3,1,4)}{2} \\ \vec{OM_2} &= (-\frac{1}{3}, -\frac{1}{6}, -\frac{8}{6}) \end{align}$$

This is wrong since the answer is: $$(\frac{5}{4}, \frac{1}{2}, 2)$$

What have I done wrong? I'd really appreciate tips / explanations exactly where I am wrong.

Answer 1

Fact. Let $\{v_1,v_2,v_3,v_4\}$ be the vertices of a tetrahedron $T$. Then the center of mass of $T$ is $$ \overline x=\frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr) $$ Details of this proof can be found here.

Applying the result to your tetrahedron gives \begin{align*} \overline x &= \frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr) \\ &= \frac{1}{4}\bigl(2-1+1+3,0+1+0+1,1+1+2+4\bigr) \\ &= \frac{1}{4}(5,2,8) \end{align*} which matches your desired result.

Essentially, you need to weight each vertex evenly and you seem to have taken a weighted sum.

Answer 2

suppose first we locate the origin of coordinates at the centroid. assume the position of the centroid is a linear function of the vertices. by symmetry such a function must be

$$ \tau(x_1+x_2+x_3+x_4) $$

if we uniformly expand the tetrahedron by a factor $\lambda$ by linearity the coordinates of the centroid are also multiplied by $\lambda$. however the centroid remains at the origin. hence in this system of coordinates we must have $$ \sum_1^4 x_k=0 \tag{1} $$ now shift the origin by $a$. in the new system of coordinates the centroid is $-a$, and the vertices are $x_k-a$

hence $$ \tau\sum_1^4 (x_k-a) =-a \tag{2} $$ (1) and (2) imply $\tau=\frac14$