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.
Best Answer
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.