Vector algebra- centroid of tetrahedron

[mathisfun1]

How to find out the position vector of the centroid of tetrahedron , the position vectors of whose vertices are a,b,c,d respectively.
I am familiar with the result, namely a+b+c+d/4 but want to know how to derive it without using the 3:1 ratio property.
Any help would be appreciated. Thank you.

 

[Evgeny.Makarov]

Which definition of centroid are you using? According to Wikipedia, "the centroid of a plane figure or two-dimensional shape is the arithmetic mean ("average") position of all the points in the shape", so the fact that the position vector of the centroid is $(a+b+c+d)/4$ holds by definition in that case.

 

[mathisfun1]

Which definition of centroid are you using? According to Wikipedia, "the centroid of a plane figure or two-dimensional shape is the arithmetic mean ("average") position of all the points in the shape", so the fact that the position vector of the centroid is $(a+b+c+d)/4$ holds by definition in that case.

The centroid of a tetrahedron is the intersection of all line segments that connect each vertex to the centroid of the opposite face.

 

[Evgeny.Makarov]

The centroid of a tetrahedron is the intersection of all line segments that connect each vertex to the centroid of the opposite face.

Not the easiest definition to work with, but OK. I assume then that the centroid of a triangle is the intersection of its medians. I prefer to work from weighted points and prove that the center of mass, or barycenter, is the intersection of medians.

A weighted point is an ordered pair $(A,x)$ where $A$ is a point and $x$ is a real number. The barycenter of a set $\{(A_1,x_1),\dots,(A_n,x_n)\}$ of weighted points is a point $A$ such that
\[
\overrightarrow{OA}=\frac{1}{S}\left(x_1\overrightarrow{OA}_1+\dots+x_n\overrightarrow{OA}_n\right)
\]
where $S=x_1+\dots+x_n$ and $O$ is any point. One can show that the resulting point $A$ does not depend on the choice of $O$. Another property is that the barycenter of two points lies on the line passing through these points.

According to this definition, the barycenter of $\{(A,1),(B,1),(C,1)\}$ is a point $M$ such that
\[
\overrightarrow{OM}=\frac{1}{3}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right).
\]
But the right-hand side equals
\[
\frac{1}{3}\left(\overrightarrow{OA}+2\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right)\right).
\]
Thus, $M$ is the barycenter of $(A,1)$ and $(A',2)$ where $A'$ is the barycenter of $(B,1)$ and $(C,1)$, i.e., the center of $BC$. Therefore, $M$ lies on the median $AA'$. Similarly, $A$ lies on the other two medians, so it is the centroid of $\triangle ABC$ according to your definition.

The barycenter of a tetrahedron $ABCD$ is a point $N$ such that
\[
\overrightarrow{ON}=\frac{1}{4}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\right).
\]
But the right-hand side equals
\[
\frac{1}{4}\left(\overrightarrow{OA}+3\frac{1}{3}\left(\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\right)\right)
\]
Thus, $N$ is the barycenter of $(A,1)$ and $(A',3)$ where $A'$ is the barycenter and the centroid of $(B,1)$, $(C,1)$ and $(D,1)$. Therefore, $N$ lies on $AA'$. Similarly, $A$ lies on the other three segments connecting vertices to the centroids of the opposite faces, so it is the centroid of the tetrahedron $ABCD$ according to your definition.

 

[blue_raver22]

The position vector of the centroid of a tetrahedron can be found by taking the average of the position vectors of its four vertices. This can be derived using vector algebra by considering the centroid as the point where the three medians of the tetrahedron intersect.

To find the position vector of the centroid, we can first define the position vectors of the four vertices as r1, r2, r3, and r4 respectively. Then, we can use the midpoint formula to find the midpoint of each of the three medians, which are formed by connecting two vertices of the tetrahedron.

For example, the midpoint of the median connecting vertices r1 and r2 can be found as (r1 + r2)/2. Similarly, the midpoints of the other two medians can be found as (r1 + r3)/2 and (r1 + r4)/2.

Next, we can find the position vector of the centroid by taking the average of these three midpoints. This can be expressed as:

rC = (r1 + r2)/2 + (r1 + r3)/2 + (r1 + r4)/2

Expanding this expression and simplifying, we get:

rC = (r1 + r2 + r3 + r4)/4

This is the same result as the one you mentioned, a+b+c+d/4, which is the average of the position vectors of all four vertices. Therefore, the position vector of the centroid of a tetrahedron can be found by taking the average of the position vectors of its vertices, without using the 3:1 ratio property.

I hope this helps clarify how to derive the position vector of the centroid of a tetrahedron using vector algebra. If you have any further questions, please let me know. Thank you.