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I have 5 points and I want the affine space of those. I also want to describe the convex envelope of the set of points. I searched on google and found how to do the affine space with 3 points. However I'm not so sure about 5 points.

$$IE = { A(-4; 2;-8); B(1;-1/2; 2); C(1;-1; 2); D(2;-3/2; 4);E(-1; 0;-2) }$$

Now I'm wondering if it looks like $$ A = (A-B)+(B-C)+(C-D)+(D-E)$$

Or should I made a bunch of vector, I mean, a vector between each points ? $$ A = (A-B)+(A-C)+(A-D)+(A-E)+(B-C)+(B-D)+(B-E)+(C-D)+(C-E)$$

Then for the barrycenter, I add all my vectors together and / 5. Right ?

Finally, I have to find if the convex envelope to a plan or a line by finding a base of my set A. Any tip ?


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up vote 1 down vote accepted

Note first: your equality $A=(A-B)+(B-C)+(C-D)+(D-E)$ doesn't make any sense! The same goes for your next equation. So I will try to explain you the terms you have used:

By "affine space", I assume you mean an affine plane in $\mathbb{R}^3$. Note that in 3-dimensional space any affine plane is determined by 3 non-colinear points (meaning they don't all lie on one straight line). It follows that if you have 5 points, you can in general not find a plane in which they all lie.

In contrast, we may always find the convex envelope of a given finite number of points in 3-space. The convex envelope of the points $P_1,\dots,P_n$ consists of all points of the form

$$t_1\cdot P_1+\cdots+t_n\cdot P_n$$

where $t_1,\dots,t_n$ are real non-negative parameters such that $t_1+t_2+\cdots+t_n=1$. This is the smallest convex set that contains the points $P_1,\dots,P_n$.

The barycenter of the points $P_1,\dots,P_n$ is given by $$C=\frac{P_1+\dots+P_n}{n}.$$

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