# Visualizing euclidean distance of high dimension line in 2D

I have distances $$d_1, d_2, d_3 ... d_n$$ corresponding to each of the $$n$$ dimensions. I compute the euclidean distance as $$D = \sqrt{ d_1^2 + d_2^2 + d_3^2 + ... + d_n^2 }$$

I want to present this information in a drawing.

If $$n=2$$, I can simply represent this as a right triangle with the $$d_1$$ and $$d_2$$ being the two sides and $$D$$ being the hypotenuse.

I want to extend this for $$n>2$$ but still represent this in a $$2D$$ drawing, with a straight line from $$(0,0)$$ to $$(D,0)$$ representing the total distance. I want to place a $$2D$$ line for each of $$d_1, d_2 ...$$ of the same length such that they start at $$(0,0)$$ and end at $$(D,0)$$ and are all connected.

I am looking for any references that could help in coming up with an algorithm to find the start and end co-ordinates for the n lines corresponding to $$d_1, d_2 ... d_n$$ .

In other words, it is like we have $$n$$ sticks of length $$d_1, d_2 ... d_n$$. We want to arrange them one after another so that the distance between the starting point and end point is $$D$$.

Another way of stating the problem: Imagine we want to form a closed polygon. One of the segment runs from $$(0,0)$$ to $$(D,0)$$. We want to place the $$n$$ segments of length $$d_1, d_2 ... d_n$$ such that the polygon gets closed.

• Is there any restriction on the angles between $d_k$ and $d_{k+1}$? Apr 17 '20 at 14:15
• @Aretino No restriction. I was thinking there will be only one unique solution. But any solution is fine. Apr 17 '20 at 14:17
• Hence the resulting polygon needn't be convex, right? Apr 17 '20 at 14:20
• Yes. But it will have to be a simple polygon i.e. without segments crossing. I think the solution will be a convex polygon. I will add a drawing so that there is no confusion. Apr 17 '20 at 14:28
• Probably this paper could be of help: researchgate.net/publication/… Apr 17 '20 at 14:36