Point location And Distance between two 3D skew lines

Question

I have read this forum post

**Location** of shortest distance between two skew lines in 3D?

multiple times but I'm stuck.

I have two lines defined resp. by

Point $P_1 (2,4,0)$ and vector $V_1 [2,-2,4]$

Point $P_2 (4,4,1)$ and vector $V_2 [-3,4,-2]$

Step 1) Define the normal vector to the two Vectors $V_3 = V_1 \times V_2 = [12,8,-2].$

Step 2) Define the Vector $V_4 = P_2-P_1 = [2,0,1].$

Step 3) Define the abs(Dot Product) : $|V_3.V_4| = 22.$

Step 4) Define the norm : $|[V_3]| = \sqrt{12^2+8^2+2^2} = 14.56$

Step 5) Divide abs(Dot Product ) / $|[V_3]|$

Conclusion : Minimum Distance = $ 22/14.56 = 1.51.$

This is correct but how do I get the location of the two points realizing this minimum distance using my example values step by step ?

Thanks in advance Marc

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Aug 28 '19 at 08:41
This is relatively unreadable. Please typeset using MathJax. — Allawonder, Aug 28 '19 at 08:48
Out of these calculations, you cannot get the location of the points achieving the smallest distance. See for that the third answer to https://math.stackexchange.com/q/1993953 — Jean Marie, Aug 28 '19 at 09:30
I have tried to modify your text, which is incorrect mathematically at some places : a point is not "on" a vector : instead, you say that a 3D line is defined by "a point and a vector" — Jean Marie, Aug 28 '19 at 09:32

0 Answers0