距离 3D 中四条线最近的点

发布于 2024-11-04 11:27:31 字数 300 浏览 3 评论 0原文

给定 3D 中的 4 条线（表示为几个点），我想找到空间中的点，使该点与每条线之间的距离总和最小。

我正在尝试找到一种方法将其表述为最小二乘问题，但我不太确定应该如何解决。我目前正在尝试使用以下位置提供的距离定义： http://mathworld.wolfram .com/Point-LineDistance3-Dimensional.html

有什么想法吗？

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浮萍、无处依 2024-11-11 11:27:31

我在 Mathematica 中编写了一个程序来计算点坐标。
结果是一个大的代数公式。我已为您将其上传到 ideone。

如果您手头有 Mathematica，这里是程序：

(*Load package*)
Needs["VectorAnalysis`"]
(*Define four lines, by specifying 2 points in each one*)
Table[p[i, j] = {x[i, j], y[i, j], z[i, j]}, {i, 4}, {j, 2}];

(*Define the target point*)
p0 = {x0, y0, z0};

(*Define a Norm function // using Std norm squared here*)
norm[a_] := a[[1]]^2 + a[[2]]^2 + a[[3]]^2

(*Define a function for the distance from line i to point v
used http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html (11) *)
d[i_, v_] :=  norm[Cross[(v - p[i, 1]), (v - p[i, 2])]]/norm[p[i, 2] - p[i, 1]]

(*Define a function for the sum of distances*)
dt[p_] := Sum[d[i, p], {i, 4}]

(*Now take the gradient, and Solve for Gradient == 0*)
s = Solve[Grad[dt[p0], Cartesian[x0, y0, z0]] == 0, {x0, y0, z0}]

(* Result tooooo long. Here you have it for downloading
http://ideone.com/XwbJu *)

结果

I made a program in Mathematica for calculating the point coordinates.
The result is a large algebraic formula. I uploaded it to ideone for you.

Here is the program, in case you have Mathematica at hand:

(*Load package*)
Needs["VectorAnalysis`"]
(*Define four lines, by specifying 2 points in each one*)
Table[p[i, j] = {x[i, j], y[i, j], z[i, j]}, {i, 4}, {j, 2}];

(*Define the target point*)
p0 = {x0, y0, z0};

(*Define a Norm function // using Std norm squared here*)
norm[a_] := a[[1]]^2 + a[[2]]^2 + a[[3]]^2

(*Define a function for the distance from line i to point v
used http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html (11) *)
d[i_, v_] :=  norm[Cross[(v - p[i, 1]), (v - p[i, 2])]]/norm[p[i, 2] - p[i, 1]]

(*Define a function for the sum of distances*)
dt[p_] := Sum[d[i, p], {i, 4}]

(*Now take the gradient, and Solve for Gradient == 0*)
s = Solve[Grad[dt[p0], Cartesian[x0, y0, z0]] == 0, {x0, y0, z0}]

(* Result tooooo long. Here you have it for downloading
http://ideone.com/XwbJu *)

RESULT

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