特征c++版本的3D线段和点

// Return minimum distance between line segment: head--->tail and point
double MinimumDistance(Eigen::Vector3d head, Eigen::Vector3d tail,Eigen::Vector3d point)
{
    double l2 = std::pow((head - tail).norm(),2);
    if(l2 ==0.0) return (head - point).norm();// head == tail case

    // Consider the line extending the segment, parameterized as head + t (tail - point).
    // We find projection of point onto the line.
    // It falls where t = [(point-head) . (tail-head)] / |tail-head|^2
    // We clamp t from [0,1] to handle points outside the segment head--->tail.

    double t = max(0,min(1,(point-head).dot(tail-head)/l2));
    Eigen::Vector3d projection = head + t*(tail-head);

    return (point - projection).norm();
}

这是一个基于向量数学的;这个解决方案也适用于更高的维度，并报告交点(在线段上)。

def dist(x1,y1,x2,y2,px,py):
    a = np.array([[x1,y1]]).T
    b = np.array([[x2,y2]]).T
    x = np.array([[px,py]]).T
    tp = (np.dot(x.T, b) - np.dot(a.T, b)) / np.dot(b.T, b)
    tp = tp[0][0]
    tmp = x - (a + tp*b)
    d = np.sqrt(np.dot(tmp.T,tmp)[0][0])
    return d, a+tp*b

x1,y1=2.,2.
x2,y2=5.,5.
px,py=4.,1.

d, inters = dist(x1,y1, x2,y2, px,py)
print (d)
print (inters)

结果是

2.1213203435596424
[[2.5]
 [2.5]]

这里解释了数学

https://brilliant.org/wiki/distance-between-point-and-line/

该算法基于求出指定直线与包含指定点的正交直线的交点，并计算其距离。在线段的情况下，我们必须检查交点是否在线段的点之间，如果不是这样，则最小距离是指定点与线段的一个端点之间的距离。这是一个c#实现。

Double Distance(Point a, Point b)
{
    double xdiff = a.X - b.X, ydiff = a.Y - b.Y;
    return Math.Sqrt((long)xdiff * xdiff + (long)ydiff * ydiff);
}

Boolean IsBetween(double x, double a, double b)
{
    return ((a <= b && x >= a && x <= b) || (a > b && x <= a && x >= b));
}

Double GetDistance(Point pt, Point pt1, Point pt2, out Point intersection)
{
    Double a, x, y, R;

    if (pt1.X != pt2.X) {
        a = (double)(pt2.Y - pt1.Y) / (pt2.X - pt1.X);
        x = (a * (pt.Y - pt1.Y) + a * a * pt1.X + pt.X) / (a * a + 1);
        y = a * x + pt1.Y - a * pt1.X; }
    else { x = pt1.X;  y = pt.Y; }

    if (IsBetween(x, pt1.X, pt2.X) && IsBetween(y, pt1.Y, pt2.Y)) {
        intersection = new Point((int)x, (int)y);
        R = Distance(intersection, pt); }
    else {
        double d1 = Distance(pt, pt1), d2 = Distance(pt, pt2);
        if (d1 < d2) { intersection = pt1; R = d1; }
        else { intersection = pt2; R = d2; }}

    return R;
}

公认的答案行不通 (例如，0,0和(-10,2,10,2)之间的距离应为2)。

下面是工作代码:

   def dist2line2(x,y,line):
     x1,y1,x2,y2=line
     vx = x1 - x
     vy = y1 - y
     ux = x2-x1
     uy = y2-y1
     length = ux * ux + uy * uy
     det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
     if det < 0:
       return (x1 - x)**2 + (y1 - y)**2
     if det > length:
       return (x2 - x)**2 + (y2 - y)**2
     det = ux * vy - uy * vx
     return det**2 / length
   def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))

Lua: 查找线段(不是整条线)与点之间的最小距离

function solveLinearEquation(A1,B1,C1,A2,B2,C2)
--it is the implitaion of a method of solving linear equations in x and y
  local f1 = B1*C2 -B2*C1
  local f2 = A2*C1-A1*C2
  local f3 = A1*B2 -A2*B1
  return {x= f1/f3, y= f2/f3}
end


function pointLiesOnLine(x,y,x1,y1,x2,y2)
  local dx1 = x-x1
  local  dy1 = y-y1
  local dx2 = x-x2
  local  dy2 = y-y2
  local crossProduct = dy1*dx2 -dx1*dy2

if crossProduct ~= 0  then  return  false
else
  if ((x1>=x) and (x>=x2)) or ((x2>=x) and (x>=x1)) then
    if ((y1>=y) and (y>=y2)) or ((y2>=y) and (y>=y1)) then
      return true
    else return false end
  else  return false end
end
end


function dist(x1,y1,x2,y2)
  local dx = x1-x2
  local dy = y1-y2
  return math.sqrt(dx*dx + dy* dy)
 end


function findMinDistBetnPointAndLine(x1,y1,x2,y2,x3,y3)
-- finds the min  distance between (x3,y3) and line (x1,y2)--(x2,y2)
   local A2,B2,C2,A1,B1,C1
   local dx = y2-y1
   local dy = x2-x1
   if dx == 0 then A2=1 B2=0 C2=-x3 A1=0 B1=1 C1=-y1 
   elseif dy == 0 then A2=0 B2=1 C2=-y3 A1=1 B1=0 C1=-x1
   else
      local m1 = dy/dx
      local m2 = -1/m1
      A2=m2 B2=-1 C2=y3-m2*x3 A1=m1 B1=-1 C1=y1-m1*x1
   end
 local intsecPoint= solveLinearEquation(A1,B1,C1,A2,B2,C2)
if pointLiesOnLine(intsecPoint.x, intsecPoint.y,x1,y1,x2,y2) then
   return dist(intsecPoint.x, intsecPoint.y, x3,y3)
 else
   return math.min(dist(x3,y3,x1,y1),dist(x3,y3,x2,y2))
end
end

一个2D和3D的解决方案

考虑基底的变化，使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上，最短的距离由

    u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u     -> d(B, C)

(到其中一个端点或到支撑线的距离，取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)

式中，d为AB线段的长度，u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。

AB.AC ≤ 0             -> |AC|
    0 ≤ AB.AC ≤ AB²   -> |ABxAC|/|AB|
          AB² ≤ AB.AC -> |BC|