c#版本

public static FP DistanceToLineSegment(FPVector3 a, FPVector3 b, FPVector3 point)
{
  var d = b - a;
  var s = d.SqrMagnitude;
  var ds = d / s;
  var lambda = FPVector3.Dot(point - a, ds);
  var p = FPMath.Clamp01(lambda) * d;
  return (a + p - point).Magnitude;
}

快速实现http://paulbourke.net/geometry/pointlineplane/source.c

    static func magnitude(p1: CGPoint, p2: CGPoint) -> CGFloat {
        let vector = CGPoint(x: p2.x - p1.x, y: p2.y - p1.y)
        return sqrt(pow(vector.x, 2) + pow(vector.y, 2))
    }

    /// http://paulbourke.net/geometry/pointlineplane/
    /// http://paulbourke.net/geometry/pointlineplane/source.c
    static func pointDistanceToLine(point: CGPoint, lineStart: CGPoint, lineEnd: CGPoint) -> CGFloat? {

        let lineMag = magnitude(p1: lineEnd, p2: lineStart)
        let u = (((point.x - lineStart.x) * (lineEnd.x - lineStart.x)) +
                ((point.y - lineStart.y) * (lineEnd.y - lineStart.y))) /
                (lineMag * lineMag)

        if u < 0 || u > 1 {
            // closest point does not fall within the line segment
            return nil
        }

        let intersectionX = lineStart.x + u * (lineEnd.x - lineStart.x)
        let intersectionY = lineStart.y + u * (lineEnd.y - lineStart.y)

        return magnitude(p1: point, p2: CGPoint(x: intersectionX, y: intersectionY))
    }

这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好，函数名更长，当然函数语法更短，因为它是在ES6 + CoffeeScript中。

JavaScript版本(ES6)

distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));

distanceToLineSegmentSquared = (p, v, w)=> {
    l2 = distanceSquared(v, w);
    if (l2 === 0) {
        return distanceSquared(p, v);
    }
    t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    t = Math.max(0, Math.min(1, t));
    return distanceSquared(p, {
        x: v.x + t * (w.x - v.x),
        y: v.y + t * (w.y - v.y)
    });
}
distanceToLineSegment = (p, v, w)=> {
    return Math.sqrt(distanceToLineSegmentSquared(p, v));
}

CoffeeScript版本

distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))

distanceToLineSegmentSquared = (p, v, w)->
    l2 = distanceSquared(v, w)
    return distanceSquared(p, v) if l2 is 0
    t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2
    t = Math.max(0, Math.min(1, t))
    distanceSquared(p, {
        x: v.x + t * (w.x - v.x)
        y: v.y + t * (w.y - v.y)
    })

distanceToLineSegment = (p, v, w)->
    Math.sqrt(distanceToLineSegmentSquared(p, v, w))

请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。

几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。

Grumdrig的c++ /JavaScript实现对我来说非常有用，所以我提供了我正在使用的Python直接端口。完整的代码在这里。

class Point(object):
  def __init__(self, x, y):
    self.x = float(x)
    self.y = float(y)

def square(x):
  return x * x

def distance_squared(v, w):
  return square(v.x - w.x) + square(v.y - w.y)

def distance_point_segment_squared(p, v, w):
  # Segment length squared, |w-v|^2
  d2 = distance_squared(v, w) 
  if d2 == 0: 
    # v == w, return distance to v
    return distance_squared(p, v)
  # Consider the line extending the segment, parameterized as v + t (w - v).
  # We find projection of point p onto the line.
  # It falls where t = [(p-v) . (w-v)] / |w-v|^2
  t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2;
  if t < 0:
    # Beyond v end of the segment
    return distance_squared(p, v)
  elif t > 1.0:
    # Beyond w end of the segment
    return distance_squared(p, w)
  else:
    # Projection falls on the segment.
    proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))
    # print proj.x, proj.y
    return distance_squared(p, proj)

我制作了一个交互式Desmos图来演示如何实现这一点:

https://www.desmos.com/calculator/kswrm8ddum

红点是A点，绿点是B点，C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。

编辑:

有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标，AB是单位x轴，单位y轴垂直于AB。