[Math]空间点到直线垂足坐标的解算方法

keywords: [Math]空间点到直线垂足坐标的解算方法

算法1

原文：https://blog.csdn.net/zhouschina/article/details/14647587

假设空间某点O的坐标为（Xo，Yo，Zo）,空间某条直线上两点A和B的坐标为：(X1,Y1,Z1),(X2，Y2，Z2),设点O在直线AB上的垂足为点N，坐标为(Xn，Yn，Zn)。点N坐标解算过程如下： 首先求出下列向量：

把公式3式代入公式1式，式中只有一个未知数k，整理化简解出k（公式4）：

把公式4式代入公式3式即得到垂足N的坐标。

二维空间

// 二维空间点到直线的垂足
struct Point
{
　　double x,y;
}
Point GetFootOfPerpendicular(
    const Point &pt,     // 直线外一点
    const Point &begin,  // 直线开始点
    const Point &end)   // 直线结束点
{
    Point retVal;
 
    double dx = begin.x - end.x;
    double dy = begin.y - end.y;
    if(abs(dx) < 0.00000001 && abs(dy) < 0.00000001 )
    {
        retVal = begin;
        return retVal;
    }
 
    double u = (pt.x - begin.x)*(begin.x - end.x) +
        (pt.y - begin.y)*(begin.y - end.y);
    u = u/((dx*dx)+(dy*dy));
 
    retVal.x = begin.x + u*dx;
    retVal.y = begin.y + u*dy;
 
    return retVal;
}

三维空间

// 三维空间点到直线的垂足
struct Point
{
　　double x,y,z;
}
Point GetFootOfPerpendicular(
    const Point &pt,     // 直线外一点
    const Point &begin,  // 直线开始点
    const Point &end)   // 直线结束点
{
    Point retVal;
 
    double dx = begin.x - end.x;
    double dy = begin.y - end.y;
　　double dz = begin.z - end.z;
    if(abs(dx) < 0.00000001 && abs(dy) < 0.00000001 && abs(dz) < 0.00000001 )
    {
        retVal = begin;
        return retVal;
    }
 
    double u = (pt.x - begin.x)*(begin.x - end.x) +
        (pt.y - begin.y)*(begin.y - end.y) + (pt.z - begin.z)*(begin.z - end.z);
    u = u/((dx*dx)+(dy*dy)+(dz*dz));
 
    retVal.x = begin.x + u*dx;
    retVal.y = begin.y + u*dy;
　　retVal.y = begin.z + u*dz;
　　
    return retVal;
}

算法2

原文：3D Perpendicular Point on Line From 3D point
https://stackoverflow.com/questions/9368436/3d-perpendicular-point-on-line-from-3d-point

计算p1、p2连成的直线上的离 q 点最近的点 f（即 q 点到直线 p1、p2的垂足坐标）：

XNA实现

Vector3 p1 = new Vector3(x1, y1, z1);
Vector3 p2 = new Vector3(x2, y2, z2);
Vector3 q = new Vector3(x3, y3, z3);

Vector3 u = p2 - p1;
Vector3 pq = q - p1;
Vector3 w2 = pq - Vector3.Multiply(u, Vector3.Dot(pq, u) / u.LengthSquared());

Vector3 f = q - w2;

UE4实现

FVector GetPerpendicularPointToLine(const FVector& PointStart, const FVector& PointEnd, const FVector& PointPerpendicular)
{
    FVector Line = PointEnd - PointStart;
    FVector PS = PointPerpendicular - PointStart;
    FVector W2 = PS - (Line * (PS | Line) / Line.SizeSquared());
    FVector FootPoint = PointPerpendicular - W2;

    return FootPoint;
}

UE4引擎提供的工具函数：

FVector UKismetMathLibrary::FindClosestPointOnLine(FVector Point, FVector LineOrigin, FVector LineDirection)
{
    const FVector SafeDir = LineDirection.GetSafeNormal();
    const FVector ClosestPoint = LineOrigin + (SafeDir * ((Point-LineOrigin) | SafeDir));
    return ClosestPoint;
}

参考

Perpendicular on a line segment from a given point
https://stackoverflow.com/questions/10301001/perpendicular-on-a-line-segment-from-a-given-point

How do you find a point at a given perpendicular distance from a line?
https://stackoverflow.com/questions/133897/how-do-you-find-a-point-at-a-given-perpendicular-distance-from-a-line