假設空間某點O的座標爲(Xo,Yo,Zo),空間某條直線上兩點A和B的座標爲:(X1,Y1,Z1),(X2,Y2,Z2),設點O在直線AB上的垂足爲點N,座標爲(Xn,Yn,Zn)。點N座標解算過程如下:
首先求出下列向量:
由向量垂直關係:
上式記爲(1)式。
點N在直線AB上,根據向量共線:
(2)
由(2)得:
(3)
把(3)式代入(1)式,式中只有一個未知數k,整理化簡解出k:
(4)
把(4)式代入(3)式即得到垂足N的座標。
C代碼實現
// 二維空間點到直線的垂足
struct Point
{
double x,y;
}
Point GetFootOfPerpendicular(
const Point &pt, // 直線外一點
const Point &begin, // 直線開始點
const Point &end) // 直線結束點
{
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;
}
// 三維空間點到直線的垂足
struct Point
{
double x,y,z;
}
Point GetFootOfPerpendicular(
const Point &pt, // 直線外一點
const Point &begin, // 直線開始點
const Point &end) // 直線結束點
{
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;
}
下面是我用java寫的代碼,就是翻譯一下,也貼上:
//點到其他兩點垂足計算
public TPoint getFoot(TPoint pt,TPoint beginPt,TPoint endPt){
TPoint result = new TPoint();
double dx = beginPt.getJd()-endPt.getJd();
double dy = beginPt.getWd()-endPt.getWd();
if(abs(dx)<0.000001&&abs(dy)<0.000001){
result = pt;
}
double u = (pt.getJd()-beginPt.getJd())*(beginPt.getJd()-endPt.getJd())+
(pt.getWd()-beginPt.getWd())*(beginPt.getWd()-endPt.getWd());
u = u /(dx*dx+dy*dy);
result.setJd(beginPt.getJd()+u*dx);
result.setWd(beginPt.getWd()+u*dy);
return result;
}