原文章地址为: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:
由向量垂直关系,两个向量如果垂直,那么两个向量的点积(点乘,向量积)则为0,可得出。
上式记为(1)式。
点N在直线AB上,根据向量共线定理:
(2)
由(2)得:
(3)
把(3)式代入(1)式,式中只有一个未知数k,整理化简解出k:
(4)
把(4)式代入(3)式即得到垂足N的坐标。
下面是C的实现方式:
// 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;
}
接下来附上汤圆给我改的js的代码:
const GetFootOfPerpendicular = (pt, begin, end) => {
const dx = begin.x - end.x
const dy = begin.y - end.y
const dz = begin.z - end.z
const EPS = 0.00000001
// 确保两个点不是同一个点
if(Math.abs(dx) < EPS &&
Math.abs(dy) < EPS &&
Math.abs(dz) < EPS){
return begin
}
//计算斜率
let 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 / (Math.pow(dx, 2) + Math.pow(dy, 2) + Math.pow(dz, 2))
return new THREE.Vector3(begin.x + u * dx, begin.y + u * dy, begin.z + u * dz)
}