原文章地址爲:https://blog.csdn.net/zhouschina/article/details/14647587
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,兩個向量如果垂直,那麼兩個向量的點積(點乘,向量積)則爲0,可得出。
上式記爲(1)式。
點N在直線AB上,根據向量共線定理:
(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。
下面是C的實現方式:
// 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;
}
接下來附上湯圓給我改的js的代碼:
const GetFootOfPerpendicular = (pt, begin, end) => {
const dx = begin.x - end.x
const dy = begin.y - end.y
const dz = begin.z - begin.z
const EPS = 0.00000001
// 確保兩個點不是同一個點
if(Math.abs(dx) < EPS &&
Math.abs(dy) < EPS &&
Math.abs(dz) < EPS){
return begin
}
//計算斜率
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)
}