三维空间最短距离以及点到直线的映射点

要求点P到直线L的最短距离，可以使用以下方法：

1. 计算出直线L上任意一点Q和P之间的向量v。

2. 将向量v投影到与直线L垂直的方向上，得到向量w。

3. 直线L上的某一点M加上向量w，就是点P在直线L上的映射点N。

代码实现如下：

#include <iostream>
#include <cmath>

using namespace std;

struct Point {
    double x, y, z;
};

struct Line {
    Point p1, p2;
};

double distance(Point p, Line l) {
    double dx = l.p2.x - l.p1.x;
    double dy = l.p2.y - l.p1.y;
    double dz = l.p2.z - l.p1.z;
    double t = ((p.x - l.p1.x) * dx + (p.y - l.p1.y) * dy + (p.z - l.p1.z) * dz) / (dx * dx + dy * dy + dz * dz);
    Point projection = {l.p1.x + t * dx, l.p1.y + t * dy, l.p1.z + t * dz};
    return sqrt((projection.x - p.x) * (projection.x - p.x) + (projection.y - p.y) * (projection.y - p.y) + (projection.z - p.z) * (projection.z - p.z));
}

Point projection(Point p, Line l) {
    double dx = l.p2.x - l.p1.x;
    double dy = l.p2.y - l.p1.y;
    double dz = l.p2.z - l.p1.z;
    double t = ((p.x - l.p1.x) * dx + (p.y - l.p1.y) * dy + (p.z - l.p1.z) * dz) / (dx * dx + dy * dy + dz * dz);
    Point projection = {l.p1.x + t * dx, l.p1.y + t * dy, l.p1.z + t * dz};
    return projection;
}

int main() {
    Point p = {1, 2, 3};
    Line l = {{0, 0, 0}, {1, 1, 1}};
    cout << "The distance between point P and line L is: " << distance(p, l) << endl;
    Point projection_point = projection(p, l);
    cout << "The projection point of point P on line L is: (" << projection_point.x << ", " << projection_point.y << ", " << projection_point.z << ")" << endl;
    return 0;
}

在上述代码中，distance函数计算了点P到直线L的最短距离，而projection函数计算了点P在直线L上的映射点。