求 3D 平面中一点的垂脚

给定 3-D 中的点 (x1, y1, z1) 和平面方程的系数，我们必须找到 3D 平面中点的垂线脚。
例子：

Input: a = 1, b = -2, c = 0, d = 0, x = -1, y = 3, z = 4 
Output: x2 = 0.4 y2 = 0.2 z2 = 4.0
Input: a = 2, b = -1, c = 1, d = 3, x = 1, y = 3, z = 4 
Output: x2 = -1.0 y2 = 4.0 z2 = 3.0

方法：平面方程为ax + by + cz + d = 0。因此，平面法线的方向比为(a, b, c) 。设 N 为从给定点到给定平面的垂线的脚，因此，线 PN 具有有向比 (a, b, c) 并且它通过 P(x1, y1, z1)。
直线 PN 的方程为：-

(x – x1) / a = (y – y1) / b = (z – z1) / c = k

因此，直线 PN 上的任何点都可以写成：-

x = a * k + x1 
y = b * k + y1 
z = c * k + z1

由于 N 位于直线和平面上，因此将满足（ax + by + cz + d = 0）。

=>a * (a * k + x1) + b * (b * k + y1) + c * (c * k + z1) + d = 0. 
=>a * a * k + a * x1 + b * b * k + b * y1 + c * c * k + c * z1 + d = 0. 
=>(a * a + b * b + c * c)k = -a * x1 – b * y1 – c * z1 – d. 
=>k = (-a * x1 – b * y1 – c * z1 – d) / (a * a + b * b + c * c).

现在，点 N 在 k 方面的坐标将是：-

x2 = a * k + x1 
y2 = b * k + y1 
z2 = c * k + z1

下面是上面的实现：

C++

// C++ program to find
// foot of perpendicular
// of a point in a 3 D plane.
#include 
#include 
#include 
#include 
using namespace std;
 
// Function to find foot of perpendicular
void foot(float a, float b,
          float c, float d,
          float x1, float y1,
          float z1)
{
    float k = (-a * x1 - b * y1 - c * z1 - d) / (float)(a * a + b * b + c * c);
    float x2 = a * k + x1;
    float y2 = b * k + y1;
    float z2 = c * k + z1;
 
    std::cout << std::fixed;
    std::cout << std::setprecision(1);
    cout << " x2 = " << x2;
    cout << " y2 = " << y2;
    cout << " z2 = " << z2;
}
 
// Driver Code
int main()
{
    float a = 1;
    float b = -2;
    float c = 0;
    float d = 0;
    float x1 = -1;
    float y1 = 3;
    float z1 = 4;
 
    // function call
    foot(a, b, c, d, x1, y1, z1);
    return 0;
}
// This code is contributed  by Amber_Saxena.

Java

// Java program to find
// foot of perpendicular
// of a point in a 3 D plane.
import java.util.*;
import java.text.*;
 
class solution
{
 
// Function to find foot of perpendicular
static void foot(float a, float b,
        float c, float d,
        float x1, float y1,
        float z1)
{
    float k = (-a * x1 - b * y1 - c * z1 - d) / (float)(a * a + b * b + c * c);
    float x2 = a * k + x1;
    float y2 = b * k + y1;
    float z2 = c * k + z1;
    DecimalFormat form = new DecimalFormat("0.0");
    System.out.print(" x2 = " +form.format(x2));
    System.out.print(" y2 = " +form.format(y2));
    System.out.print( " z2 = " +form.format(z2));
}
 
// Driver Code
public static void main(String arr[])
{
    float a = 1;
    float b = -2;
    float c = 0;
    float d = 0;
    float x1 = -1;
    float y1 = 3;
    float z1 = 4;
 
    // function call
    foot(a, b, c, d, x1, y1, z1);
 
}
}

Python3

# Python3 program to find
# foot of perpendicular
# of a point in a 3 D plane.
 
# Function to find foot of perpendicular
def foot(a, b, c, d, x1, y1, z1) :
 
    k = (-a * x1 - b * y1 - c * z1 - d) / (a * a + b * b + c * c);
    x2 = a * k + x1;
    y2 = b * k + y1;
    z2 = c * k + z1;
 
    print("x2 =",round(x2,1))
    print("y2 =",round(y2,1))
    print("z2 =",round(z2,1))
 
 
# Driver Code
if __name__ == "__main__" :
 
    a = 1
    b = -2
    c = 0
    d = 0
    x1 = -1
    y1 = 3
    z1 = 4
 
    # function call
    foot(a, b, c, d, x1, y1, z1)
 
# This code is contributed by Ryuga

C#

// C# program to find
// foot of perpendicular
// of a point in a 3 D plane.
using System;
using System.Globalization;
 
class GFG
{
 
// Function to find foot of perpendicular
static void foot(float a, float b,
        float c, float d,
        float x1, float y1,
        float z1)
{
    float k = (-a * x1 - b * y1 - c * z1 - d) /
                (float)(a * a + b * b + c * c);
    float x2 = a * k + x1;
    float y2 = b * k + y1;
    float z2 = c * k + z1;
    NumberFormatInfo form = new NumberFormatInfo();
    form.NumberDecimalSeparator = ".";
    Console.Write(" x2 = " + x2.ToString(form));
    Console.Write(" y2 = " + y2.ToString(form));
    Console.Write( " z2 = " + z2.ToString(form));
}
 
// Driver Code
public static void Main(String []arr)
{
    float a = 1;
    float b = -2;
    float c = 0;
    float d = 0;
    float x1 = -1;
    float y1 = 3;
    float z1 = 4;
 
    // function call
    foot(a, b, c, d, x1, y1, z1);
}
}
 
// This code contributed by Rajput-Ji

Javascript

      // JavaScript program to find
      // foot of perpendicular
      // of a point in a 3 D plane.
 
      // Function to find foot of perpendicular
      function foot(a, b, c, d, x1, y1, z1) {
        var k = (-a * x1 - b * y1 - c * z1 - d) / (a * a + b * b + c * c);
        var x2 = a * k + x1;
        var y2 = b * k + y1;
        var z2 = c * k + z1;
 
        document.write("x2 =" + x2.toFixed(1) + " ");
        document.write("y2 =" + y2.toFixed(1) + " ");
        document.write("z2 =" + z2.toFixed(1) + " ");
      }
 
      // Driver Code
 
      var a = 1;
      var b = -2;
      var c = 0;
      var d = 0;
      var x1 = -1;
      var y1 = 3;
      var z1 = 4;
 
      // function call
      foot(a, b, c, d, x1, y1, z1);
    

输出：

x2 = 0.4 y2 = 0.2 z2 = 4.0