求一个点在三维平面上的垂线脚
给定三维中的一个点(x1，y1，z1)和平面方程的系数，我们必须找到三维平面中点的垂线偏差。 例:
输入: a = 1，b = -2，c = 0，d = 0，x = -1，y = 3，z = 4 T3】输出:x2 = 0.4 y2 = 0.2 z2 = 4.0 T6】输入: a = 2，b = -1，c = 1，d = 3，x = 1，y = 3，z = 4 T9】输出:x2 =-1.0 y2 = 4.0 z2 = 0
逼近:平面方程为 ax + by + cz + d = 0。因此，法线与平面的方向比是 (a，b，c) 。假设 N 是从给定点到给定平面的垂线的脚，那么，线 PN 有向比(a，b，c)，它穿过 P(x1，y1，z1)。 PN 线的方程式如下:-
(x–x1)/a =(y–y1)/b =(z–Z1)/c = k
因此，PN 线上的任何点都可以写成:-
x = a * k+x1 y = b * k+y1 z = c * k+Z1
因为 N 位于直线和平面上，所以将满足(ax + by + cz + d = 0)。
= > a (a * k+x1)+b (b * k+y1)+c *(c * k+Z1)+d = 0 .>a * k+a * x1+b * k+b * y 1+c * k+c * Z1+d = 0。 =>(a * a+b * b+c * c)k =-a * x1–b * y 1–c * Z1–d。 =>k =(-a * x1–b * y 1–c * Z1–d)/(a * a+b * b+c * c)。
现在，N 点在 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 <bits/stdc++.h>
#include <iomanip>
#include <iostream>
#include <math.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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);
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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)
{
$k = (-$a * $x1 - $b * $y1 - $c * $z1 - $d) /
($a * $a + $b * $b + $c * $c);
$x2 = $a * $k + $x1;
$y2 = $b * $k + $y1;
$z2 = $c * $k + $z1;
echo "x2 = " . round($x2, 1);
echo " y2 = " . round($y2, 1);
echo " z2 = " . round($z2, 1);
}
// Driver Code
$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 ita_c
?>
java 描述语言
<script>
// 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);
</script>
Output:
x2 = 0.4 y2 = 0.2 z2 = 4.0
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