Find the foot of perpendicular of a point in a 3 D plane - GeeksforGeeks

11 May, 2021
Given a point (x1, y1, z1) in 3-D and coefficients of the equation of plane, we have to find the foot of perpendicular of a point in a 3 D plane.Examples:  
Input: a = 1, b = -2, c = 0, d = 0, x = -1, y = 3, z = 4 Output: x2 = 0.4 y2 = 0.2 z2 = 4.0Input: a = 2, b = -1, c = 1, d = 3, x = 1, y = 3, z = 4 Output: x2 = -1.0 y2 = 4.0 z2 = 3.0
 
 
Approach: Equation of plane is given as ax + by + cz + d = 0. Therefore, the direction ratios of the normal to the plane are (a, b, c). Let N be the foot of perpendicular from given point to the given plane so, line PN has directed ratios (a, b, c) and it passes through P(x1, y1, z1). The equation of line PN will be as:- 
(x – x1) / a = (y – y1) / b = (z – z1) / c = k
Hence any point on line PN can be written as:- 
x = a * k + x1 y = b * k + y1 z = c * k + z1
since N lies in both line and plane so will satisfy(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).
Now, the coordinates of Point N in terms of k will be:-  
x2 = a * k + x1 y2 = b * k + y1 z2 = c * k + z1
Below is the implementation of the above:  
C++
Java
Python3
C#
PHP
Javascript
// 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 perpendicularvoid 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 Codeint 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 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 perpendicularstatic 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 Codepublic 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 program to find# foot of perpendicular# of a point in a 3 D plane. # Function to find foot of perpendiculardef 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 Codeif __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# 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 perpendicularstatic 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 Codepublic 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
<?php// PHP program to find foot of perpendicular// of a point in a 3 D plane. // Function to find foot of perpendicularfunction 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; $z1 = 4; // function callfoot($a, $b, $c, $d, $x1, $y1, $z1); // This code is contributed by ita_c?>
<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>
x2 = 0.4 y2 = 0.2 z2 = 4.0
 
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