You are given two planes P1: a1 * x + b1 * y + c1 * z + d1 = 0 and P2: a2 * x + b2 * y + c2 * z + d2 = 0. The task is to write a program to find distance between these two Planes.
Examples :
Input: a1 = 1, b1 = 2, c1 = -1, d1 = 1, a2 = 3, b2 = 6, c2 = -3, d2 = -4 Output: Distance is 0.952579344416 Input: a1 = 1, b1 = 2, c1 = -1, d1 = 1, a2 = 1, b2 = 6, c2 = -3, d2 = -4 Output: Planes are not parallel
Approach :Consider two planes are given by the equations:-
P1 : a1 * x + b1 * y + c1 * z + d1 = 0, where a1, b1 and c1, d1 are real constants and
P2 : a2 * x + b2 * y + c2 * z + d2 = 0, where a2, b2 and c2, d2 are real constants.
The condition for two planes to be parallel is:-
=> a1 / a2 = b1 / b2 = c1 / c2
Find a point in any one plane such that the distance from that point to the other plane that will be the distance between those two planes. The distance can be calculated by using the formulae:
Distance = (| a*x1 + b*y1 + c*z1 + d |) / (sqrt( a*a + b*b + c*c))
Let a point in Plane P1 be P(x1, y1, z1),
put x = y = 0 in equation a1 * x + b1 * y + c1 * z + d1 = 0 and find z.
=> z = -d1 / c1
Now we have coordinates of P(0, 0, z) = P(x1, y1, z1).
Distance of point P to Plane P2 will be:-
Distance = (| a2*x1 + b2*y1 + c2*z1 + d2 |) / (sqrt( a2*a2 + b2*b2 + c2*c2))
= (| a2*0 + b2*0 + c2*z1 + d2 |) / (sqrt( a2*a2 + b2*b2 + c2*c2))
= (| c2*z1 + d2 |) / (sqrt( a2*a2 + b2*b2 + c2*c2))
Below is the implementation of the above formulae:
C++
// C++ program to find the Distance // between two parallel Planes in 3 D.#include <bits/stdc++.h> #include<math.h>using namespace std;// Function to find distancevoid distance(float a1, float b1, float c1, float d1, float a2, float b2, float c2, float d2) { float x1, y1, z1, d; if (a1 / a2 == b1 / b2 && b1 / b2 == c1 / c2) { x1 = y1 = 0; z1 = -d1 / c1; d = fabs(( c2 * z1 + d2)) / (sqrt(a2 * a2 + b2 * b2 + c2 * c2)); cout << "Perpendicular distance is " << d << endl; } else cout << "Planes are not parallel"; return; } // Driver Code int main() { float a1 = 1; float b1 = 2; float c1 = -1; float d1 = 1; float a2 = 3; float b2 = 6; float c2 = -3; float d2 = -4; distance(a1, b1, c1, d1, a2, b2, c2, d2); // Fxn cal return 0; } // This code is contributed // by Akanksha Rai(Abby_akku) |
C
// C program to find the Distance between// two parallel Planes in 3 D. #include <stdio.h> #include<math.h> // Function to find distancevoid distance(float a1, float b1, float c1, float d1, float a2, float b2, float c2, float d2) { float x1,y1,z1,d; if (a1 / a2 == b1 / b2 && b1 / b2 == c1 / c2) { x1 = y1 = 0; z1 =-d1 / c1; d = fabs(( c2 * z1 + d2)) / (sqrt(a2 * a2 + b2 * b2 + c2 * c2)); printf("Perpendicular distance is %f\n", d); } else printf("Planes are not parallel"); return; } // Driver Code int main() { float a1 = 1; float b1 = 2; float c1 = -1; float d1 = 1; float a2 = 3; float b2 = 6; float c2 = -3; float d2 = -4; distance(a1, b1, c1, d1, a2, b2, c2, d2); // Fxn cal return 0; } // This code is contributed // by Amber_Saxena. |
Java
// Java program to find the Distance // between two parallel Planes in 3 D. import java .io.*; import java.lang.Math; class GFG { // Function to find distancestatic void distance(float a1, float b1, float c1, float d1, float a2, float b2, float c2, float d2) { float x1,y1,z1,d; if (a1 / a2 == b1 / b2 && b1 / b2 == c1 / c2) { x1 = y1 = 0; z1 =-d1 / c1; d = Math.abs(( c2 * z1 + d2)) / (float)(Math.sqrt(a2 * a2 + b2 * b2 + c2 * c2)); System.out.println("Perpendicular distance is "+ d); } else System.out.println("Planes are not parallel"); } // Driver code public static void main(String[] args) { float a1 = 1; float b1 = 2; float c1 = -1; float d1 = 1; float a2 = 3; float b2 = 6; float c2 = -3; float d2 = -4; distance(a1, b1, c1, d1, a2, b2, c2, d2);// Fxn cal} } // This code is contributed // by Amber_Saxena. |
Python
# Python program to find the Distance between# two parallel Planes in 3 D.import math# Function to find distancedef distance(a1, b1, c1, d1, a2, b2, c2, d2): if (a1 / a2 == b1 / b2 and b1 / b2 == c1 / c2): x1 = y1 = 0 z1 =-d1 / c1 d = abs(( c2 * z1 + d2)) / (math.sqrt(a2 * a2 + b2 * b2 + c2 * c2)) print("Perpendicular distance is"), d else: print("Planes are not parallel")# Driver Code a1 = 1b1 = 2c1 = -1d1 = 1a2 = 3b2 = 6c2 = -3d2 = -4distance(a1, b1, c1, d1, a2, b2, c2, d2) # Fxn cal |
C#
// C# program to find the Distance // between two parallel Planes in 3 D. using System; class GFG { // Function to find distancestatic void distance(float a1, float b1, float c1, float d1, float a2, float b2, float c2, float d2) { float z1, d; if (a1 / a2 == b1 / b2 && b1 / b2 == c1 / c2) { z1 =-d1 / c1; d = Math.Abs((c2 * z1 + d2)) / (float)(Math.Sqrt(a2 * a2 + b2 * b2 + c2 * c2)); Console.Write("Perpendicular distance is " + d); } else Console.Write("Planes are not parallel"); } // Driver code public static void Main() { float a1 = 1; float b1 = 2; float c1 = -1; float d1 = 1; float a2 = 3; float b2 = 6; float c2 = -3; float d2 = -4; distance(a1, b1, c1, d1, a2, b2, c2, d2);// Fxn cal} } // This code is contributed// by ChitraNayal |
PHP
<?php // PHP program to find the Distance // between two parallel Planes in 3 D// Function to find distancefunction distance($a1, $b1, $c1, $d1, $a2, $b2, $c2, $d2) { if ($a1 / $a2 == $b1 / $b2 && $b1 / $b2 == $c1 / $c2) { $x1 = $y1 = 0; $z1 =- $d1 / $c1; $d = abs(($c2 * $z1 + $d2)) / (sqrt($a2 * $a2 + $b2 * $b2 + $c2 * $c2)); echo "Perpendicular distance is ", $d; } else echo "Planes are not parallel";} // Driver Code $a1 = 1;$b1 = 2;$c1 = -1;$d1 = 1;$a2 = 3;$b2 = 6;$c2 = -3;$d2 = -4;distance($a1, $b1, $c1, $d1, $a2, $b2, $c2, $d2); // This code is contributed // by Amber_Saxena.?> |
Javascript
<script>// Javascript program to find the Distance // between two parallel Planes in 3 D. // Function to find distancefunction distance(a1, b1, c1, d1, a2, b2, c2, d2) { let x1,y1,z1,d; if (a1 / a2 == b1 / b2 && b1 / b2 == c1 / c2) { x1 = y1 = 0; z1 =-d1 / c1; d = Math.abs(( c2 * z1 + d2)) / (Math.sqrt(a2 * a2 + b2 * b2 + c2 * c2)); document.write("Perpendicular distance is "+ d); } else document.write("Planes are not parallel"); } // Driver Code let a1 = 1; let b1 = 2; let c1 = -1; let d1 = 1; let a2 = 3; let b2 = 6; let c2 = -3; let d2 = -4; distance(a1, b1, c1, d1, a2, b2, c2, d2);// Fxn cal </script> |
Output
Perpendicular distance is 0.952579
Time Complexity: O(logn) as it is using inbuilt sqrt function
Auxiliary Space: O(1)
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