Given two points A(x1, y1, z1) and B(x2, y2, z2) and a set of points (a, b, c) which represent the axis (ai + bj + ck), the task is to find the equation of plane which passes through the given points A and B and parallel to the given axis.
Examples:
Input: x1 = 1, y1 = 2, z1 = 3, x2 = 3, y2 = 4, z2 = 5, a= 6, b = 7, c = 8
Output: 2x + 4y + 2z + 0 = 0Input: x1 = 2, y1 = 3, z1 = 5, x2 = 6, y2 = 7, z2 = 8, a= 11, b = 23, c = 10.
Output: -29x + 7y + 48z + 0= 0
Approach:
From the given two points on plane A and B, The directions ratios a vector equation of line AB is given by:
direction ratio = (x2 – x1, y2 – y1, z2 – z1)
Since the line
is parallel to the given axis
. Therefore, the cross-product of
and
is 0 which is given by:
where,
d, e, and f are the coefficient of vector equation of line AB i.e.,
d = (x2 – x1),
e = (y2 – y1), and
f = (z2 – z1)
and a, b, and c are the coefficient of given axis.
The equation formed by the above determinant is given by:
(Equation 1)
Equation 1 is perpendicular to the line AB which means it is perpendicular to the required plane.
Let the Equation of the plane is given by 
(Equation 2)
where A, B, and C are the direction ratio of the plane perpendicular to the plane.
Since Equation 1 is Equation 2 are perpendicular to each other, therefore the value of the direction ratio of Equation 1 & 2 are parallel. Then the coefficient of the plane is given by:
A = (b*f – c*e),
B = (a*f – c*d), and
C = (a*e – b*d)
Now dot product of plane and vector line AB gives the value of D as
D = -(A * d – B * e + C * f)
Below is the implementation of the above approach:
C++
// C++ implementation to find the// equation of plane which passes// through two points and parallel// to a given axis#include <bits/stdc++.h>using namespace std;void findEquation(int x1, int y1, int z1, int x2, int y2, int z2, int d, int e, int f){ // Find direction vector // of points (x1, y1, z1) // and (x2, y2, z2) double a = x2 - x1; double b = y2 - y1; double c = z2 - z1; // Values that are calculated // and simplified from the // cross product int A = (b * f - c * e); int B = (a * f - c * d); int C = (a * e - b * d); int D = -(A * d - B * e + C * f); // Print the equation of plane cout << A << "x + " << B << "y + " << C << "z + " << D << "= 0";}// Driver Codeint main(){ // Point A int x1 = 2, y1 = 3, z1 = 5; // Point B int x2 = 6, y2 = 7, z2 = 8; // Given axis int a = 11, b = 23, c = 10; // Function Call findEquation(x1, y1, z1, x2, y2, z2, a, b, c); return 0;} |
Java
// Java implementation to find the // equation of plane which passes // through two points and parallel // to a given axis import java.util.*; class GFG{ static void findEquation(int x1, int y1, int z1, int x2, int y2, int z2, int d, int e, int f) { // Find direction vector // of points (x1, y1, z1) // and (x2, y2, z2) double a = x2 - x1; double b = y2 - y1; double c = z2 - z1; // Values that are calculated // and simplified from the // cross product int A = (int)(b * f - c * e); int B = (int)(a * f - c * d); int C = (int)(a * e - b * d); int D = -(int)(A * d - B * e + C * f); // Print the equation of plane System.out.println(A + "x + " + B + "y + " + C + "z + " + D + "= 0 "); } // Driver code public static void main(String[] args) { // Point A int x1 = 2, y1 = 3, z1 = 5; // Point B int x2 = 6, y2 = 7, z2 = 8; // Given axis int a = 11, b = 23, c = 10; // Function Call findEquation(x1, y1, z1, x2, y2, z2, a, b, c); } } // This code is contributed by Pratima Pandey |
Python3
# Python3 implementation # to find the equation # of plane which passes# through two points and # parallel to a given axisdef findEquation(x1, y1, z1, x2, y2, z2, d, e, f): # Find direction vector # of points (x1, y1, z1) # and (x2, y2, z2) a = x2 - x1 b = y2 - y1 c = z2 - z1 # Values that are calculated # and simplified from the # cross product A = (b * f - c * e) B = (a * f - c * d) C = (a * e - b * d) D = -(A * d - B * e + C * f) # Print the equation of plane print (A, "x + ", B, "y + ", C, "z + ", D, "= 0")# Driver Codeif __name__ == "__main__": # Point A x1 = 2 y1 = 3 z1 = 5; # Point B x2 = 6 y2 = 7 z2 = 8 # Given axis a = 11 b = 23 c = 10 # Function Call findEquation(x1, y1, z1, x2, y2, z2, a, b, c)# This code is contributed by Chitranayal |
C#
// C# implementation to find the // equation of plane which passes // through two points and parallel // to a given axis using System;class GFG{ static void findEquation(int x1, int y1, int z1, int x2, int y2, int z2, int d, int e, int f) { // Find direction vector // of points (x1, y1, z1) // and (x2, y2, z2) double a = x2 - x1; double b = y2 - y1; double c = z2 - z1; // Values that are calculated // and simplified from the // cross product int A = (int)(b * f - c * e); int B = (int)(a * f - c * d); int C = (int)(a * e - b * d); int D = -(int)(A * d - B * e + C * f); // Print the equation of plane Console.Write(A + "x + " + B + "y + " + C + "z + " + D + "= 0 "); } // Driver code public static void Main() { // Point A int x1 = 2, y1 = 3, z1 = 5; // Point B int x2 = 6, y2 = 7, z2 = 8; // Given axis int a = 11, b = 23, c = 10; // Function Call findEquation(x1, y1, z1, x2, y2, z2, a, b, c); } } // This code is contributed by Code_Mech |
Javascript
<script>// javascript implementation to find the // equation of plane which passes // through two points and parallel // to a given axis function findEquation(x1 , y1 , z1 , x2 , y2 , z2 , d , e , f) { // Find direction vector // of points (x1, y1, z1) // and (x2, y2, z2) var a = x2 - x1; var b = y2 - y1; var c = z2 - z1; // Values that are calculated // and simplified from the // cross product var A = parseInt( (b * f - c * e)); var B = parseInt( (a * f - c * d)); var C = parseInt( (a * e - b * d)); var D = -parseInt( (A * d - B * e + C * f)); // Print the equation of plane document.write(A + "x + " + B + "y + " + C + "z + " + D + "= 0 "); } // Driver code // Point A var x1 = 2, y1 = 3, z1 = 5; // Point B var x2 = 6, y2 = 7, z2 = 8; // Given axis var a = 11, b = 23, c = 10; // Function Call findEquation(x1, y1, z1, x2, y2, z2, a, b, c);// This code is contributed by Rajput-Ji</script> |
-29x + 7y + 48z + 0= 0
Time Complexity: O(1)
Auxiliary Space: O(1)
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