Find the equation of plane which passes through two points and parallel to a given axis

Last Updated : 27 Mar, 2022

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 = 0

Input: 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)

\overrightarrow{AB} = (x2-x1) i + (y2-y1) j + (z2 - z1) k

Since the line

\overrightarrow{AB}

is parallel to the given axis

(ai + bj + ck)

. Therefore, the cross-product of

\overrightarrow{AB}

and

(ai + bj + ck)

is 0 which is given by:

\begin{vmatrix} i & j & k\\ d & e & f \\ a & b & c \end{vmatrix} = 0

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:

(b*f - c*e) i - (a * f - c * d) j + (a * e - b * d) k = 0

(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
Ax + By + Cz = D
(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 Code int 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 axis def 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 Code if __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# 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>

Output:

-29x + 7y + 48z + 0= 0

Time Complexity: O(1)

Auxiliary Space: O(1)

Find the equation of plane which passes through two points and parallel to a given axis

yogesh shukla 1

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Find the equation of plane which passes through two points and parallel to a given axis

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