Given a point(x, y, z) in 3-D and coefficients of the equation of a plane, the task is to find the mirror image of that point through the given plane.
Examples:
Input: a = 1, b = -2, c = 0, d = 0, x = -1, y = 3, z = 4
Output: x3 = 1.7999999999999998, y3 = -2.5999999999999996, z3 = 4.0
Input: a = 2, b = -1, c = 1, d = 3, x = 1, y = 3, z = 4
Output: x3 = -3.0, y3 = 5.0, z3 = 2.0
Approach: Equation of plane is as ax + by + cz + d = 0. Therefore, direction ratios of the normal to the plane are (a, b, c). Let N be the foot of perpendicular from a 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
Since, Point N(x2, y2, z2) is midpoint of point P(x1, y1, z1) and point Q(x3, y3, z3), coordinates of Point Q are:-
=> x3 = 2 * x2 - x1 => y3 = 2 * y2 - y1 => z3 = 2 * z2 - z1
C++
// C++ program to find // Mirror of a point // through a 3 D plane#include <bits/stdc++.h> #include<math.h>#include <iostream>#include <iomanip>using namespace std;// Function to mirror image void mirror_point(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; float x3 = 2 * x2 - x1; float y3 = 2 * y2 - y1; float z3 = 2 * z2 - z1; std::cout << std::fixed; std::cout << std::setprecision(1); cout << " x3 = " << x3; cout << " y3 = " << y3; cout << " z3 = " << z3;}// 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 mirror_point(a, b, c, d, x1, y1, z1); return 0;}// This code is contributed // by Amber_Saxena. |
C
// C program to find// Mirror of a point // through a 3 D plane#include<stdio.h> // Function to mirror imagevoid mirror_point(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; float x3 = 2 * x2 - x1; float y3 = 2 * y2 - y1; float z3 = 2 * z2 - z1; printf("x3 = %.1f ", x3); printf("y3 = %.1f ", y3); printf("z3 = %.1f ", z3);}// 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 mirror_point(a, b, c, d, x1, y1, z1);}// This code is contributed // by Amber_Saxena. |
Java
// Java program to find// Mirror of a point // through a 3 D planeimport java.io.*;class GFG { // Function to mirror imagestatic void mirror_point(int a, int b, int c, int d, int x1, int y1, int 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; float x3 = 2 * x2 - x1; float y3 = 2 * y2 - y1; float z3 = 2 * z2 - z1; System.out.print("x3 = " + x3 + " "); System.out.print("y3 = " + y3 + " "); System.out.print("z3 = " + z3 + " ");}// Driver Code public static void main(String[] args){ int a = 1; int b = -2; int c = 0; int d = 0; int x1 = -1; int y1 = 3; int z1 = 4; // function call mirror_point(a, b, c, d, x1, y1, z1) ;}}// This code is contributed// by inder_verma |
Python
# Function to mirror imagedef mirror_point(a, b, c, d, x1, y1, z1): k =(-a * x1-b * y1-c * z1-d)/float((a * a + b * b + c * c)) x2 = a * k + x1 y2 = b * k + y1 z2 = c * k + z1 x3 = 2 * x2-x1 y3 = 2 * y2-y1 z3 = 2 * z2-z1 print "x3 =", x3, print "y3 =", y3, print "z3 =", z3,# Driver Code a = 1b = -2c = 0d = 0x1 = -1y1 = 3z1 = 4# function callmirror_point(a, b, c, d, x1, y1, z1) |
C#
// C# program to find Mirror of // a point through a 3 D plane using System;class GFG{ // Function to mirror image static void mirror_point(int a, int b, int c, int d, int x1, int y1, int 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; float x3 = 2 * x2 - x1; float y3 = 2 * y2 - y1; float z3 = 2 * z2 - z1; Console.Write("x3 = " + x3 + " "); Console.Write("y3 = " + y3 + " "); Console.Write("z3 = " + z3 + " "); } // Driver Code static public void Main (){ int a = 1; int b = -2; int c = 0; int d = 0; int x1 = -1; int y1 = 3; int z1 = 4; // function call mirror_point(a, b, c, d, x1, y1, z1); }}// This code is contributed by jit_t |
PHP
<?php // PHP program to find Mirror of// a point through a 3 D plane// Function to mirror imagefunction mirror_point($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; $x3 = 2 * $x2 - $x1; $y3 = 2 * $y2 - $y1; $z3 = 2 * $z2 - $z1; echo sprintf("x3 = %.1f ", $x3); echo sprintf("y3 = %.1f ", $y3); echo sprintf("z3 = %.1f ", $z3); } // Driver Code $a = 1;$b = -2;$c = 0;$d = 0;$x1 = -1;$y1 = 3;$z1 = 4;// function callmirror_point($a, $b, $c, $d, $x1, $y1, $z1);// This code is contributed // by Amber_Saxena.?> |
Javascript
<script> // JavaScript program to find // Mirror of a point // through a 3 D plane // Function to mirror image function mirror_point(a, b, c, d, x1, y1, z1) { var k = parseFloat( (-a * x1 - b * y1 - c * z1 - d) / parseFloat(a * a + b * b + c * c) ); var x2 = parseFloat(a * k + x1); var y2 = parseFloat(b * k + y1); var z2 = parseFloat(c * k + z1); var x3 = parseFloat(2 * x2 - x1).toFixed(1); var y3 = parseFloat(2 * y2 - y1).toFixed(1); var z3 = parseFloat(2 * z2 - z1).toFixed(1); document.write(" x3 = " + x3); document.write(" y3 = " + y3); document.write(" z3 = " + z3); } // 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 mirror_point(a, b, c, d, x1, y1, z1); </script> |
Output:
x3 = 1.8 y3 = -2.6 z3 = 4.0
Time Complexity: O(1)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
