Given two force vectors, find out whether they are parallel, perpendicular or neither. Let the first vector be A = a1 i +a2 j + a3 k and the second vector be B = b1 i + b2 j + b3 k.
A.B = a1 * b1 + a2 * b2 + a3 * b3
A x B = (a2 * b3 – a3 * b2) i – (a1 * b3 – b1* a3) j + (a1 * b2 – a2 * b1) k
|A|2 = a12 + a22 + a32
If A.B = 0, then A and B are perpendicular.
If |A X B|2 = 0, then A and B are parallel.
Return 1 if they are parallel to each other, 2 if they are perpendicular to each other or 0 otherwise.
Examples:
Input: A = {3, 2, 1}, B = {6, 4, 2}
Output: 1
Explanation: |A X B|2 = 0Input: A = {4, 6, 1}, B = {1, -1, 2}
Output: 2
Explanation: A.B = 0
Approach: Follow the steps to solve this problem:
- Find A.B and A X B
- If A.B = 0, then return 2.
- Else if, |A X B| = 0, then |A X B|2 is also 0, then return 1.
- Else, return 0.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find out whether they// are parallel, perpendicular or neitherint find(vector<int> A, vector<int> B){ // Find A.B int per = A[0] * B[0] + A[1] * B[1] + A[2] * B[2]; // Find A X B int par = (A[1] * B[2] - A[2] * B[1]) * (A[1] * B[2] - A[2] * B[1]) + (A[0] * B[2] - B[0] * A[2]) * (A[0] * B[2] - B[0] * A[2]) + (A[0] * B[1] - A[1] * B[0]) * (A[0] * B[1] - A[1] * B[0]); if (per == 0) { return 2; } else if (par == 0) { return 1; } else { return 0; }}// Driver Codeint main(){ vector<int> a = { 3, 2, 1 }; vector<int> b = { 6, 4, 2 }; // Function call cout << find(a, b) << endl; return 0;} |
Java
// Java code for the above approachimport java.io.*;class GFG { // Function to find out whether they are parallel, // perpendicular or neither static int find(int[] A, int[] B) { // Find A.B int per = A[0] * B[0] + A[1] * B[1] + A[2] * B[2]; // Find A X B int par = (A[1] * B[2] - A[2] * B[1]) * (A[1] * B[2] - A[2] * B[1]) + (A[0] * B[2] - B[0] * A[2]) * (A[0] * B[2] - B[0] * A[2]) + (A[0] * B[1] - A[1] * B[0]) * (A[0] * B[1] - A[1] * B[0]); if (per == 0) { return 2; } else if (par == 0) { return 1; } else { return 0; } } public static void main(String[] args) { int[] a = { 3, 2, 1 }; int[] b = { 6, 4, 2 }; // Function call System.out.println(find(a, b)); }}// This code is contributed by lokeshmvs21. |
Python3
# Python3 code to implement the approach# Function to find out whether they# are parallel, perpendicular or neitherdef find(A, B): # Find A.B per = A[0] * B[0] + A[1] * B[1] + A[2] * B[2] # Find A X B par = (A[1] * B[2] - A[2] * B[1]) * (A[1] * B[2] - A[2] * B[1]) + (A[0] * B[2] - B[0] * A[2]) * \ (A[0] * B[2] - B[0] * A[2]) + (A[0] * B[1] - A[1] * B[0]) * (A[0] * B[1] - A[1] * B[0]) if (per == 0): return 2 elif (par == 0): return 1 else: return 0# Driver Codeif __name__ == "__main__": a = [3, 2, 1] b = [6, 4, 2] # Function call print(find(a, b)) # This code is contributed by rakeshsahni |
C#
using System;public class GFG{ // Function to find out whether they // are parallel, perpendicular or neither public static int find(int[] A, int[] B) { // Find A.B int per = A[0] * B[0] + A[1] * B[1] + A[2] * B[2]; // Find A X B int par = (A[1] * B[2] - A[2] * B[1]) * (A[1] * B[2] - A[2] * B[1]) + (A[0] * B[2] - B[0] * A[2]) * (A[0] * B[2] - B[0] * A[2]) + (A[0] * B[1] - A[1] * B[0]) * (A[0] * B[1] - A[1] * B[0]); if (per == 0) { return 2; } else if (par == 0) { return 1; } else { return 0; } } static public void Main() { int[] a = { 3, 2, 1 }; int[] b = { 6, 4, 2 }; // Function call Console.WriteLine(find(a, b)); }}// This code is contributed by akashish__ |
Javascript
<script> // JavaScript code for the above approach // Function to find out whether they // are parallel, perpendicular or neither function find(A, B) { // Find A.B let per = A[0] * B[0] + A[1] * B[1] + A[2] * B[2]; // Find A X B let par = (A[1] * B[2] - A[2] * B[1]) * (A[1] * B[2] - A[2] * B[1]) + (A[0] * B[2] - B[0] * A[2]) * (A[0] * B[2] - B[0] * A[2]) + (A[0] * B[1] - A[1] * B[0]) * (A[0] * B[1] - A[1] * B[0]); if (per == 0) { return 2; } else if (par == 0) { return 1; } else { return 0; } } // Driver Code let a = [3, 2, 1]; let b = [6, 4, 2]; // Function call document.write(find(a, b)); // This code is contributed by Potta Lokesh </script> |
Output
1
Time Complexity: O(1)
Auxiliary Space: O(1)
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