Orthogonal Vectors: Two vectors are orthogonal to each other when their dot product is 0. How do we define the dot product? Dot product(scalar product) of two n-dimensional vectors A and B, is given by this expression.
Thus the vectors A and B are orthogonal to each other if and only if
Note: In a compact form the above expression can be written as (A^T)B. Example: Consider the vectors v1 and v2 in 3D space.![v_{1}=\left[\begin{array}{c} 1 \\ -2 \\ 4 \end{array}\right], v_{2}=\left[\begin{array}{l} 2 \\ 5 \\ 2 \end{array}\right]](https://geeksforgeeks.org/wp-content/uploads/2023/10/quicklatex.com-9ac97949a7cacb52e0abba17bd3889b9_l3.png "Rendered by QuickLaTeX.com")
Taking the dot product of the vectors.![v_{1}, v_{2}=V_{1}^{T} V_{2}=[1-24]\left[\begin{array}{l} 2 \\ 5 \\ 2 \end{array}\right]=0](https://geeksforgeeks.org/wp-content/uploads/2023/10/quicklatex.com-b26cf19294eadd7f235e7bd08898a7ae_l3.png "Rendered by QuickLaTeX.com")
Hence the vectors are orthogonal to each other. Code: Python program to illustrate orthogonal vectors.
C++14
// C++ program to calculate the dot product of two vectors #include <iostream> #include <vector> using namespace std; // Function to calculate the transpose of a matrix vector<vector<int> > transpose(vector<vector<int> > matrix) { vector<vector<int> > transpose(matrix[0].size(), vector<int>(matrix.size())); for (int i = 0; i < matrix.size(); i++) { for (int j = 0; j < matrix[0].size(); j++) { transpose[j][i] = matrix[i][j]; } } return transpose; } // Function to calculate the dot product of two vectors int dot(vector<vector<int> > vect_A, vector<vector<int> > vect_B) { int product = 0; int n = vect_A[0].size(); // Loop to calculate the dot product for (int i = 0; i < n; i++) { product = product + vect_A[0][i] * vect_B[i][0]; } return product; } int main() { vector<vector<int> > v1 = {{1, -2, 4}}; vector<vector<int> > v2 = {{2, 5, 2}}; vector<vector<int> > transposeOfV1 = transpose(v1); int result = dot(v2, transposeOfV1); cout << "Result = " << result << endl; return 0; } |
Java
import java.util.Arrays; public class OrthogonalVector { public static void main(String[] args) { int[][] v1 = {{1, -2, 4}}; int[][] v2 = {{2, 5, 2}}; int[][] transposeOfV1 = transpose(v1); int result = dot(v2, transposeOfV1); System.out.println("Result = " + result); } public static int[][] transpose(int[][] matrix) { int[][] transpose = new int[(matrix[0].length)][(matrix.length)]; for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[0].length; j++) { transpose[j][i] = matrix[i][j]; } } return transpose; } static int dot(int vect_A[][], int vect_B[][]) { int product = 0; int n = vect_A[0].length; // Loop for calculate dot product for (int i = 0; i < n; i++) product = product + vect_A[0][i] * vect_B[i][0]; return product; } } |
Python3
# A python program to illustrate orthogonal vector # Import numpy module import numpy # Taking two vectors v1 = [[1, -2, 4]] v2 = [[2, 5, 2]] # Transpose of v1 transposeOfV1 = numpy.transpose(v1) # Matrix multiplication of both vectors result = numpy.dot(v2, transposeOfV1) print("Result = ", result) # This code is contributed by Amiya Rout |
C#
using System; public class OrthogonalVector { public static void Main() { int[][] v1 = { new[] { 1, -2, 4 } }; int[][] v2 = { new[] { 2, 5, 2 } }; int[][] transposeOfV1 = Transpose(v1); int result = Dot(v2, transposeOfV1); Console.WriteLine("Result = " + result); } public static int[][] Transpose(int[][] matrix) { int[][] transpose = new int[(matrix[0].Length)][]; for (int i = 0; i < matrix[0].Length; i++) { transpose[i] = new int[(matrix.Length)]; for (int j = 0; j < matrix.Length; j++) { transpose[i][j] = matrix[j][i]; } } return transpose; } static int Dot(int[][] vect_A, int[][] vect_B) { int product = 0; int n = vect_A[0].Length; // Loop for calculate dot product for (int i = 0; i < n; i++) { product = product + vect_A[0][i] * vect_B[i][0]; } return product; } } |
Result = 0
Unit Vector: Let’s consider a vector A. The unit vector of the vector A may be defined as
Let’s understand this by taking an example. Consider a vector A in 2D space.![A=\left[\begin{array}{l} 3 \\ 4 \end{array}\right]](https://geeksforgeeks.org/wp-content/uploads/2023/10/quicklatex.com-a1ef7581cebaeb2c295b99b492eae515_l3.png "Rendered by QuickLaTeX.com")
The magnitude of A is given by
So the unit vector of A can be calculated as![\hat{a}=\frac{A}{|A|}=\left[\begin{array}{l} 3 / 5 \\ 4 / 5 \end{array}\right]](https://geeksforgeeks.org/wp-content/uploads/2023/10/quicklatex.com-a3951a4b8839d585e223f7e2a03d5807_l3.png "Rendered by QuickLaTeX.com")
Properties of unit vector:
- Unit vectors are used to define directions in a coordinate system.
- Any vectors can be written as a product of a unit vector and a scalar magnitude.
Orthonormal vectors: These are the vectors with unit magnitude. Now, take the same 2 vectors which are orthogonal to each other and you know that when I take a dot product between these 2 vectors it is going to 0. So If we also impose the condition that we want each of these vectors to have unit magnitude then what we could possibly do is by taking this vector and then divide this vector by the magnitude of this vector as we see in the unit vector. Now we can write v1 and v2 as![v_{1}=\left[\begin{array}{c} 1 \\ -2 \\ 4 \end{array}\right] / \sqrt{1^{2}+(-2)^{2}+4^{2}} \quad v_{2}=\left[\begin{array}{l} 2 \\ 5 \\ 2 \end{array}\right] / \sqrt{2^{2}+5^{2}+2^{2}}](https://geeksforgeeks.org/wp-content/uploads/2023/10/quicklatex.com-4ddd95dea274e6498a6406f0323ad34f_l3.png "Rendered by QuickLaTeX.com")
So what we do is we have taken the vectors from the previous example and converted them into unit vectors by dividing them by their magnitudes. So, these vectors will still be orthogonal to each other and now individually they also have unit magnitude. Such vectors are known as orthonormal vectors. Note: All orthonormal vectors are orthogonal by the definition itself.
