There are two vector A and B and we have to find the dot product and cross product of two vector array. Dot product is also known as scalar product and cross product also known as vector product.
Dot Product – Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. Where i, j and k are the unit vector along the x, y and z directions. Then dot product is calculated as dot product = a1 * b1 + a2 * b2 + a3 * b3
Example –
A = 3 * i + 5 * j + 4 * k B = 2 * i + 7 * j + 5 * k dot product = 3 * 2 + 5 * 7 + 4 * 5 = 6 + 35 + 20 = 61
Cross Product – Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k. Then cross product is calculated as cross product = (a2 * b3 – a3 * b2) * i + (a3 * b1 – a1 * b3) * j + (a1 * b2 – a2 * b1) * k, where [(a2 * b3 – a3 * b2), (a3 * b1 – a1 * b3), (a1 * b2 – a2 * b1)] are the coefficient of unit vector along i, j and k directions.
Example –
A = 3 * i + 5 * j + 4 * k B = 2 * i + 7 * j + 5 * k cross product = (5 * 5 - 4 * 7) * i + (4 * 2 - 3 * 5) * j + (3 * 7 - 5 * 2) * k = (-3)*i + (-7)*j + (11)*k
Example –
Input: vect_A[] = {3, -5, 4} vect_B[] = {2, 6, 5} Output: Dot product: -4 Cross product = -49 -7 28
Code-
C++
// C++ implementation for dot product // and cross product of two vector. #include <bits/stdc++.h> #define n 3 using namespace std; // Function that return // dot product of two vector array. int dotProduct( int vect_A[], int vect_B[]) { int product = 0; // Loop for calculate dot product for ( int i = 0; i < n; i++) product = product + vect_A[i] * vect_B[i]; return product; } // Function to find // cross product of two vector array. void crossProduct( int vect_A[], int vect_B[], int cross_P[]) { cross_P[0] = vect_A[1] * vect_B[2] - vect_A[2] * vect_B[1]; cross_P[1] = vect_A[2] * vect_B[0] - vect_A[0] * vect_B[2]; cross_P[2] = vect_A[0] * vect_B[1] - vect_A[1] * vect_B[0]; } // Driver function int main() { int vect_A[] = { 3, -5, 4 }; int vect_B[] = { 2, 6, 5 }; int cross_P[n]; // dotProduct function call cout << "Dot product:" ; cout << dotProduct(vect_A, vect_B) << endl; // crossProduct function call cout << "Cross product:" ; crossProduct(vect_A, vect_B, cross_P); // Loop that print // cross product of two vector array. for ( int i = 0; i < n; i++) cout << cross_P[i] << " " ; return 0; } |
Java
// java implementation for dot product // and cross product of two vector. import java.io.*; class GFG { static int n = 3 ; // Function that return // dot product of two vector array. static int dotProduct( int vect_A[], int vect_B[]) { int product = 0 ; // Loop for calculate dot product for ( int i = 0 ; i < n; i++) product = product + vect_A[i] * vect_B[i]; return product; } // Function to find // cross product of two vector array. static void crossProduct( int vect_A[], int vect_B[], int cross_P[]) { cross_P[ 0 ] = vect_A[ 1 ] * vect_B[ 2 ] - vect_A[ 2 ] * vect_B[ 1 ]; cross_P[ 1 ] = vect_A[ 2 ] * vect_B[ 0 ] - vect_A[ 0 ] * vect_B[ 2 ]; cross_P[ 2 ] = vect_A[ 0 ] * vect_B[ 1 ] - vect_A[ 1 ] * vect_B[ 0 ]; } // Driver code public static void main(String[] args) { int vect_A[] = { 3 , - 5 , 4 }; int vect_B[] = { 2 , 6 , 5 }; int cross_P[] = new int [n]; // dotProduct function call System.out.print( "Dot product:" ); System.out.println(dotProduct(vect_A, vect_B)); // crossProduct function call System.out.print( "Cross product:" ); crossProduct(vect_A, vect_B, cross_P); // Loop that print // cross product of two vector array. for ( int i = 0 ; i < n; i++) System.out.print(cross_P[i] + " " ); } } // This code is contributed by vt_m |
Python3
# Python3 implementation for dot product # and cross product of two vector. n = 3 # Function that return # dot product of two vector array. def dotProduct(vect_A, vect_B): product = 0 # Loop for calculate dot product for i in range ( 0 , n): product = product + vect_A[i] * vect_B[i] return product # Function to find # cross product of two vector array. def crossProduct(vect_A, vect_B, cross_P): cross_P.append(vect_A[ 1 ] * vect_B[ 2 ] - vect_A[ 2 ] * vect_B[ 1 ]) cross_P.append(vect_A[ 2 ] * vect_B[ 0 ] - vect_A[ 0 ] * vect_B[ 2 ]) cross_P.append(vect_A[ 0 ] * vect_B[ 1 ] - vect_A[ 1 ] * vect_B[ 0 ]) # Driver function if __name__ = = '__main__' : vect_A = [ 3 , - 5 , 4 ] vect_B = [ 2 , 6 , 5 ] cross_P = [] # dotProduct function call print ( "Dot product:" , end = " " ) print (dotProduct(vect_A, vect_B)) # crossProduct function call print ( "Cross product:" , end = " " ) crossProduct(vect_A, vect_B, cross_P) # Loop that print # cross product of two vector array. for i in range ( 0 , n): print (cross_P[i], end = " " ) # This code is contributed by # Sanjit_Prasad |
C#
// C# implementation for dot product // and cross product of two vector. using System; class GFG { static int n = 3; // Function that return dot // product of two vector array. static int dotProduct( int [] vect_A, int [] vect_B) { int product = 0; // Loop for calculate dot product for ( int i = 0; i < n; i++) product = product + vect_A[i] * vect_B[i]; return product; } // Function to find cross product // of two vector array. static void crossProduct( int [] vect_A, int [] vect_B, int [] cross_P) { cross_P[0] = vect_A[1] * vect_B[2] - vect_A[2] * vect_B[1]; cross_P[1] = vect_A[2] * vect_B[0] - vect_A[0] * vect_B[2]; cross_P[2] = vect_A[0] * vect_B[1] - vect_A[1] * vect_B[0]; } // Driver code public static void Main() { int [] vect_A = { 3, -5, 4 }; int [] vect_B = { 2, 6, 5 }; int [] cross_P = new int [n]; // dotProduct function call Console.Write( "Dot product:" ); Console.WriteLine( dotProduct(vect_A, vect_B)); // crossProduct function call Console.Write( "Cross product:" ); crossProduct(vect_A, vect_B, cross_P); // Loop that print // cross product of two vector array. for ( int i = 0; i < n; i++) Console.Write(cross_P[i] + " " ); } } // This code is contributed by vt_m. |
PHP
<?php // PHP implementation for dot // product and cross product // of two vector. $n = 3; // Function that return // dot product of two // vector array. function dotproduct( $vect_A , $vect_B ) { global $n ; $product = 0; // Loop for calculate // dot product for ( $i = 0; $i < $n ; $i ++) $product = $product + $vect_A [ $i ] * $vect_B [ $i ]; return $product ; } // Function to find // cross product of // two vector array. function crossproduct( $vect_A , $vect_B , $cross_P ) { $cross_P [0] = $vect_A [1] * $vect_B [2] - $vect_A [2] * $vect_B [1]; $cross_P [1] = $vect_A [2] * $vect_B [0] - $vect_A [0] * $vect_B [2]; $cross_P [2] = $vect_A [0] * $vect_B [1] - $vect_A [1] * $vect_B [0]; return $cross_P ; } // Driver Code $vect_A = array ( 3, -5, 4 ); $vect_B = array ( 2, 6, 5 ); $cross_P = array_fill (0, $n , 0); // dotproduct function call echo "Dot product:" ; echo dotproduct( $vect_A , $vect_B ); // crossproduct function call echo "\nCross product:" ; $cross_P = crossproduct( $vect_A , $vect_B , $cross_P ); // Loop that print // cross product of // two vector array. for ( $i = 0; $i < $n ; $i ++) echo $cross_P [ $i ] . " " ; // This code is contributed by mits ?> |
Javascript
<script> // Javascript implementation for dot product // and cross product of two vector. let n = 3; // Function that return // dot product of two vector array. function dotProduct(vect_A, vect_B) { let product = 0; // Loop for calculate dot product for (let i = 0; i < n; i++) product = product + vect_A[i] * vect_B[i]; return product; } // Function to find // cross product of two vector array. function crossProduct(vect_A, vect_B, cross_P) { cross_P[0] = vect_A[1] * vect_B[2] - vect_A[2] * vect_B[1]; cross_P[1] = vect_A[2] * vect_B[0] - vect_A[0] * vect_B[2]; cross_P[2] = vect_A[0] * vect_B[1] - vect_A[1] * vect_B[0]; } // Driver code let vect_A = [ 3, -5, 4 ]; let vect_B = [ 2, 6, 5 ]; let cross_P = []; // dotProduct function call document.write( "Dot product:" ); document.write(dotProduct(vect_A, vect_B) + "<br/>" ); // crossProduct function call document.write( "Cross product:" ); crossProduct(vect_A, vect_B, cross_P); // Loop that print // cross product of two vector array. for (let i = 0; i < n; i++) document.write(cross_P[i] + " " ); // This code is contributed by sanjoy_62. </script> |
Dot product:-4 Cross product:-49 -7 28
Time Complexity: O(3), the code will run in a constant time because the size of the arrays will be always 3.
Auxiliary Space: O(3), no extra space is required, so it is a constant.
This article is contributed by Dharmendra Kumar. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!