Program for dot product and cross product of two vectors
Last Updated :
29 Jul, 2024
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 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 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# 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.
<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>
<?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
?>
OutputDot 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.
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