给定一个由两个 N 维向量A和B的大小组成的数组arr[] ,任务是找到两个向量之间的角度。
例子:
Input: arr[] = {-0.5, -2, 1}, brr[] = {-1, -1, -0.3}
Output: 0.845289
Explanation:
Placing the values in the formula , the required result is obtained.
Input: arr[] = {1, -2, 3}, brr[] = {2, 3, -1}
Output: -0.5
方法:该思想基于求两个向量的点积并将其除以向量 A、B 的大小的乘积的数学公式。
Formula:
Considering the two vectors to be separated by angle θ. the dot product of the two vectors is given by the equation:
Therefore,
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the magnitude
// of the given vector
double magnitude(double arr[], int N)
{
// Stores the final magnitude
double magnitude = 0;
// Traverse the array
for (int i = 0; i < N; i++)
magnitude += arr[i] * arr[i];
// Return square root of magnitude
return sqrt(magnitude);
}
// Function to find the dot
// product of two vectors
double dotProduct(double arr[],
double brr[], int N)
{
// Stores dot product
double product = 0;
// Traverse the array
for (int i = 0; i < N; i++)
product = product + arr[i] * brr[i];
// Return the product
return product;
}
void angleBetweenVectors(double arr[],
double brr[], int N)
{
// Stores dot product of two vectors
double dotProductOfVectors
= dotProduct(arr, brr, N);
// Stores magnitude of vector A
double magnitudeOfA
= magnitude(arr, N);
// Stores magnitude of vector B
double magnitudeOfB
= magnitude(brr, N);
// Stores angle between given vectors
double angle = dotProductOfVectors
/ (magnitudeOfA * magnitudeOfB);
// Print the angle
cout << angle;
}
// Driver Code
int main()
{
// Given magnitude arrays
double arr[] = { -0.5, -2, 1 };
double brr[] = { -1, -1, 0.3 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Function call to find the
// angle between two vectors
angleBetweenVectors(arr, brr, N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find the magnitude
// of the given vector
static double magnitude(double arr[], int N)
{
// Stores the final magnitude
double magnitude = 0;
// Traverse the array
for(int i = 0; i < N; i++)
magnitude += arr[i] * arr[i];
// Return square root of magnitude
return Math.sqrt(magnitude);
}
// Function to find the dot
// product of two vectors
static double dotProduct(double[] arr,
double[] brr, int N)
{
// Stores dot product
double product = 0;
// Traverse the array
for(int i = 0; i < N; i++)
product = product + arr[i] * brr[i];
// Return the product
return product;
}
static void angleBetweenVectors(double[] arr,
double[] brr, int N)
{
// Stores dot product of two vectors
double dotProductOfVectors = dotProduct(arr, brr, N);
// Stores magnitude of vector A
double magnitudeOfA = magnitude(arr, N);
// Stores magnitude of vector B
double magnitudeOfB = magnitude(brr, N);
// Stores angle between given vectors
double angle = dotProductOfVectors /
(magnitudeOfA * magnitudeOfB);
// Print the angle
System.out.println(angle);
}
// Driver Code
public static void main(String[] args)
{
// Given magnitude arrays
double[] arr = { -0.5, -2, 1 };
double[] brr = { -1, -1, 0.3 };
// Size of the array
int N = arr.length;
// Function call to find the
// angle between two vectors
angleBetweenVectors(arr, brr, N);
}
}
// This code is contributed by user_qa7r
Python3
# Python3 program for the above approach
import math
# Function to find the magnitude
# of the given vector
def magnitude(arr, N):
# Stores the final magnitude
magnitude = 0
# Traverse the array
for i in range(N):
magnitude += arr[i] * arr[i]
# Return square root of magnitude
return math.sqrt(magnitude)
# Function to find the dot
# product of two vectors
def dotProduct(arr, brr, N):
# Stores dot product
product = 0
# Traverse the array
for i in range(N):
product = product + arr[i] * brr[i]
# Return the product
return product
def angleBetweenVectors(arr, brr, N):
# Stores dot product of two vectors
dotProductOfVectors = dotProduct(arr, brr, N)
# Stores magnitude of vector A
magnitudeOfA = magnitude(arr, N)
# Stores magnitude of vector B
magnitudeOfB = magnitude(brr, N)
# Stores angle between given vectors
angle = (dotProductOfVectors
/ (magnitudeOfA * magnitudeOfB))
# Print the angle
print('%.5f'%angle)
# Driver Code
if __name__ == "__main__":
# Given magnitude arrays
arr = [-0.5, -2, 1]
brr = [-1, -1, 0.3]
# Size of the array
N = len(arr)
# Function call to find the
# angle between two vectors
angleBetweenVectors(arr, brr, N)
# This code is contributed by ukasp.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the magnitude
// of the given vector
static double magnitude(double []arr, int N)
{
// Stores the final magnitude
double magnitude = 0;
// Traverse the array
for(int i = 0; i < N; i++)
magnitude += arr[i] * arr[i];
// Return square root of magnitude
return Math.Sqrt(magnitude);
}
// Function to find the dot
// product of two vectors
static double dotProduct(double []arr,
double []brr, int N)
{
// Stores dot product
double product = 0;
// Traverse the array
for(int i = 0; i < N; i++)
product = product + arr[i] * brr[i];
// Return the product
return product;
}
static void angleBetweenVectors(double []arr,
double []brr, int N)
{
// Stores dot product of two vectors
double dotProductOfVectors = dotProduct(arr, brr, N);
// Stores magnitude of vector A
double magnitudeOfA = magnitude(arr, N);
// Stores magnitude of vector B
double magnitudeOfB = magnitude(brr, N);
// Stores angle between given vectors
double angle = dotProductOfVectors /
(magnitudeOfA * magnitudeOfB);
// Print the angle
Console.Write(angle);
}
// Driver Code
public static void Main()
{
// Given magnitude arrays
double []arr = { -0.5, -2, 1 };
double []brr = { -1, -1, 0.3 };
// Size of the array
int N = arr.Length;
// Function call to find the
// angle between two vectors
angleBetweenVectors(arr, brr, N);
}
}
// This code is contributed by bgangwar59
Javascript
输出:
0.845289
时间复杂度: O(1)
辅助空间: O(1)
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