3D 中一对线之间的角度

给定 3D 平面中三个点A(x1, y1, z1)、B(x2, y2, z2)和 C(x3, y3, z3)的坐标，其中B是线AB和BC的交点，则任务是找到线AB 和 BC之间的角度。

例子：

Input: x1 = 1, y1 = 3, z1 = 3; x2 = 3, y2 = 4, z2 = 5; x3 = 5, y3 = 6, z3 = 9; 
Output: 54.6065
Input: x1 = 10, y1 = 10, z1 = 10; x2 = 0, y2 = 0, z2 = 0; x3 = 15, y3 = 10, z3 = 15; 
Output: 56.4496  

方法：

1. 用给定坐标根据方向比求直线AB和BC的方程为：

AB = (x1 – x2)i + (y1 – y2)j + (z1 – z2)k 
BC = (x3 – x2)i + (y3 – y2)j + (z3 – z2)k 
 

2. 对AB线和BC线的两个方向比使用cosθ的公式，求AB线和BC线夹角的余弦为：

where, 
AB.BC is the dot product of direction ratios AB and BC. 
|AB| is the magnitude of line AB 
|BC| is the magnitude of line BC 

3. 假设有两个方向比率：

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

然后

Dot Product(A.B) = a*x + b*y + c*z 
magnitude of A = |A| = $\sqrt{a^{2} + b^{2} + c^{2}}$
magnitude of B = |B| = $\sqrt{x^{2} + y^{2} + z^{2}}$ 

4. 计算出的角度的余弦给出了以弧度为单位的余弦值。要找到角度，请将余弦值乘以(180/ Π ) 。

下面是上述方法的实现：

C++

A = ai + bj + ck
B = xi + yj + zk

Java

// C++ program for the above approach
 
#include "bits/stdc++.h"
#define PI 3.14
using namespace std;
 
// Function to find the angle between
// the two lines
void calculateAngle(
    int x1, int y1, int z1,
    int x2, int y2, int z2,
    int x3, int y3, int z3)
{
    // Find direction ratio of line AB
    int ABx = x1 - x2;
    int ABy = y1 - y2;
    int ABz = z1 - z2;
 
    // Find direction ratio of line BC
    int BCx = x3 - x2;
    int BCy = y3 - y2;
    int BCz = z3 - z2;
 
    // Find the dotProduct
    // of lines AB & BC
    double dotProduct
        = ABx * BCx
          + ABy * BCy
          + ABz * BCz;
 
    // Find magnitude of
    // line AB and BC
    double magnitudeAB
        = ABx * ABx
          + ABy * ABy
          + ABz * ABz;
    double magnitudeBC
        = BCx * BCx
          + BCy * BCy
          + BCz * BCz;
 
    // Find the cosine of
    // the angle formed
    // by line AB and BC
    double angle = dotProduct;
    angle /= sqrt(
        magnitudeAB * magnitudeBC);
 
    // Find angle in radian
    angle = (angle * 180) / PI;
 
    // Print the angle
    cout << abs(angle) << endl;
}
 
// Driver Code
int main()
{
 
    // Given coordinates
    // Points A
    int x1 = 1, y1 = 3, z1 = 3;
 
    // Points B
    int x2 = 3, y2 = 4, z2 = 5;
 
    // Points C
    int x3 = 5, y3 = 6, z3 = 9;
 
    // Function Call
    calculateAngle(x1, y1, z1,
                   x2, y2, z2,
                   x3, y3, z3);
 
    return 0;
}

Python3

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find the angle
// between the two lines
static void calculateAngle(int x1, int y1, int z1,
                           int x2, int y2, int z2,
                           int x3, int y3, int z3)
{
     
    // Find direction ratio of line AB
    int ABx = x1 - x2;
    int ABy = y1 - y2;
    int ABz = z1 - z2;
 
    // Find direction ratio of line BC
    int BCx = x3 - x2;
    int BCy = y3 - y2;
    int BCz = z3 - z2;
 
    // Find the dotProduct
    // of lines AB & BC
    double dotProduct = ABx * BCx +
                        ABy * BCy +
                        ABz * BCz;
 
    // Find magnitude of
    // line AB and BC
    double magnitudeAB = ABx * ABx +
                         ABy * ABy +
                         ABz * ABz;
    double magnitudeBC = BCx * BCx +
                         BCy * BCy +
                         BCz * BCz;
 
    // Find the cosine of the
    // angle formed by line
    // AB and BC
    double angle = dotProduct;
    angle /= Math.sqrt(magnitudeAB * magnitudeBC);
 
    // Find angle in radian
    angle = (angle * 180) / 3.14;
 
    // Print the angle
    System.out.printf("%.4f", Math.abs(angle));
}
 
// Driver code
public static void main(String[] args)
{
 
    // Given coordinates
    // Points A
    int x1 = 1, y1 = 3, z1 = 3;
 
    // Points B
    int x2 = 3, y2 = 4, z2 = 5;
 
    // Points C
    int x3 = 5, y3 = 6, z3 = 9;
 
    // Function Call
    calculateAngle(x1, y1, z1,
                   x2, y2, z2,
                   x3, y3, z3);
}
}
 
// This code is contributed by offbeat

C#

# Python3 program for the above approach
import math
 
# Function to find the angle
# between the two lines
def calculateAngle(x1, y1, z1,
                   x2, y2, z2,
                   x3, y3, z3):
                        
    # Find direction ratio of line AB
    ABx = x1 - x2;
    ABy = y1 - y2;
    ABz = z1 - z2;
 
    # Find direction ratio of line BC
    BCx = x3 - x2;
    BCy = y3 - y2;
    BCz = z3 - z2;
 
    # Find the dotProduct
    # of lines AB & BC
    dotProduct = (ABx * BCx +
                  ABy * BCy +
                  ABz * BCz);
 
    # Find magnitude of
    # line AB and BC
    magnitudeAB = (ABx * ABx +
                   ABy * ABy +
                   ABz * ABz);
    magnitudeBC = (BCx * BCx +
                   BCy * BCy +
                   BCz * BCz);
 
    # Find the cosine of
    # the angle formed
    # by line AB and BC
    angle = dotProduct;
    angle /= math.sqrt(magnitudeAB *
                       magnitudeBC);
 
    # Find angle in radian
    angle = (angle * 180) / 3.14;
 
    # Print angle
    print(round(abs(angle), 4))
 
# Driver Code
if __name__=='__main__':
 
    # Given coordinates
    # Points A
    x1, y1, z1 = 1, 3, 3;
 
    # Points B
    x2, y2, z2 = 3, 4, 5;
 
    # Points C
    x3, y3, z3 = 5, 6, 9;
  
    # Function Call
    calculateAngle(x1, y1, z1,
                   x2, y2, z2,
                   x3, y3, z3);
 
# This code is contributed by AbhiThakur

Javascript

// C# program for the above approach
using System;
class GFG{
     
// Function to find the angle
// between the two lines
static void calculateAngle(int x1, int y1,
                           int z1, int x2,
                           int y2, int z2,
                           int x3, int y3,
                           int z3)
{
     
    // Find direction ratio of line AB
    int ABx = x1 - x2;
    int ABy = y1 - y2;
    int ABz = z1 - z2;
 
    // Find direction ratio of line BC
    int BCx = x3 - x2;
    int BCy = y3 - y2;
    int BCz = z3 - z2;
 
    // Find the dotProduct
    // of lines AB & BC
    double dotProduct = ABx * BCx +
                        ABy * BCy +
                        ABz * BCz;
 
    // Find magnitude of
    // line AB and BC
    double magnitudeAB = ABx * ABx +
                         ABy * ABy +
                         ABz * ABz;
    double magnitudeBC = BCx * BCx +
                         BCy * BCy +
                         BCz * BCz;
 
    // Find the cosine of the
    // angle formed by line
    // AB and BC
    double angle = dotProduct;
    angle /= Math.Sqrt(magnitudeAB *
                       magnitudeBC);
 
    // Find angle in radian
    angle = (angle * 180) / 3.14;
 
    // Print the angle
    Console.Write(String.Format("{0:F4}", Math.Abs(angle)));
}
 
// Driver code
public static void Main()
{
 
    // Given coordinates
    // Points A
    int x1 = 1, y1 = 3, z1 = 3;
 
    // Points B
    int x2 = 3, y2 = 4, z2 = 5;
 
    // Points C
    int x3 = 5, y3 = 6, z3 = 9;
 
    // Function Call
    calculateAngle(x1, y1, z1,
                   x2, y2, z2,
                   x3, y3, z3);
}
}
 
// This code is contributed by Code_Mech

输出：

 
 
// Javascript program for the above approach
 
var PI = 3.14;
 
// Function to find the angle between
// the two lines
function calculateAngle(
    x1, y1, z1,
    x2, y2, z2,
    x3, y3, z3)
{
    // Find direction ratio of line AB
    var ABx = x1 - x2;
    var ABy = y1 - y2;
    var ABz = z1 - z2;
 
    // Find direction ratio of line BC
    var BCx = x3 - x2;
    var BCy = y3 - y2;
    var BCz = z3 - z2;
 
    // Find the dotProduct
    // of lines AB & BC
    var dotProduct
        = ABx * BCx
          + ABy * BCy
          + ABz * BCz;
 
    // Find magnitude of
    // line AB and BC
    var magnitudeAB
        = ABx * ABx
          + ABy * ABy
          + ABz * ABz;
    var magnitudeBC
        = BCx * BCx
          + BCy * BCy
          + BCz * BCz;
 
    // Find the cosine of
    // the angle formed
    // by line AB and BC
    var angle = dotProduct;
    angle /= Math.sqrt(
        magnitudeAB * magnitudeBC);
 
    // Find angle in radian
    angle = (angle * 180) / PI;
 
    // Prvar the angle
    document.write(Math.abs(angle).toFixed(4));
}
 
// Driver Code
// Given coordinates
// Povars A
var x1 = 1, y1 = 3, z1 = 3;
// Povars B
var x2 = 3, y2 = 4, z2 = 5;
// Povars C
var x3 = 5, y3 = 6, z3 = 9;
// Function Call
calculateAngle(x1, y1, z1,
               x2, y2, z2,
               x3, y3, z3);