三维几何中两条线的共面性
给定两条直线 L1 和 L2 ，每条直线都经过一个点，该点的位置向量为 (X，Y，Z) ，并且与方向比为 (a，b，c) 的直线平行，任务是检查直线 L1 和 L2 是否共面。
共面:如果两条线在同一个平面上，那么线可以称为共面。
示例:
输入: L1: (x1，y1，z1) = (-3，1，5)和(a1，b1，c1) = (-3，1，5) L2: (x1，y1，z1) = (-1，2，5)和(a1，b1，c1) = (-1，2，5) 输出:线路共面
输入: L1: (x1，y1，z1) = (1，2，3)和(a1，b1，c1) = (2，4，6) L2: (x1，y1，z1) = (-1，2，-3)和(a1，b1，c1) = (3，4，5) 输出:线路不共面
进场:
有两种方法可以在三维空间中表示一条线:
向量形式: 两条直线的方程，其共面度以向量形式确定。
在上面的直线方程中，矢量是三维平面中的一个点，给定的直线从该点穿过，称为位置矢量 a 和 b 矢量是三维平面中的矢量线，我们的给定直线与该点平行。所以可以说线(1)通过点，说 A ，位置向量 a1 平行于向量B1；线(2)通过点，说 B 位置向量 a2 平行于向量 b2 。因此:
当且仅当 AB 矢量垂直于矢量 b1 和 b2 的叉积时，给定的直线为共面，即，
这里向量 b1 和 b2 的叉积将给出另一条向量线，该向量线将垂直于 b1 和 b2 向量线。 AB 是连接两条给定直线的位置向量 a1 和 a2 的直线向量。现在，通过确定上面的点积是否为零来检查两条线是否共面。****
*笛卡尔形式:* 让 (x1，y1，z1) 和 (x2，y2，z2) 分别为点 A 和 B 的坐标。 设 a1、b1、c1 和 a2、b2、c2 分别为向量 b1 和 b2 的方向比。然后
给定的线是共面的当且仅当:
在笛卡尔形式中，它可以表示为:
因此，对于这两种类型的表单，都需要输入位置向量 a1 和 a2 分别作为 (x1，y1，z1) 和 (x2，y2，z2) ，向量 b1 和 b2 的方向比分别作为 (a1，b1，c1) 和 (a2，b2，c2) 。 按照以下步骤解决问题:
- 初始化一个 3×3 矩阵来存储上面显示的行列式的元素。
- 计算 b2 和 b1 的叉积和的点积(a2–a1)。
- 如果行列式的值为 0，则直线共面。否则，它们不共面。
下面是上述方法的实现:
C++
// C++ program implement
// the above approach
#include <iostream>
using namespace std;
// Function to generate determinant
int det(int d[][3])
{
int Sum = d[0][0] * ((d[1][1] * d[2][2]) - (d[2][1] * d[1][2]));
Sum -= d[0][1] * ((d[1][0] * d[2][2]) - (d[1][2] * d[2][0]));
Sum += d[0][2] * ((d[0][1] * d[1][2]) -(d[0][2] * d[1][1]));
// Return the sum
return Sum;
}
// Driver Code
int main()
{
// Position vector of first line
int x1 = -3, y1 = 1, z1 = 5;
// Direction ratios of line to
// which first line is parallel
int a1 = -3, b1 = 1, c1 = 5;
// Position vectors of second line
int x2 = -1, y2 = 2, z2 = 5;
// Direction ratios of line to
// which second line is parallel
int a2 = -1, b2 = 2, c2 = 5;
// Determinant to check coplanarity
int det_list[3][3] = { {x2 - x1, y2 - y1, z2 - z1},
{a1, b1, c1}, {a2, b2, c2}};
// If determinant is zero
if(det(det_list) == 0)
{
cout << "Lines are coplanar" << endl;
}
// Otherwise
else
{
cout << "Lines are non coplanar" << endl;
}
return 0;
}
// This code is contributed by avanitrachhadiya2155
Java 语言(一种计算机语言，尤用于创建网站)
// Java program implement
// the above approach
import java.io.*;
class GFG{
// Function to generate determinant
static int det(int[][] d)
{
int Sum = d[0][0] * ((d[1][1] * d[2][2]) -
(d[2][1] * d[1][2]));
Sum -= d[0][1] * ((d[1][0] * d[2][2]) -
(d[1][2] * d[2][0]));
Sum += d[0][2] * ((d[0][1] * d[1][2]) -
(d[0][2] * d[1][1]));
// Return the sum
return Sum;
}
// Driver Code
public static void main (String[] args)
{
// Position vector of first line
int x1 = -3, y1 = 1, z1 = 5;
// Direction ratios of line to
// which first line is parallel
int a1 = -3, b1 = 1, c1 = 5;
// Position vectors of second line
int x2 = -1, y2 = 2, z2 = 5;
// Direction ratios of line to
// which second line is parallel
int a2 = -1, b2 = 2, c2 = 5;
// Determinant to check coplanarity
int[][] det_list = { {x2 - x1, y2 - y1, z2 - z1},
{a1, b1, c1}, {a2, b2, c2}};
// If determinant is zero
if(det(det_list) == 0)
System.out.print("Lines are coplanar");
// Otherwise
else
System.out.print("Lines are non coplanar");
}
}
// This code is contributed by offbeat
Python 3
# Python Program implement
# the above approach
# Function to generate determinant
def det(d):
Sum = d[0][0] * ((d[1][1] * d[2][2])
- (d[2][1] * d[1][2]))
Sum -= d[0][1] * ((d[1][0] * d[2][2])
- (d[1][2] * d[2][0]))
Sum += d[0][2] * ((d[0][1] * d[1][2])
- (d[0][2] * d[1][1]))
# Return the sum
return Sum
# Driver Code
if __name__ == '__main__':
# Position vector of first line
x1, y1, z1 = -3, 1, 5
# Direction ratios of line to
# which first line is parallel
a1, b1, c1 = -3, 1, 5
# Position vectors of second line
x2, y2, z2 = -1, 2, 5
# Direction ratios of line to
# which second line is parallel
a2, b2, c2 = -1, 2, 5
# Determinant to check coplanarity
det_list = [[x2-x1, y2-y1, z2-z1],
[a1, b1, c1], [a2, b2, c2]]
# If determinant is zero
if(det(det_list) == 0):
print("Lines are coplanar")
# Otherwise
else:
print("Lines are non coplanar")
C#
// C# program implement
// the above approach
using System;
class GFG{
// Function to generate determinant
static int det(int[,] d)
{
int Sum = d[0, 0] * ((d[1, 1] * d[2, 2]) -
(d[2, 1] * d[1, 2]));
Sum -= d[0, 1] * ((d[1, 0] * d[2, 2]) -
(d[1, 2] * d[2, 0]));
Sum += d[0, 2] * ((d[0, 1] * d[1, 2]) -
(d[0, 2] * d[1, 1]));
// Return the sum
return Sum;
}
// Driver Code
public static void Main()
{
// Position vector of first line
int x1 = -3, y1 = 1, z1 = 5;
// Direction ratios of line to
// which first line is parallel
int a1 = -3, b1 = 1, c1 = 5;
// Position vectors of second line
int x2 = -1, y2 = 2, z2 = 5;
// Direction ratios of line to
// which second line is parallel
int a2 = -1, b2 = 2, c2 = 5;
// Determinant to check coplanarity
int[,] det_list = { {x2 - x1, y2 - y1, z2 - z1},
{a1, b1, c1}, {a2, b2, c2}};
// If determinant is zero
if (det(det_list) == 0)
Console.Write("Lines are coplanar");
// Otherwise
else
Console.Write("Lines are non coplanar");
}
}
// This code is contributed by sanjoy_62
java 描述语言
<script>
// JavaScript program for the above approach
// Function to generate determinant
function det(d)
{
let Sum = d[0][0] * ((d[1][1] * d[2][2]) -
(d[2][1] * d[1][2]));
Sum -= d[0][1] * ((d[1][0] * d[2][2]) -
(d[1][2] * d[2][0]));
Sum += d[0][2] * ((d[0][1] * d[1][2]) -
(d[0][2] * d[1][1]));
// Return the sum
return Sum;
}
// Driver Code
// Position vector of first line
let x1 = -3, y1 = 1, z1 = 5;
// Direction ratios of line to
// which first line is parallel
let a1 = -3, b1 = 1, c1 = 5;
// Position vectors of second line
let x2 = -1, y2 = 2, z2 = 5;
// Direction ratios of line to
// which second line is parallel
let a2 = -1, b2 = 2, c2 = 5;
// Determinant to check coplanarity
let det_list = [[x2 - x1, y2 - y1, z2 - z1],
[a1, b1, c1], [a2, b2, c2]];
// If determinant is zero
if(det(det_list) == 0)
document.write("Lines are coplanar");
// Otherwise
else
document.write("Lines are non coplanar");
</script>
**Output:
Lines are coplanar**
*时间复杂度: O(1) 辅助空间: O(1)***
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