检查是否可以绘制具有给定方向余弦的直线

给定3-D平面的三个方向余弦l ， m和n ，任务是检查是否可以用它们绘制一条直线。打印是如果可能的话其他打印无。

例子：

Input: l = 0.258, m = 0.80, n = 0.23
Output: No

Input: l = 0.70710678, m = 0.5, n = 0.5
Output: Yes

方法：如果与正X轴，角度b与正Y轴和角度c。与正Z轴的直线形式的角度则其方向余弦是COS的(a)，COS(B)和cos(c)中。
对于一条直线， cos ² (a)+ cos ² (b)+ cos ² (c)= 1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function that returns true
// if a straight line is possible
bool isPossible(float x, float y, float z)
{
    float a = x * x + y * y + z * z;
    if (ceil(a) == 1 && floor(a) == 1)
        return true;
    return false;
}
  
// Driver code
int main()
{
    float l = 0.70710678, m = 0.5, n = 0.5;
  
    if (isPossible(l, m, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function that returns true
// if a straight line is possible
static boolean isPossible(float x, float y, float z)
{
    float a = x * x + y * y + z * z;
    if (Math.ceil(a) == 1 && Math.floor(a) == 1)
        return true;
    return false;
}
  
// Driver code
public static void main(String args[])
{
    float l = 0.70710678f, m = 0.5f, n = 0.5f;
  
    if (isPossible(l, m, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by
// Shashank_Sharma

Python3

# Python3 implementation of the approach 
from math import ceil, floor
  
# Function that returns true 
# if a straight line is possible 
def isPossible(x, y, z) :
  
    a = x * x + y * y + z * z
    a = round(a, 8)
      
    if (ceil(a) == 1 & floor(a) == 1) :
        return True
    return False
  
# Driver code 
if __name__ == "__main__" :
      
    l = 0.70710678
    m = 0.5
    n = 0.5
  
    if (isPossible(l, m, n)): 
        print("Yes") 
    else :
        print("No")
  
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function that returns true
// if a straight line is possible
static bool isPossible(float x, float y, float z)
{
    float a = x * x + y * y + z * z;
    if (Math.Ceiling(a) == 1 && Math.Floor(a) == 1)
        return true;
    return false;
}
  
// Driver code
public static void Main()
{
    float l = 0.70710678f, m = 0.5f, n = 0.5f;
    if (isPossible(l, m, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by Ita_c.

---

输出：

Yes