检查是否可以用给定的方向余弦画一条直线

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画一条直线。如果可能，打印Yes否则打印No 。
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.

PHP




Javascript


输出：

Yes