Check if it is possible to draw a straight line with the given direction cosines

Given three direction cosines l, m and n of a 3-D plane, the task is to check if it is possible to draw a straight line with them or not. Print Yes if possible else print No.

Examples:

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

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

Approach: If a straight line forms angle a with positive X-axis, angle b with positive Y-axis and angle c with positive Z-axis then its direction cosines are cos(a), cos(b) and cos(c). 
For a straight line, cos²(a) + cos²(b) + cos²(c) = 1.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

<?php
// PHP implementation of the approach

// Function that returns true
// if a straight line is possible
function isPossible($x, $y, $z)
{
    $a = round($x * $x + $y * $y + $z * $z);
    if (ceil($a) == 1 && floor($a) == 1)
        return true;
    return false;
}

// Driver code
$l = 0.70710678; $m = 0.5; $n = 0.5;

if (isPossible($l, $m, $n))
    echo("Yes");
else
    echo("No");
// This code is contributed by mukul singh.

Javascript

<script>
// Javascript implementation of the approach

// Function that returns true
// if a straight line is possible
function isPossible(x,y,z)
{
    let a = Math.round(x * x + y * y + z * z);

    if (Math.ceil(a) == 1 && Math.floor(a) == 1)
        return true;
    return false;
}

// Driver code
let l = 0.70710678, m = 0.5, n = 0.5;
if (isPossible(l, m, n))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by rag2127
</script>

Output: 

Yes

Time Complexity: O(1)
Auxiliary Space: O(1)