Program to find the shortest distance between diagonal and edge skew of a Cube

Given an integer A, denoting the length of a cube, the task is to find the shortest distance between the diagonal of a cube and an edge skew to it i.e. KL in the below figure.

Examples :

Input: A = 2
Output: 1.4142
Explanation:
Length of KL = A / ?2
Length of KL = 2 / 1.41 = 1.4142

Input: A = 3
Output: 2.1213

Approach: The idea to solve the problem is based on the following mathematical formula:

Let's draw a perpendicular from K to down face of cube as Q.
Using Pythagorean Theorem in triangle QKL,

KL² = QK² + QL²

l² = (a/2)² + (a/2)² 

l² = 2 * (a/2)²

l = a / sqrt(2)

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the shortest distance
// between the diagonal of a cube and
// an edge skew to it
float diagonalLength(float a)
{
    // Stores the required distance
    float L = a / sqrt(2);

    // Print the required distance
    cout << L;
}

// Driver Code
int main()
{
    // Given side of the cube
    float a = 2;

    // Function call to find the shortest
    // distance between the diagonal of a
    // cube and an edge skew to it
    diagonalLength(a);

    return 0;
}

// Java program for the above approach

class GFG {

    // Function to find the shortest
    // distance between the diagonal of a
    // cube and an edge skew to it
    static void diagonalLength(float a)
    {
        // Stores the required distance
        float L = a / (float)Math.sqrt(2);

        // Print the required distance
        System.out.println(L);
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given side of the cube
        float a = 2;

        // Function call to find the shortest
        // distance between the diagonal of a
        // cube and an edge skew to it
        diagonalLength(a);
    }
}

# Python3 program for the above approach

from math import sqrt

# Function to find the shortest 
# distance between the diagonal of a 
# cube and an edge skew to it
def diagonalLength(a):
  
    # Stores the required distance
    L = a / sqrt(2)

    # Print the required distance
    print(L)


# Given side of the cube
a = 2

# Function call to find the shortest 
# distance between the diagonal of a 
# cube and an edge skew to it
diagonalLength(a)

// C# program for the above approach

using System;
class GFG {

    // Function to find the shortest
    // distance between the diagonal of a
    // cube and an edge skew to it
    static void diagonalLength(float a)
    {
        // Stores the required distance
        float L = a / (float)Math.Sqrt(2);

        // Print the required distance
        Console.Write(L);
    }

    // Driver Code
    public static void Main()
    {
        // Given side of the cube
        float a = 2;

        // Function call to find the shortest
        // distance between the diagonal of a
        // cube and an edge skew to it
        diagonalLength(a);
    }
}

<?php 
// PHP program for the above approach
  
// Function to find the shortest 
// distance between the diagonal of a 
// cube and an edge skew to it
function diagonalLength($a) 
{ 
    // Stores the required distance
    $L = $a / sqrt(2); 

    # Print the required distance
    echo $L; 
} 

// Given side of the cube
$a = 2; 

// Function call to find the shortest 
// distance between the diagonal of a 
// cube and an edge skew to it
diagonalLength($a); 

?> 

<script>
// javascript program for the above approach

// Function to find the shortest distance
// between the diagonal of a cube and
// an edge skew to it
function diagonalLength( a)
{

    // Stores the required distance
    let L = a / Math.sqrt(2);

    // Print the required distance
     document.write( L.toFixed(5));
}

// Driver Code

    // Given side of the cube
    let a = 2;

    // Function call to find the shortest
    // distance between the diagonal of a
    // cube and an edge skew to it
    diagonalLength(a);

// This code is contributed by todaysgaurav 

</script>

1.41421

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