Shortest distance between a Line and a Point in a 3-D plane

Given a line passing through two points A and B and an arbitrary point C in a 3-D plane, the task is to find the shortest distance between the point C and the line passing through the points A and B.

Examples:

Input: A = (5, 2, 1), B = (3, 1, -1), C = (0, 2, 3)
Output: Shortest Distance is 5

Input: A = (4, 2, 1), B = (3, 2, 1), C = (0, 2, 0)
Output: Shortest Distance is 1

Consider a point C and a line that passes through A and B as shown in the below figure.



Now Consider the vectors, AB and AC and the shortest distance as CD. The Shortest Distance is always the perpendicular distance. The point D is taken on AB such that CD is perpendicular to AB.

Construct BP and CP as shown in the figure to form a Parallelogram. Now C is a vertex of parallelogram ABPC and CD is perpendicular to Side AB. Hence CD is the height of the parallelogram.

Note: In the case when D does not fall on line segment AB there will be a point D’ such that PD’ is perpendicular to AB and D’ lies on line segment AB with CD = PD’.

The magnitude of cross product AB and AC gives the Area of the parallelogram. Also, the area of a parallelogram is Base * Height = AB * CD. So,

CD = |ABxAC| / |AB|

Below is the CPP program to find the shortest distance:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the Shortest
// Distance Between A line and a
// Given point.
#include<bits/stdc++.h>
using namespace std;
  
class Vector {
private:
    int x, y, z;
    // 3D Coordinates of the Vector
  
public:
    Vector(int x, int y, int z)
    {
        // Constructor
        this->x = x;
        this->y = y;
        this->z = z;
    }
    Vector operator+(Vector v); // ADD 2 Vectors
    Vector operator-(Vector v); // Subtraction
    int operator^(Vector v); // Dot Product
    Vector operator*(Vector v); // Cross Product
    float magnitude()
    {
        return sqrt(pow(x, 2) + pow(y, 2) + pow(z, 2));
    }
    friend ostream& operator<<(ostream& out, const Vector& v);
    // To output the Vector
};
  
// ADD 2 Vectors
Vector Vector::operator+(Vector v)
{
    int x1, y1, z1;
    x1 = x + v.x;
    y1 = y + v.y;
    z1 = z + v.z;
    return Vector(x1, y1, z1);
}
  
// Subtract 2 vectors
Vector Vector::operator-(Vector v)
{
    int x1, y1, z1;
    x1 = x - v.x;
    y1 = y - v.y;
    z1 = z - v.z;
    return Vector(x1, y1, z1);
}
  
// Dot product of 2 vectors
int Vector::operator^(Vector v)
{
    int x1, y1, z1;
    x1 = x * v.x;
    y1 = y * v.y;
    z1 = z * v.z;
    return (x1 + y1 + z1);
}
  
// Cross product of 2 vectors
Vector Vector::operator*(Vector v)
{
    int x1, y1, z1;
    x1 = y * v.z - z * v.y;
    y1 = z * v.x - x * v.z;
    z1 = x * v.y - y * v.x;
    return Vector(x1, y1, z1);
}
  
// Display Vector
ostream& operator<<(ostream& out,
                    const Vector& v)
{
    out << v.x << "i ";
    if (v.y >= 0)
        out << "+ ";
    out << v.y << "j ";
    if (v.z >= 0)
        out << "+ ";
    out << v.z << "k" << endl;
    return out;
}
  
// calculate shortest dist. from point to line
float shortDistance(Vector line_point1, Vector line_point2,
                    Vector point)
{
    Vector AB = line_point2 - line_point1;
    Vector AC = point - line_point1;
    float area = Vector(AB * AC).magnitude();
    float CD = area / AB.magnitude();
    return CD;
}
  
// Driver program
int main()
{
    // Taking point C as (2, 2, 2)
    // Line Passes through A(4, 2, 1)
    // and B(8, 4, 2).
    Vector line_point1(4, 2, 1), line_point2(8, 4, 2);
    Vector point(2, 2, 2);
  
    cout << "Shortest Distance is : "
         << shortDistance(line_point1, line_point2, point);
  
  return 0;
}

chevron_right


Output:

Shortest Distance is : 1.63299


My Personal Notes arrow_drop_up

Final year BTech IT student at DTU, Upcoming Technology Analyst at Morgan Stanley

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.