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Shortest distance between a Line and a Point in a 3-D plane
• Difficulty Level : Easy
• Last Updated : 08 Feb, 2021

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:

## CPP

 // C++ program to find the Shortest// Distance Between A line and a// Given point.#includeusing 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 VectorsVector 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 vectorsVector 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 vectorsint 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 vectorsVector 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 Vectorostream& 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 linefloat 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 programint 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;}
Output:
Shortest Distance is : 1.63299

Time Complexity: O(1)

Auxiliary Space: O(1)

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