This note describes the technique and gives the solution to finding the shortest distance from a point to a line or line segment. The equation of a line defined through two points P1 (x1,y1) and P2 (x2,y2) is

The point P3 (x3,y3) is closest to the line at the tangent to the line which passes through P3, that is, the dot product of the tangent and line is 0, thus

Substituting the equation of the line gives

Solving this gives the value of u

Substituting this into the equation of the line gives the point of intersection (x,y) of the tangent as

The distance therefore between the point P3 and the line is the distance between (x,y) above and P3.

• The only special testing for a software implementation is to ensure that P1 and P2 are not coincident (denominator in the equation for u is 0)

• If the distance of the point to a line segment is required then it is only necessary to test that u lies between 0 and 1.

• The solution is similar in higher dimensions.

Let P_a = (x_a, y_a, z_a) be the point in question.

A plane can be defined by its normal n = (A, B, C) and any point on the plane P_b = (x_b, y_b, z_b)

Any point P = (x,y,z) lies on the plane if it satisfes the following

The minimum distance between P_a and the plane is given by the absolute value of

(A xa + B ya + C za + D) / sqrt(A2 + B2 + C2)

. . . 1

To derive this result consider the projection of the line (P_a - P_b) onto the normal of the plane n, that is just ||P_a - P_b|| cos(theta), where theta is the angle between (P_a - P_b) and the normal n. This projection is the minimum distance of P_a to the plane.

This can be written in terms of the dot product as

That is

minimum distance = (A (xa - xb) + B (ya - yb) + C (za - zb)) / sqrt(A2 + B2 + C2)

. . . 2

Since point (x_b, y_b, z_b) is a point on the plane

A xb + B yb + C zb + D = 0

. . . 3

Substituting equation 3 into equation 2 gives the result shown in equation 1.