EUCLIDEAN DISTANCE: CONCEPT, FORMULA, CALCULATION, EXAMPLE - MATHS - 2025

The Euclidean distance is a positive number that indicates the separation between two points in a space where the axioms and theorems of Euclid's geometry are fulfilled.

The distance between two points A and B in a Euclidean space is the length of the vector AB belonging to the only line that passes through these points.

Figure 1. One-dimensional Euclidean space formed by the line (OX). Several points are shown on said space, their coordinates and distances. (Prepared by Ricardo Pérez).

The space that humans perceive and where we move is a three-dimensional (3-D) space, where the axioms and theorems of Euclid's geometry are fulfilled. Two-dimensional subspaces (planes) and one-dimensional subspaces (lines) are contained in this space.

Euclidean spaces can be one-dimensional (1-D), two-dimensional (2-D), three-dimensional (3-D), or n-dimensional (nD).

Points in the one-dimensional space X are those that belong to the oriented line (OX), the direction from O to X is the positive direction. To locate the points on this line, the Cartesian system is used, which consists of assigning a number to each point of the line.

Formula

The Euclidean distance d (A, B) between points A and B, located on a line, is defined as the square root of the square of the differences in their X coordinates:

d (A, B) = √ ((XB - XA) ^ 2)

This definition guarantees that: the distance between two points is always a positive quantity. And that the distance between A and B is equal to the distance between B and A.

Figure 1 shows the one-dimensional Euclidean space formed by the line (OX) and several points on said line. Each point has a coordinate:

Point A has coordinate XA = 2.5, point B coordinate XB = 4 and point C coordinate XC = -2.5

d (A, B) = √ ((4 - 2.5) 2) = 1.5

d (B, A) = √ ((2.5 - 4) 2) = 1.5

d (A, C) = √ ((- 2.5 - 2.5) 2) = 5.0

Euclidean distance in two dimensions

Two-dimensional Euclidean space is a plane. The points of a Euclidean plane fulfill the axioms of Euclidean geometry, for example:

- A single line passes through two points.

- Three points on the plane form a triangle whose internal angles always add up to 180º.

- In a right triangle, the square of the hypotenuse is equal to the sum of the squares of its legs.

In two dimensions, a point has X and Y coordinates.

For example a point P has coordinates (XP, YP) ​​and a point Q coordinates (XQ, YQ).

The Euclidean distance between point P and Q is defined with the following formula:

d (P, Q) = √ ((XQ - XP) ^ 2 + (YQ - YP) ^ 2)

It should be noted that this formula is equivalent to the Pythagorean theorem, as shown in Figure 2.

Figure 2. The distance between two points P and Q in the plane fulfills the Pythagorean theorem. (Prepared by Ricardo Pérez).

Non-Euclidean surfaces

Not all two-dimensional spaces conform to Euclidean geometry. The surface of a sphere is a two-dimensional space.

The angles of a triangle on a spherical surface do not add up to 180º and with this the Pythagorean theorem is not fulfilled, therefore a spherical surface does not fulfill Euclid's axioms.

Euclidean distance in n dimensions

The concept of coordinates can be extended to larger dimensions:

- In 2-D point P has coordinates (XP, YP)

- In 3-D a point Q has coordinates (XQ, YQ, ZQ)

- In 4-D the point R will have coordinates (XR, YR, ZR, WR)

- In nD a point P will have coordinates (P1, P2, P3,….., Pn)

The distance between two points P and Q of an n-dimensional Euclidean space is calculated with the following formula:

d (P, Q) = √ ((Q1 - P1) ^ 2 + (Q2 - P2) ^ 2 + …….. + (Qn - Pn) ^ 2)

The locus of all points Q in an n-dimensional Euclidean space equidistant from another fixed point P (the center) form an n-dimensional hypersphere.

How to calculate Euclidean distance

The following shows how the distance between two points located in the Euclidean three-dimensional space is calculated.

Suppose point A of Cartesian coordinates x, y, z given by A:(2, 3, 1) and point B of coordinates B:(-3, 2, 2).

We want to determine the distance between these points, for which use is made of the general relationship:

d (A, B) = √ ((-3 - 2) 2 + (2 - 3) 2 + (2 - 1) 2) = √ ((-5) 2 + (-1) 2 + (1) 2)

d (A, B) = √ (25 + 1 + 1) = √ (27) = √ (9 * 3) = 3 √ (3) = 5,196

Example

There are two points P and Q. The point P of Cartesian coordinates x, y, z given by P:(2, 3, 1) and the point Q of coordinates Q:(-3, 2, 1).

It is asked to find the coordinates of the midpoint M of the segment connecting the two points.

The unknown point M is assumed to have coordinates (X, Y, Z).

Since M is the midpoint of, it must be true that d (P, M) = d (Q, M), so d (P, M) ^ 2 = d (Q, M) ^ 2 must also be true:

(X - 2) ^ 2 + (Y - 3) ^ 2 + (Z - 1) ^ 2 = (X - (-3)) ^ 2 + (Y - 2) ^ 2 + (Z - 1) ^ 2

As in this case, the third term is equal in both members, the previous expression simplifies to:

(X - 2) ^ 2 + (Y - 3) ^ 2 = (X + 3) ^ 2 + (Y - 2) ^ 2

We then have an equation with two unknowns X and Y. Another equation is required to solve the problem.

Point M belongs to the line that passes through points P and Q, which we can calculate as follows:

First we find the director vector PQ of the line: PQ = <-3-2, 2-3, 1-1> = <-5, -1, 0>.

Then PM = OP + a PQ, where OP is the position vector of the point P and is a parameter that belongs to the real numbers.

The above equation is known as the vector equation of the line, which in Cartesian coordinates takes the following form:

<X-2, Y-3, Z-1> = <2, 3, 1> + a <-5, -1, 0> = <2 - 5a, 3 - a, 0>

Equating the corresponding components we have:

X - 2 = 2-5 a; Y - 3 = 3 -a; Z - 1 = 0

That is, X = 4 - 5a, Y = 6 - a, finally Z = 1.

It is substituted in the quadratic expression that relates X to Y:

(4 - 5a - 2) ^ 2 + (6 - a - 3) ^ 2 = (4 - 5a + 3) ^ 2 + (6 - a - 2) ^ 2

It is simplified:

(2 - 5a) ^ 2 + (3 -a) ^ 2 = (7 - 5a) ^ 2 + (4 - a) ^ 2

Now unfolds:

4 + 25 a ^ 2 - 20a + 9 + a ^ 2 - 6a = 49 + 25 a ^ 2 - 70a + 16 + a ^ 2 - 8a

It is simplified, canceling like terms in both members:

4 - 20a + 9 - 6a = 49 - 70a + 16 - 8a

The parameter a is cleared:

52 a = 49 + 16 - 4 - 9 = 52 resulting in a = 1.

That is, X = 4 - 5, Y = 6 - 1, finally Z = 1.

Finally we obtain the Cartesian coordinates of the midpoint M of the segment:

M: (-1, 5, 1).

References

1. Lehmann C. (1972) Analytical Geometry. UTEHA.
2. Superprof. Distance between two points. Recovered from: superprof.es
3. UNAM. Distance between affine sublinear manifolds. Recovered from: prometeo.matem.unam.mx/
4. wikipedia. Euclidean distance. Recovered from: es.wikipedia.com
5. wikipedia. Euclidean space. Recovered from: es.wikipedia.com