Last updated: 2026-07-01

The distance between two points in 3D

The distance between two points in space is the length of the straight line segment connecting them. Given points P₁ = (x₁, y₁, z₁) and P₂ = (x₂, y₂, z₂), the Euclidean distance formula in three dimensions is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

It is the natural extension of the 2D distance formula: a third squared difference, for the z-coordinate, joins the two already under the radical. The measure appears throughout solid geometry, physics, 3D computer graphics, robotics, and data science.

Derivation from the Pythagorean theorem

The 3D distance formula is the Pythagorean theorem applied twice. Place P₁ and P₂ at opposite corners of a rectangular box whose edges run parallel to the axes:

• The edges of the box have lengths |Δx| = |x₂ − x₁|, |Δy| = |y₂ − y₁|, and |Δz| = |z₂ − z₁|.
• First, the diagonal across the base (in the xy-plane) has length $\sqrt{\Delta x^2 + \Delta y^2}$, by the 2D Pythagorean theorem.
• That base diagonal and the vertical edge Δz form a second right triangle. Its hypotenuse is the space diagonal of the box — the distance d.

Applying the Pythagorean theorem to the second triangle gives:

$d^2 = \left(\sqrt{\Delta x^2 + \Delta y^2}\right)^2 + \Delta z^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$

so $d = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$. Squaring removes the absolute values, so the sign of any difference never affects the result.

Worked Example

Problem: Find the distance between A = (1, 2, 3) and B = (4, 6, 15).

Step 1 — compute Δx, Δy, and Δz: $\Delta x = 4 - 1 = 3 \qquad \Delta y = 6 - 2 = 4 \qquad \Delta z = 15 - 3 = 12$

Step 2 — apply the formula: $d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

The two points are exactly 13 units apart. The triple (3, 4, 12, 13) is one of the well-known "Pythagorean quadruples," where three integer edges produce an integer space diagonal.

The midpoint in 3D

This calculator also reports the midpoint of the segment P₁P₂ — the point lying exactly halfway between the two endpoints. Each coordinate is the average of the corresponding coordinates:

$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2},\ \frac{z_1 + z_2}{2}\right)$

For A = (1, 2, 3) and B = (4, 6, 15), the midpoint is M = (2.5, 4, 9). The midpoint is symmetric in the two points: it does not depend on which point is labelled P₁.

Δx, Δy, Δz: Direction vs. Distance

The three differences reported alongside the distance describe the signed displacement from P₁ to P₂:

• Δx = x₂ − x₁ — positive if P₂ lies in the +x direction from P₁.
• Δy = y₂ − y₁ — positive if P₂ lies in the +y direction.
• Δz = z₂ − z₁ — positive if P₂ lies in the +z direction.

Together (Δx, Δy, Δz) form the displacement vector from P₁ to P₂, and the distance d is the magnitude (length) of that vector. The displacements can be negative; the distance d is always non-negative.

Order of the points

Distance is symmetric: swapping P₁ and P₂ flips the sign of each displacement but not the distance, because squaring converts every term to a positive value. The distance from A to B equals the distance from B to A, and the midpoint is the same either way.

Extension to higher dimensions

The same pattern continues beyond three dimensions. In n-dimensional space, with points P₁ = (a₁, …, aₙ) and P₂ = (b₁, …, bₙ):

$d = \sqrt{\sum_{i=1}^{n} (b_i - a_i)^2}$

Each additional coordinate contributes one more squared difference under the radical. This generalization underlies distance computations in machine learning, where data points often live in spaces of dozens or hundreds of dimensions.

Common Applications

Solid geometry. Measuring the length of an edge or space diagonal, verifying that four points are coplanar, or finding the radius of a sphere given its center and a surface point.

Physics and engineering. The separation between two particles, the length of a force or velocity vector, and the magnitude of any displacement in space are all 3D Euclidean distances.

3D computer graphics. Collision detection, level-of-detail selection, and lighting all rely on point-to-point distances. As in 2D, the squared distance d² is often used directly to avoid the cost of the square root when only comparisons are needed.

Robotics and navigation. Path planning and obstacle avoidance use 3D distances between a robot, its target, and surrounding objects to decide where it can safely move.