Distance Formula in Three Dimensions

The distance between two points in space follows directly from Pythagoras’ theorem applied twice. If $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are any two points, the distance $PQ$ is the length of the straight segment joining them.

This is exactly the 2D distance formula with one extra squared term — the $z$-difference — added under the root. Compare the two side by side.

Distance from the origin: Putting $P = O(0, 0, 0)$ gives a handy special case.

What the formula lets you prove: Because distance is unambiguous, you can settle a great many geometric questions purely by computing lengths. Three points are collinear when the sum of two of the three pairwise distances equals the third. A triangle is isosceles if two sides are equal, equilateral if all three are equal, and right-angled when the three side-lengths satisfy $a^2 + b^2 = c^2$.

Deeper Insight — why one extra term is all you need: The 3D distance formula is not a new idea so much as Pythagoras’ theorem used twice. First imagine the box (rectangular parallelepiped) whose opposite corners are $P$ and $Q$ and whose edges are parallel to the axes; the diagonal of its base has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ by the ordinary plane theorem. That base diagonal and the vertical edge of length $|\Delta z|$ are themselves perpendicular, so a second application of Pythagoras gives the space diagonal $\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$. This "Pythagoras twice" argument is why every coordinate dimension contributes its own squared difference, and it generalises cleanly to any number of dimensions. The practical payoff is huge: shape problems that would be hard to picture in space — is this triangle right-angled? are these four points the vertices of a square? — collapse into routine arithmetic the moment you compute a few distances.