Distance Formula in 3D Space

The 3D distance formula extends the 2D Pythagorean theorem by adding a third dimension:

$d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}}$

For $(1,2,3)$ to $(4,6,3)$: $d = \sqrt{9 + 16 + 0} = \sqrt{25} = 5$.

The distance is the space diagonal of a rectangular box with sides $\Delta x = 3$, $\Delta y = 4$, $\Delta z = 0$. In this case, since $\Delta z = 0$, the two points lie in the same horizontal plane and the 3D distance equals the 2D distance. But try changing the z-coordinates to see the full 3D effect.

Try it in the visualization

Adjust the coordinates of both points and watch the distance update. The rectangular box shows how the space diagonal is built from three perpendicular components. Rotate the 3D view to see the geometry from different angles.