Coordinates of a Point and Distance Between Two Points

Coordinates of a point. Once the three coordinate axes and origin $O$ are fixed (Section 2), every point $P$ of space corresponds to a unique ordered triple of real numbers $(x, y, z)$ -- its coordinates -- and conversely every ordered triple corresponds to exactly one point. Geometrically, $(x,y,z)$ are obtained by drawing a rectangular box with one corner at the origin $O$ and the opposite corner at $P$, with edges parallel to the three axes; $x$, $y$, $z$ are then the (signed) lengths of the three edges meeting at $O$.

Distance of a point from the origin. Let $P(x,y,z)$ be any point, and let $A$ be the foot of the perpendicular from $P$ to the $XY$-plane, so $A=(x,y,0)$ and $AP$ is a segment of length $|z|$ parallel to the $z$-axis. In the $XY$-plane, by the ordinary 2D distance formula, $OA=\sqrt{x^2+y^2}$. Since $OA$ lies in the $XY$-plane and $AP$ is perpendicular to that plane, the triangle $OAP$ has a right angle at $A$, so by the Pythagorean theorem,

This is the distance of $P(x,y,z)$ from the origin.

Distance between two general points -- derivation. Let $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ be any two points of space. Draw a rectangular box with $P_1P_2$ as one diagonal and edges parallel to the coordinate axes; the three edges meeting at $P_1$ have lengths $|x_2-x_1|$, $|y_2-y_1|$, $|z_2-z_1|$. Applying the Pythagorean theorem twice -- once in the base face of the box to find the diagonal of that face, and once more between that face-diagonal and the vertical edge to find the box's main diagonal $P_1P_2$ -- gives

This is the distance formula in three dimensions; setting $P_1$ at the origin recovers the special case above. …