How Three Numbers Pin Down Any Point in a Room

A point on a sheet of paper needs two numbers, but a fan on a ceiling, a bird in flight or an atom in a crystal needs three. Adding a third axis extends coordinate geometry into space, and most formulae carry over with one extra term.

This lesson covers axes, planes and octants, the distance formula in three dimensions, and the section and midpoint formulae.

**Three mutually perpendicular lines through the origin — the x-, y- and z-axes — determine three coordinate planes, the xy-, yz- and zx-planes, which divide space into eight octants, and the point $(x,y,z)$ is reached by moving x along the x-axis, then y parallel to the y-axis, then z parallel to the z-axis.

Coordinate planes and axes:

- xy-plane** — every point has $z=0$
- yz-plane — every point has $x=0$
- zx-plane — every point has $y=0$
- x-axis — points of the form $(a,0,0)$

Octants. The signs of x, y and z decide the octant: $(+,+,+)$ is the first octant, and $(-,-,-)$ is the octant diagonally opposite.

Worked example. For $P(3,-2,5)$:

- Signs $(+,-,+)$, so P lies in the octant XOY'Z
- Distance from the xy-plane: $|z|=5$
- Distance from the z-axis: $\sqrt{x^2+y^2}=\sqrt{9+4}=\sqrt{13}\approx 3.61$
- Foot of the perpendicular on the xy-plane: $(3,-2,0)$

An everyday example. Locating a ceiling light in a classroom needs its distance from two walls and its height above the floor, with one corner of the room as the origin.

The substance. Distance from a coordinate plane uses one coordinate, but distance from an axis uses two — from the x-axis it is $\sqrt{y^2+z^2}$.

**The distance between $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is $PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$, obtained by applying Pythagoras' theorem twice.

Derivation.** Build a cuboid with PQ as a diagonal. The diagonal of its base has length $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$; it is perpendicular to the vertical edge of length $|z_2-z_1|$, so a second use of Pythagoras gives the formula.

Worked example. Find the distance between $(1,-3,4)$ and $(-4,1,2)$:

$$PQ=\sqrt{(-5{)}^2+4^2+(-2{)}^2}=\sqrt{25+16+4}=\sqrt{45}=3\sqrt{5}\approx 6.71$$

Worked example 2. Show that $A(0,7,-10)$, $B(1,6,-6)$ and $C(4,9,-6)$ form an isosceles triangle.

- $AB=\sqrt{1+1+16}=\sqrt{18}$
- $BC=\sqrt{9+9+0}=\sqrt{18}$
- $AC=\sqrt{16+4+16}=6$

Since $AB=BC$, the triangle is isosceles.

An everyday example. The longest pole that fits in a hall 12 m long, 4 m wide and 3 m high lies along the space diagonal, $\sqrt{144+16+9}=13$ m.

The substance. **The distance from the origin is $\sqrt{x^2+y^2+z^2}$** — the familiar 2-D formula with one more square added.

**The point dividing the join of $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ internally in the ratio $m:n$ is $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$, the midpoint is the average of the coordinates, and external division replaces the plus signs with minus signs.

Worked example (internal).** Find the point dividing $A(1,-2,3)$ and $B(3,4,-5)$ in the ratio $2:3$:

$$\left(\frac{6+3}{5},\frac{8-6}{5},\frac{-10+9}{5}\right)=(1.8,0.4,-0.2)$$

Worked example (ratio from a plane). In what ratio does the yz-plane divide the join of $(-2,4,7)$ and $(3,-5,8)$? On the yz-plane $x=0$, so $\frac{3m-2n}{m+n}=0$, giving $m:n=2:3$.

Midpoint and centroid. The midpoint of $(2,3,5)$ and $(4,-1,1)$ is $(3,1,3)$, and the centroid of a triangle is the average of its three vertices, coordinate by coordinate.

An everyday example. Hanging a lamp on a straight cable between two poles of different heights, one-third of the way along, uses the section formula with the ratio $1:2$.

The substance. A negative ratio signals external division — if solving gives a negative $m:n$, the dividing point lies outside the segment.