On a room’s flat plan, (2;1) might mean a mark on the floor or a light above it. A third number removes the ambiguity. We will use one simple scene to describe a point, displacement, plane and sphere without repeating the entire two-dimensional course.

A third coordinate and straight-line distance

Prerequisites: plane coordinates, vectors, plane isometries, spatial lines, angles and distances.

Choose three mutually perpendicular axes x,y,z with common origin O and the same unit, a metre. Choose right-handed axis orientation. x and y lie in the floor; positive z points up. Triple P(x;y;z) specifies position; z can be negative below the floor. Coordinate planes xy,xz,yz have equations z=0,y=0,x=0 respectively. A coordinate axis requires two zero coordinates, not one.

Let A(1;1;1), B(3;2;3). Coordinate differences are 2,1,2. First, Pythagoras on the floor gives squared horizontal displacement 2²+1²=5; then add the squared height change 2². Thus AB=√9=3 m. In general:

d = √((x_(B) − x_(A))^(2) + (y_(B) − y_(A))^(2) + (z_(B) − z_(A))^(2)).

A midpoint averages each coordinate: M(2;1.5;2). Point P=A+t(B−A), 0≤t≤1, divides the segment in ratio AP:PB=t:(1−t). At t=1/3, P(5/3;4/3;5/3) gives ratio 1:2. t is a fraction of the journey, not a length in metres.

Vectors: the same operations, another component

Vector u from A to B is u=(2;1;2). It describes position change, not point B itself. Addition, subtraction and scalar multiplication act componentwise: with v=(1;−2;0), u+v=(3;−1;2) and 2u=(4;2;4). The length of u is √(2²+1²+2²)=3.

The dot product gains a third term too:

u ⋅ v = u_(x)v_(x) + u_(y)v_(y) + u_(z)v_(z) = ∣u∣ ∣v∣cos⁡θ.

Here 2·1+1·(−2)+2·0=0, so nonzero u and v are perpendicular. A zero vector has no defined angle, although its dot product with every vector is zero. For nonzero line directions, the smaller angle φ satisfies cos φ=|u·v|/(|u||v|); omit the absolute value for the angle between oriented vectors.

Third coordinate and midpoint M

xyzA(1;1;1)B(3;2;3)MuO

Normal: Q−H=n

H(1;2;2)Q(2;4;4)nα: x + 2y + 2z = 9

Sphere: a section through H and Q

HQ3 mnα
Panels 1 and 2 are schematic parallel projections computed from our 3D coordinates; do not read pixel lengths or angles. In panel 1 dashes show displacement components and B lowered to xy, M=(A+B)/2; AB=3 m. In panel 2 the triangle is only a patch of infinite α, H∈α, Q=H+n, |n|=3. The angle marker projects an actual spatial right angle. Panel 3 is the section plane containing HQ and perpendicular to α: at uniform 40 px/m it shows a great circle of the sphere centered at H, radius 3 m, and α’s trace through H. It is a section, not a perspective silhouette.

Lines and planes: distinguish their equations

The line through A in direction u uses one shared parameter t: x=1+2t, y=1+t, z=1+2t, t∈ℝ. At t=0 it is A; at t=1 it is B. Restrict t to 0 through 1 for the segment. A zero direction component makes the corresponding coordinate constant: for v, z=const. Never divide by zero in a symmetric line equation.

One linear equation ax+by+cz=d with n=(a;b;c)≠0 defines a plane, not a spatial line. Vector n is its normal. Why? Subtract the equations for two plane points X,X₀: n·(X−X₀)=0. Every displacement within the plane is perpendicular to n. Two independent linear constraints can specify a line as the intersection of planes.

Take α: x+2y+2z=9, n=(1;2;2). H(1;2;2) belongs to α. Set Q(2;4;4)=H+n. Length QH=|n|=3 m; QH is perpendicular to the plane, hence the shortest distance from Q to α. Substituting Q gives 18 rather than 9: Q is indeed outside the plane.

The general distance from Q(x₀;y₀;z₀) to ax+by+cz=d is |ax₀+by₀+cz₀−d|/√(a²+b²+c²). Indeed, for X₀ in the plane and perpendicular foot H, n·(Q−X₀)=n·(Q−H), since n·(H−X₀)=0. Dividing by |n| projects onto a unit normal. The absolute value removes the choice of side, and the denominator normalizes the normal. Our example gives |18−9|/3=3 m. Proportional normals mean parallel or coincident planes; check the equations’ right-hand sides to distinguish them strictly.

A sphere: constant distance

All points at distance r>0 from C(a;b;c) form a sphere. The distance formula immediately gives:

(x − a)^(2) + (y − b)^(2) + (z − c)^(2) = (r)^(2).

For example, (x−1)²+(y−2)²+(z−2)²=9 is a sphere centered at H, radius 3 m; Q belongs to it. Replacing = by ≤ gives the closed ball, including its boundary. In plane z=2 this sphere gives circle (x−1)²+(y−2)²=9; in space the circle’s equation must also include z=2. The sphere lesson will further distinguish surface, solid and plane circle.

Additional coordinate examplesThe point, line, plane and sphere definitions above are already complete.

Branch · Spatial transformations

Translation by (p;q;r) gives (x+p;y+q;z+r). Reflection in xy gives (x;y;−z); central symmetry about O gives (−x;−y;−z). Symmetry about the z-axis is the half-turn (−x;−y;z), not reflection in a plane. A 90° rotation about positive z by the right-hand rule gives (−y;x;z). Each preserves distances: substitute coordinate differences into the sum of three squares. We move points in the coordinate frame, not the axes.

A homothety centered at O with k≠0 gives (kx;ky;kz), scaling lengths by |k|, areas by k² and volumes by |k|³. At k=±1 it is an isometry; other k change scale. This extends G023 into space, rather than projecting away a coordinate.

Sources and limits

School definitions and transformations: Ister’s grade-10 geometry, §§13–19. Additionally, AMSI, Three-dimensional coordinate geometry covers coordinates, midpoint, lines and planes; Purdue, Lines and Planes covers directions, normals and symmetric-form restrictions. Our points and calculations were checked separately. All metric formulas require a rectangular system with equal units; an arbitrary 3D-editor grid does not guarantee these conditions.