Coordinate axes and coordinate planes
In a plane, two numbers fix a point. In space we need three. We take three lines that meet at one point O (the origin) and are at right angles to each other: the x-axis, the y-axis and the z-axis.
Each pair of axes forms a coordinate plane:
- XY-plane (holds x and y axes): every point has z = 0.
- YZ-plane: every point has x = 0.
- ZX-plane: every point has y = 0.
Octants
The three planes divide space into eight octants. The signs of the coordinates tell the octant:
| Octant | I | II | III | IV | V | VI | VII | VIII |
|---|---|---|---|---|---|---|---|---|
| x | + | − | − | + | + | − | − | + |
| y | + | + | − | − | + | + | − | − |
| z | + | + | + | + | − | − | − | − |
The first four octants lie above the XY-plane, the last four below it.
Coordinates of a point in space
Point P is written P(x, y, z). Starting at O, move x units along the x-axis, then y units parallel to the y-axis, then z units parallel to the z-axis.
Another way to see it: x is the distance of P from the YZ-plane, y from the ZX-plane, and z from the XY-plane (with sign).
- Origin: (0, 0, 0).
- On the x-axis: (x, 0, 0); y-axis: (0, y, 0); z-axis: (0, 0, z).
- On the XY-plane: (x, y, 0); YZ-plane: (0, y, z); ZX-plane: (x, 0, z).
- The foot of the perpendicular from P(x, y, z) to the XY-plane is (x, y, 0).
Distance between two points
Take P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). Build a box whose edges are parallel to the axes, with P and Q at opposite corners. The edges are |x₂ − x₁|, |y₂ − y₁| and |z₂ − z₁|.
On the floor: PN² = (x₂ − x₁)² + (y₂ − y₁)². Going up: PQ² = PN² + (z₂ − z₁)². So
PQ = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
Distance from the origin: OP = √(x² + y² + z²). The formula also helps test whether three points are collinear (AB + BC = AC) or whether a triangle is right-angled or isosceles.
Key formulas and definitions
- XY-plane: z = 0; YZ-plane: x = 0; ZX-plane: y = 0
- x-axis: (x, 0, 0); y-axis: (0, y, 0); z-axis: (0, 0, z)
- PQ = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
- OP = √(x² + y² + z²)
Worked examples
1. In which octant does (−3, 1, −2) lie?
Step 1: signs are x −, y +, z −. Step 2: (−, +, −) is octant VI.
2. Find the distance between P(1, −3, 4) and Q(−4, 1, 2).
Step 1: Δx = −5, Δy = 4, Δz = −2. Step 2: PQ² = 25 + 16 + 4 = 45. Step 3: PQ = √45 = 3√5 ≈ 6.71.
3. Find the distance of P(2, 3, 6) from the origin.
Step 1: OP² = 4 + 9 + 36 = 49. Step 2: OP = 7.
4. Show that A(−2, 3, 5), B(1, 2, 3) and C(7, 0, −1) are collinear.
Step 1: AB = √(9 + 1 + 4) = √14. Step 2: BC = √(36 + 4 + 16) = √56 = 2√14; AC = √(81 + 9 + 36) = √126 = 3√14. Step 3: AB + BC = √14 + 2√14 = 3√14 = AC, so they are collinear.
5. Show that (0, 7, −10), (1, 6, −6) and (4, 9, −6) form an isosceles triangle.
Step 1: AB = √(1 + 1 + 16) = √18. Step 2: BC = √(9 + 9 + 0) = √18. Step 3: AC = √(16 + 4 + 16) = 6. Two sides equal → isosceles.
6. Find the point on the x-axis that is equally far from A(1, 2, 3) and B(3, 5, −2).
Step 1: let the point be (x, 0, 0). Step 2: (x − 1)² + 4 + 9 = (x − 3)² + 25 + 4. Step 3: −2x + 14 = −6x + 38 → 4x = 24 → x = 6. Point (6, 0, 0).
7. Find the equation of the set of points P such that PA² + PB² = 2k², where A(3, 4, 5) and B(−1, 3, −7).
Step 1: PA² = (x − 3)² + (y − 4)² + (z − 5)²; PB² = (x + 1)² + (y − 3)² + (z + 7)². Step 2: add: 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = 2k². Step 3: 2(x² + y² + z²) − 4x − 14y + 4z + 109 − 2k² = 0.
Common mistakes
- Mixing up which plane has which coordinate zero: the XY-plane has z = 0 (not x = 0).
- Forgetting the third square and using the 2D distance formula in space.
- Getting the sign of a coordinate wrong when naming the octant; check x, then y, then z.
- Squaring a negative difference and writing a negative answer: (−5)² = 25, not −25.