Coordinate axes and planes (3D)

Every point you have ever plotted lived on a flat page, located by two numbers $(x,y)$ . Step off the page into the room around you and two numbers are no longer enough — you also need to say how high or how deep. Three-dimensional geometry supplies the third measurement by adding a third axis, mutually perpendicular to the other two, so every point in space gets a unique triple $(x,y,z)$ . This article sets up that scaffolding: the three coordinate axes, the three coordinate planes they generate, and the eight octants these planes carve out of space. It also asks a natural follow-up question — how far is a given point from each of these planes and axes — and answers it with a short, fully verified formula.

The core insight

Three mutually perpendicular axes through a fixed origin generate three coordinate planes that divide space into eight octants, and a point's perpendicular distance from any coordinate plane is just the absolute value of the coordinate that plane forces to zero.

The core idea

To locate a point in space, take three mutually perpendicular lines through a fixed point $O$ , the origin. These are the $x$ -axis, $y$ -axis, and $z$ -axis, together the coordinate axes. Any two axes determine a coordinate plane: the $x$ -axis and $y$ -axis give the $XY$ -plane; the $y$ -axis and $z$ -axis give the $YZ$ -plane; the $z$ -axis and $x$ -axis give the $ZX$ -plane. These three planes divide the whole of space into eight octants — the 3-D analogue of how the $x$ -axis and $y$ -axis split the plane into four quadrants. A short list of facts pins down the geometry precisely:

Every point on the $XY$ -plane has $z=0$ ; every point on the $YZ$ -plane has $x=0$ ; every point on the $ZX$ -plane has $y=0$ . In each case, the coordinate belonging to the missing axis is zero.
Every point on the $x$ -axis has $y=0$ and $z=0$ (similarly for the other axes) — two coordinates vanish, not one.
The origin has coordinates $(0,0,0)$ , common to all three planes and all three axes.
In the first octant ( $OXYZ$ ), all three coordinates are positive; this is the octant most worked examples use, just as the first quadrant is favoured in 2-D.

The eight octants are labelled by which axis directions are used, a prime marking the negative direction:

OXYZ,\; OX'YZ,\; OX'Y'Z,\; OXY'Z,\; OXYZ',\; OX'YZ',\; OX'Y'Z',\; OXY'Z'.

where: an unprimed letter means the positive direction of that axis, a primed letter the negative direction, read off from the sign of the corresponding coordinate of a point in that octant.

Figure

Three perpendicular coordinate axes meeting at the origin with the three coordinate planes shaded and the eight octants they create labelled around the origin

Distance of a point from the coordinate planes and axes

The natural next question, once the planes and axes are set up, is: given a point $P(x,y,z)$ , how far is it from each coordinate plane and each coordinate axis? Deriving this rigorously, using nothing beyond the fact that a perpendicular from a point to a plane meets it at the point where one coordinate is forced to zero, gives compact formulas that also explain the "foot of perpendicular" idea used in the worked example below.

Derivation ADistance of a point from the coordinate planes and axes

The goal is to show, starting only from the definitions above (a coordinate plane is where one coordinate is zero, a coordinate axis is where two coordinates are zero), that the perpendicular distance from a point $P(x,y,z)$ to the $XY$ -plane is $|z|$ , to the $YZ$ -plane is $|x|$ , to the $ZX$ -plane is $|y|$ , and to the $x$ -axis is $\sqrt{y^2+z^2}$ — with the $y$ -axis and $z$ -axis distances following by the same pattern.

Step 1. Drop a perpendicular from $P(x,y,z)$ onto the $XY$ -plane. Since every point of the $XY$ -plane has $z=0$ , and the coordinate axes are mutually perpendicular, moving from $P$ straight down to the $XY$ -plane changes only the $z$ -coordinate, landing at the foot $F(x,y,0)$ . The segment $PF$ is therefore vertical, of length $|z-0| = |z|$ .

Step 2. Repeat for the other two planes by the same reasoning: the foot of perpendicular from $P$ onto the $YZ$ -plane is $(0,y,z)$ , giving distance $|x|$ ; the foot onto the $ZX$ -plane is $(x,0,z)$ , giving distance $|y|$ . In each case the distance equals the absolute value of exactly the coordinate that plane's equation sets to zero.

Step 3. Now find the distance from $P$ to the $x$ -axis. The foot of perpendicular from $P(x,y,z)$ onto the $x$ -axis is the point $M(x,0,0)$ , because moving along the axis only the $x$ -coordinate can vary; $y$ and $z$ must both become $0$ to land on the axis. The segment $PM$ has $\Delta x = 0$ , $\Delta y = y-0=y$ , $\Delta z = z-0=z$ , so by the ordinary Pythagorean theorem applied twice (once in each perpendicular direction),

PM = \sqrt{y^2+z^2}.

where: $y$ and $z$ are the two coordinates of $P$ that are not measured along the $x$ -axis itself.

Step 4. By the identical argument, swapping the role of the axis, the distance from $P$ to the $y$ -axis is $\sqrt{x^2+z^2}$ , and the distance from $P$ to the $z$ -axis is $\sqrt{x^2+y^2}$ . Each formula uses the two coordinates other than the one measured along that axis — a direct consequence of the axes being mutually perpendicular.

Worked example

Question: A point $P$ has coordinates $(-3, 5, -2)$ . Find (a) the octant in which $P$ lies, (b) the foot of perpendicular and the distance of $P$ from the $XY$ -plane, and (c) the distance of $P$ from the $z$ -axis.

Step 1. Determine the octant. The signs are $x=-3$ (negative), $y=5$ (positive), $z=-2$ (negative). Priming the axes with negative coordinates gives octant $OX'YZ'$ .

Step 2. Find the foot of perpendicular on the $XY$ -plane. Every point of the $XY$ -plane has $z=0$ , so the foot keeps $x$ and $y$ unchanged and sets $z=0$ : the foot is $(-3,5,0)$ .

Step 3. Apply the plane-distance result from the derivation: distance from $P$ to the $XY$ -plane is $|z| = |-2| = 2$ units.

Step 4. Apply the axis-distance result for the $z$ -axis, $\sqrt{x^2+y^2}$ : distance from $P$ to the $z$ -axis is $\sqrt{(-3)^2+5^2} = \sqrt{9+25} = \sqrt{34}$ units.

$P$ lies in octant $OX'YZ'$ , its foot of perpendicular on the $XY$ -plane is $(-3,5,0)$ at distance $2$ units, and its distance from the $z$ -axis is $\sqrt{34}$ units.

Common traps

Confusing "coordinate plane" with "octant": the coordinate planes are flat 2-D surfaces where one coordinate is zero, while octants are the eight solid regions these planes carve out of space.
Assuming a point with one zero coordinate lies "in an octant" — a zero coordinate always puts the point on a coordinate plane (or an axis, if two coordinates are zero), never inside an octant.
Mixing up which coordinate is zero on which plane — remembering that the missing letter's coordinate is zero (as in the $XY$ -plane meaning $z=0$ ) avoids the error.
Using the plane-distance formula $|x|$ , $|y|$ , or $|z|$ when the axis-distance formula (a square root of a sum of two squares) is needed, or vice versa — the plane distance uses one coordinate, the axis distance uses the other two.

Ask the tutor: if the octant-naming convention or the plane-versus-axis distance formulas feel shaky, ask the AI tutor to walk through a few more sample points with you.

Why does a point on a coordinate axis have two zero coordinates instead of one?

An axis is where two coordinate planes intersect. The $x$ -axis, for instance, is where the $XY$ -plane (which forces $z=0$ ) and the $ZX$ -plane (which forces $y=0$ ) meet, so both conditions hold simultaneously: $y=0$ and $z=0$ together, leaving only $x$ free to vary along the axis.

How is the octant of a point actually decided from its coordinates?

Look at the sign of each coordinate in turn. A positive coordinate keeps the corresponding letter unprimed; a negative coordinate primes it. So a point with $x>0$ , $y<0$ , $z>0$ sits in octant $OXY'Z$ . This is a direct generalisation of how the sign pattern of $(x,y)$ decides the quadrant in 2-D geometry, just extended to three signs instead of two.

Is the distance-from-axis formula the same as the distance-between-two-points formula?

It is a special case of it, though this chapter introduces it directly from the perpendicular-foot idea rather than the general distance formula (which is a separate, later topic). The distance from $P(x,y,z)$ to the $x$ -axis is really the distance from $P$ to its foot of perpendicular $M(x,0,0)$ on that axis, and computing that distance with the Pythagorean theorem happens to reproduce $\sqrt{y^2+z^2}$ — consistent with, but not requiring, the general two-point distance formula.

Coordinate axes and planes (3D)

The core idea

Distance of a point from the coordinate planes and axes

Worked example

Common traps

Where this leads