Coordinate system in 3D — x, y, z axes, octants: Worked examples

Notation and the octant map we will reuse

Before any example, let us pin down two things the rest of the page leans on constantly: the segment notation and the octant numbering table.

Formula The 8 octants and their sign patterns (the "standard table")

This is the numbering scheme every example below refers to. Read it as: signs of $(x, y, z)$ .

Octant	$x$	$y$	$z$	Where it sits
I	$+$	$+$	$+$	top layer, above quadrant I
II	$-$	$+$	$+$	top layer, above quadrant II
III	$-$	$-$	$+$	top layer, above quadrant III
IV	$+$	$-$	$+$	top layer, above quadrant IV
V	$+$	$+$	$-$	basement, below quadrant I
VI	$-$	$+$	$-$	basement, below quadrant II
VII	$-$	$-$	$-$	basement, below quadrant III
VIII	$+$	$-$	$-$	basement, below quadrant IV

The pattern: the top four ( $z > 0$ ) copy the four 2D quadrants of the floor in order; the bottom four ( $z < 0$ ) sit directly beneath them, continuing V, VI, VII, VIII.

The picture below shows this "upstairs / basement" split so you can see the numbering rather than memorise it cold.

Figure — Coordinate system in 3D — x, y, z axes, octants

How to read the figure: the blue floor is the xy-plane carrying quadrants I–IV; the orange labels above it (I–IV) are the top-layer octants, and the gray labels below (V–VIII) are the basement octants stacked directly under their matching quadrant. Trace any vertical dashed line: it links a top octant to the basement octant that shares its $(x, y)$ signs and only flips $z$ .

The scenario matrix

Now the checklist of every distinct kind of question this topic can produce. Each worked example is tagged with the cell it covers, so you can see nothing is missing.

Cell	Case class	What makes it different	Covered by
A	All three coords non-zero, signs $(+, +, +)$	The "easy" upstairs octant	Ex 1
B	Mixed signs, $z > 0$ (top layer II/III/IV)	Choose octant from sign pattern	Ex 2
C	Mixed signs, $z < 0$ (basement V–VIII)	"Go to the basement" reasoning	Ex 3
D	Exactly one coordinate zero	Point lies in a plane	Ex 4
E	Exactly two coordinates zero	Point lies on an axis	Ex 5
F	All coordinates zero (degenerate)	The origin itself	Ex 5
G	Distance from origin, negative coords	Squaring kills the sign	Ex 6
H	Real-world word problem	Translate words → triple	Ex 7
I	Exam twist: reflection / mirror image	Sign flip across a plane	Ex 8

Example 1 — Cell A: the friendly octant

Step 1 — read the sign pattern. $(2, 3, 4) \Rightarrow (+, +, +)$ . Why this step? An octant is defined only by the signs of $(x, y, z)$ — the magnitudes never change which box you're in. From the standard table, $(+, +, +)$ is Octant I.

Step 2 — set up the distance from the origin. $O P = x^{2} + y^{2} + z^{2} = 2^{2} + 3^{2} + 4^{2} .$ Why this step? Recall $O P$ is the length of the segment from origin $O$ to $P$ ; the parent note derived this by Pythagoras applied twice, and length is what "how far" means.

Step 3 — compute. $O P = 4 + 9 + 16 = 29 \approx 5.39.$

Verify: $29$ must exceed the largest single coordinate ( $4$ ), because the diagonal of a box is longer than any edge — and $5.39 > 4$ . ✓ Units: if coordinates are metres, $O P$ is in metres.

Example 2 — Cell B: top layer, mixed signs

Step 1 — extract signs. $(- 2, 3, 4) \Rightarrow (-, +, +)$ . Why this step? Same reason as before — octant lives in the signs.

Step 2 — match to the standard table above. Top layer ( $z > 0$ ) mirrors the 2D quadrants in the xy-plane. Here $x < 0, y > 0$ is quadrant II of the floor, and $z > 0$ keeps us upstairs ⟹ Octant II. Why this step? The octant table is organised precisely by $(x, y, z)$ sign pattern, so matching our signs to a table row is the identification — no guessing, just a lookup.

Verify: In the standard table, Octant II's defining pattern is $(-, +, +)$ — exactly ours. ✓

Example 3 — Cell C: into the basement

Step 1 — signs. $(-, -, -)$ . Why this step? Reading signs is always the first move for an octant question.

Step 2 — locate. $x < 0, y < 0$ is quadrant III of the floor; $z < 0$ sends us downstairs ⟹ Octant VII. Why this step? "Top floor copies the quadrants, basement sits below" — from the standard table, the basement of quadrant III is Octant VII.

Step 3 — the relationship. Every coordinate of $R$ is the negative of $P (2, 3, 4)$ from Example 1. So $R$ is the point you reach by shooting the segment from $P$ straight through the origin and out the same distance the other side. Octant VII is the antipode (diametric opposite) of Octant I — this is exactly the vertical-plus-horizontal flip the octant figure above illustrates.

Verify: Distance of $R$ from origin $= (- 2)^{2} + (- 3)^{2} + (- 4)^{2} = 29$ — identical to $O P$ , as an antipode must be. ✓

Example 4 — Cell D: one zero ⟹ stuck in a plane

Step 1 — count the zeros. Exactly one coordinate is zero ( $y = 0$ ). Why this step? The parent's quick rule keys off how many zeros: one zero ⟹ in a plane, two ⟹ on an axis.

Step 2 — name the plane. $y = 0$ means the point is "flat" in the $y$ -direction. A plane is named by the axes inside it; the axes with non-zero freedom here are $x$ and $z$ ⟹ the xz-plane. Why this step? This is the exact trap from the parent's mistake section — the missing coordinate names the plane, not the coordinate you see. (Naming convention: we list the two in-plane axes in the order $x$ then $z$ , so "xz-plane"; some books write "zx-plane" — same plane, just the axes named in the other order.)

Step 3 — why no octant. Octants are the open regions between the planes. A point sitting on a coordinate plane is on the boundary wall itself, belonging to no single octant.

Verify: Plug into the plane equations: $y = 0$ ✓ (xz-plane), while $x = 4 \neq = 0$ and $z = 7 \neq = 0$ rule out the yz- and xy-planes. ✓

Example 5 — Cells E & F: two zeros (axis) and three zeros (origin)

Step 1 — case (a): count zeros. For $A (0, 0, 5)$ : two coordinates zero ( $x = 0, y = 0$ ). Why this step? Two zeros ⟹ on an axis by the quick rule.

Step 2 — which axis, and which direction along it? The surviving coordinate is $z = 5 > 0$ , so $A$ rides the positive $z$ -axis, 5 units above the origin. Its partner $A^{'} (0, 0, - 5)$ has $z = - 5 < 0$ : same $z$ -axis, but 5 units below the origin, on the negative $z$ -axis. Why this step? The non-zero coordinate names the axis; its sign then tells you which half of that axis (the positive ray or the negative ray) the point sits on. Naming only the axis would leave the direction ambiguous.

Step 3 — case (b): all zeros. $O (0, 0, 0)$ : nothing along any ruler ⟹ the origin, the single point where all three axes and all three planes meet.

Verify: $A$ satisfies $x = 0$ and $y = 0$ with $z > 0$ — the positive $z$ -axis; $A^{'}$ satisfies the same two equations with $z < 0$ — the negative $z$ -axis. And $O A = 0 + 0 + 25 = 5 = O A^{'}$ , so both are 5 units out, on opposite sides. ✓ For $O$ : $0 + 0 + 0 = 0$ — zero distance from itself. ✓

Example 6 — Cell G: distance with negative coordinates

Step 1 — write the formula. $O T = x^{2} + y^{2} + z^{2} .$ Why this step? $O T$ is a length (segment from origin $O$ to $T$ ); the derivation (Pythagoras twice) never cared about signs.

Step 2 — square each coordinate carefully. $(- 3)^{2} = + 9, 4^{2} = 16, (- 12)^{2} = + 144.$ Why this step? This is the exact trap: squaring kills the sign — $(- 3)^{2}$ is $+ 9$ , never $- 9$ . A distance can never come out negative.

Step 3 — add and root. $O T = 9 + 16 + 144 = 169 = 13.$

Verify: $169 = 13$ , a whole number, because $(3, 4, 12)$ is a 3D Pythagorean triple ( $9 + 16 + 144 = 169 = 1 3^{2}$ ). The negatives vanished. ✓

Example 7 — Cell H: real-world word problem

Step 1 — assign axes to the room. Let the first floor edge be $+ x$ , the second floor edge be $+ y$ , and "up" be $+ z$ (a right-handed choice, matching the parent's right-hand rule). Why this step? Word problems have no numbers until you pick axes; the corner is the natural origin $O$ .

Step 2 — record each move as a coordinate. $6$ m along $x$ , $3$ m along $y$ , $2$ m up $\Rightarrow$ fly at $F (6, 3, 2)$ . Why this step? Each independent move adds to exactly one ruler, so the moves become the ordered triple directly.

Step 3 — straight-line distance. $O F = 6^{2} + 3^{2} + 2^{2} = 36 + 9 + 4 = 49 = 7 m .$ Why this step? $O F$ is the segment from the corner $O$ to the fly $F$ — the "as the crow flies" length, which is exactly the distance-from-origin formula.

Verify: The straight-line (diagonal) distance $7$ m must be less than the walked path $11$ m — a straight line beats a detour. $7 < 11$ ✓. Units: metres throughout. ✓ Sign check: all moves positive ⟹ Octant I. ✓

Example 8 — Cell I: exam twist (mirror image)

Step 1 — which coordinate flips? The xy-plane is $z = 0$ . Reflecting across it is like a mirror on the floor: left/right ( $x$ ) and forward/back ( $y$ ) are unchanged, only up/down ( $z$ ) flips. So $P^{'} (3, - 2, - 6)$ . Why this step? A reflection negates only the coordinate perpendicular to the mirror plane; the two in-plane coordinates survive untouched.

Step 2 — octant of each. $P (3, - 2, 6) = (+, -, +) \Rightarrow$ Octant IV (from the standard table). $P^{'} (3, - 2, - 6) = (+, -, -) \Rightarrow$ Octant VIII. Why this step? Flipping $z$ from $+$ to $-$ sends the point straight from the top layer to the basement directly beneath — IV becomes VIII, the exact vertical stacking shown in the octant map.

Step 3 — equal distances. $O P = 9 + 4 + 36 = 49 = 7, O P^{'} = 9 + 4 + 36 = 49 = 7.$ Why this step? The reflection changes only the sign of $z$ , and the distance formula squares $z$ — so $(+ 6)^{2}$ and $(- 6)^{2}$ are both $36$ . The sign flip cannot change the distance, so $P$ and its mirror image $P^{'}$ are exactly equally far from the origin.

The figure below shows $P$ above the floor and $P^{'}$ the same drop below it, joined by a vertical dashed line — the visual signature that only $z$ changed.

How to read the figure: the green translucent sheet is the mirror (the xy-plane, $z = 0$ ); the blue dot $P$ sits above it in Octant IV, the orange dot $P^{'}$ sits an equal distance below in Octant VIII, and the red dashed segment linking them is vertical — proof that $x$ and $y$ never moved.

Verify: $O P = O P^{'} = 7$ ✓, octants IV and VIII sit in the same vertical column (same $x, y$ signs, opposite $z$ ) ✓.

Recall

Recall Which coordinate flips when you reflect in each plane? (click)

Reflect in xy-plane ( $z = 0$ ) ::: only $z$ flips sign. Reflect in yz-plane ( $x = 0$ ) ::: only $x$ flips sign. Reflect in xz-plane ( $y = 0$ ) ::: only $y$ flips sign.

Recall Sanity checks that catch 90% of errors (click)

Distance ever negative? ::: Never — squaring removes signs, then the positive root is taken. One coordinate zero means the point is… ::: in a coordinate plane (named by the two non-zero axes). Two coordinates zero — does the sign of the third matter? ::: Yes — it picks the positive or negative half of that axis. Antipode of Octant I (all $+$ ) is… ::: Octant VII (all $-$ ). Straight-line distance vs walked path? ::: straight line is always $\leq$ the sum of the legs.

Connections

Distance Formula in 3D — Examples 1, 6, 7, 8 use its special case (distance from origin).
Section Formula in 3D — next step after locating points: divide the segment between them.
Vectors in 3D — each triple $P (x, y, z)$ is the position vector $O P = x \hat{i} + y \hat{j} + z \hat{k}$ .
Quadrants in 2D Coordinate Geometry — the top-layer octant reasoning in Ex 2–3 copies quadrants.
Equation of a Plane — the mirror planes in Ex 4 & 8 are the simplest planes ( $x = 0$ , etc.).
Direction Cosines and Direction Ratios — describe the direction from $O$ to any of these points.