Coordinate system in 3D — x, y, z axes, octants: Visual walkthrough

Everything below is built on top of the ideas in the parent note; the only outside tool we will borrow is Pythagoras from flat 2D geometry, which we recall in Step 3.

Step 0 — The three rulers (so no symbol is a mystery)

WHAT. Before any formula, fix what the three numbers even are. We stand in the corner of a room. Three edges shoot out from that corner, each at a right angle to the other two. We call these edges the $x$ -axis, $y$ -axis and $z$ -axis, and the corner itself the origin $O$ .

WHY. A "distance" is meaningless until we agree how a point is addressed. A point $P$ gets three numbers:

$x$ = how far you walk along the $x$ -ruler,
$y$ = how far along the $y$ -ruler,
$z$ = how far straight up the $z$ -ruler.

We write this address as the ordered triple $P (x, y, z)$ — order matters, exactly as the parent note warned.

PICTURE.

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

Step 1 — Drop $P$ onto the floor to get its shadow $M$

WHAT. From $P$ we drop a straight vertical line down to the floor (the $x y$ -plane, where $z = 0$ ). Where it lands we call $M$ . Its address is $M (x, y, 0)$ — same $x$ and $y$ as $P$ , but height flattened to $0$ .

WHY. $P$ floats in 3D, which is hard to reason about directly. But its shadow $M$ lives entirely in the flat floor, where ordinary 2D geometry works. Splitting a hard 3D problem into "a flat part" + "a vertical part" is the whole trick.

$M (same as P x, same as P y, flattened height 0)$

PICTURE — notice the dashed vertical drop from $P$ to $M$ .

Step 2 — Measure the shadow's distance $O M$ across the floor

WHAT. Inside the floor we now have a right triangle: from $O$ walk $x$ along the $x$ -axis, then $y$ parallel to the $y$ -axis, and the slanted line back is $O M$ .

WHY tan/Pythagoras and not something else? We want a length of a slanted line whose two perpendicular legs we already know ( $x$ and $y$ ). The one tool that turns two perpendicular legs into the hypotenuse is the Pythagoras theorem — that is exactly its job.

$O M = leg along x x^{2} + leg along y y^{2}$

$x^{2}$ — the horizontal leg, squared.
$y^{2}$ — the perpendicular leg, squared.
— turns "sum of squared legs" back into an honest length.

PICTURE — the red triangle lives flat in the floor.

Step 3 — Recall WHY Pythagoras is allowed here

PICTURE — the little pink square marks the right angle that unlocks Pythagoras.

Step 4 — Stand the vertical edge $P M$ back up

WHAT. Now rebuild the height we threw away. The segment $P M$ from the shadow $M$ up to $P$ has length equal to the height of $P$ , which is $∣ z ∣$ .

WHY the absolute value $∣ z ∣$ ? A length can never be negative. If $P$ is below the floor, its $z$ is negative (an octant with $z < 0$ ), yet the physical rod from $M$ up to $P$ still has a positive length. So we write $∣ z ∣$ to strip the sign — this is the same guard the parent note gives against "squaring negatives wrong".

$P M = height, always \geq 0 ∣ z ∣$

Crucially, $P M$ is perpendicular to the whole floor, so $P M ⊥ O M$ . That right angle is what lets Pythagoras run again.

PICTURE — the violet vertical rod, with the right angle where it meets the floor.

Step 5 — Pythagoras a second time in the standing triangle $O M P$

WHAT. Triangle $O M P$ has a right angle at $M$ (Step 4). Its legs are the floor-distance $O M$ and the vertical $P M$ ; its hypotenuse is exactly our target $O P$ .

WHY. We now know both legs, and they meet at $9 0^{\circ}$ — the precise setup Pythagoras answers.

$O P^{2} = = x^{2} + y^{2} O M^{2} + = ∣ z ∣^{2} = z^{2} P M^{2}$

$O M^{2}$ — substitute the floor result from Step 2, which already equals $x^{2} + y^{2}$ .
$P M^{2}$ — and $∣ z ∣^{2} = z^{2}$ , because squaring erases the sign anyway (so the $∣ \cdot ∣$ quietly disappears here).

Putting the pieces together:

$O P^{2} = (x^{2} + y^{2}) + z^{2} = x^{2} + y^{2} + z^{2}$

PICTURE — the standing right triangle, hypotenuse $O P$ glowing.

Step 6 — Take the root: length is positive

WHAT. We have $O P^{2}$ ; take the positive square root to get $O P$ itself.

WHY only the positive root? $O P$ is a distance, so we keep the $+$ root and discard the negative one.

Step 7 — Every case is already handled (the degenerate checks)

WHY this step. A formula you cannot trust at the edges is useless. Let us walk the corner cases and see the picture collapse gracefully each time.

PICTURE — three tiny frames: the point, the axis-only case, the floor-only case.

The one-picture summary

This single frame stacks both right triangles — the flat one $O M x y$ and the standing one $O M P$ — so you can watch $x^{2} + y^{2}$ become $O M^{2}$ , then join $z^{2}$ to become $O P^{2}$ , all at once.

Recall Feynman retelling — the whole walkthrough in plain words

A balloon floats somewhere in your room. To find how far it is from the corner, first look straight down and mark its shadow on the floor. Walk from the corner to that shadow — Pythagoras across the floor tells you that walk is $x^{2} + y^{2}$ . Now the balloon is just straight up from the shadow by its height $∣ z ∣$ . So the corner, the shadow, and the balloon make a second right triangle standing on its side. Pythagoras again: the corner-to-balloon length squared is (floor-walk squared) plus (height squared) $= x^{2} + y^{2} + z^{2}$ . Take the square root because a length can't be negative, and you're done: $O P = x^{2} + y^{2} + z^{2}$ . Notice we squared everything, so it never cares whether the balloon is left or right, front or back, up or down — all eight octants, one formula.

Connections

Distance Formula in 3D — repeat this exact two-triangle argument between any two points, not just from the origin.
Section Formula in 3D — uses the same coordinate addresses to cut a segment.
Vectors in 3D — $O P = ∣ O P ∣ = x^{2} + y^{2} + z^{2}$ is the length of the position vector.
Quadrants in 2D Coordinate Geometry — the flat floor-triangle of Step 2 is pure 2D.
Direction Cosines and Direction Ratios — once $O P$ is known, ratios like $x / O P$ describe direction.