Equation of a plane — normal form, intercept form, general form: Visual walkthrough

Step 1 — A point in space is just an arrow from the corner

WHAT. We label where things are. Put a corner in the room — call it the origin $O$ . To name any point $P$ , draw an arrow from $O$ straight to $P$ . That arrow is the position vector $r$ .

WHY. We want an equation — something with $x, y, z$ in it. The arrow $r$ carries exactly three numbers, its shadows on the three walls: $r = (x, y, z)$ . So "an arrow" and "three coordinates" are the same object seen two ways. Every plane equation is really a rule these three numbers must obey.

PICTURE. The lavender arrow is $r$ ; its three dotted shadows are $x$ (floor-right), $y$ (floor-back), $z$ (up).

Figure — Equation of a plane — normal form, intercept form, general form

Step 2 — The one arrow a plane cannot ignore: the normal

WHAT. Take a flat sheet (our plane). Push a pencil straight out of it. That perpendicular arrow is the normal vector $n = (a, b, c)$ .

WHY this arrow and not one lying in the sheet? Inside the sheet there are infinitely many directions — you cannot pick "the" one. But perpendicular to the sheet there is only one line of directions. That uniqueness is why we anchor the whole plane to $n$ : tilt the pencil, the sheet tilts; slide the sheet along the pencil, it just moves without turning.

PICTURE. Coral arrow $n$ stabs out of the mint sheet at $9 0^{\circ}$ . Two green arrows lie in the sheet — notice $n$ makes a right angle with both.

Step 3 — The dot product: a machine that measures "how much along"

WHAT. Before we can say "perpendicular" in symbols we need one tool. Given two arrows $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ , their dot product is $u \cdot v = x-parts u_{1} v_{1} + y-parts u_{2} v_{2} + z-parts u_{3} v_{3} .$ Here each term multiplies the two shadows on the same axis, then adds. The single number that comes out equals $∣ u ∣ ∣ v ∣ cos θ$ , where $∣ u ∣$ is the length of $u$ and $θ$ is the angle between the arrows.

WHY this tool? We need a test for perpendicular. Look at $cos θ$ : when the arrows point at $9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ , so the whole product is $0$ — no matter how long the arrows are. The dot product answers exactly the question "are these two arrows square to each other?" — and answers it with a single number being zero. (See Dot product and projection.)

PICTURE. Two arrows with the angle $θ$ marked; a caption shows $u \cdot v$ shrinking to $0$ as $θ \to 9 0^{\circ}$ .

Step 4 — Pin the sheet to a known point, then let a test point wander

WHAT. Fix one point we know is on the plane; call its arrow $r_{0} = (x_{0}, y_{0}, z_{0})$ . Let $r = (x, y, z)$ be any candidate point we are testing.

WHY. A direction alone ( $n$ ) does not locate the sheet — infinitely many parallel sheets share the same pencil. We need one nail. $r_{0}$ is that nail; it says which of the parallel sheets we mean.

PICTURE. The nail $r_{0}$ sits on the mint sheet; a roaming point $r$ hovers nearby. The difference arrow $r - r_{0}$ (drawn tip-of- $r_{0}$ to tip-of- $r$ ) is highlighted.

Step 5 — The master equation, born from one right angle

WHAT. If $r$ lies on the plane, the little arrow $r - r_{0}$ lies in the plane, so it is perpendicular to the pencil $n$ . Feed that into the perpendicular test from Step 3: $n \cdot (r - r_{0}) = 0$

WHY. This is the plane. Read it as a rule: "the test arrow, measured along the pencil, must come out to zero." A point passes the test $\Leftrightarrow$ it is flush with the sheet.

PICTURE. The right angle between coral $n$ and the in-plane arrow $r - r_{0}$ , with the equation printed beside the corner it comes from.

Now expand it, term by term, using the dot-product recipe with $n = (a, b, c)$ : $x along n a (x - x_{0}) + y along n b (y - y_{0}) + z along n c (z - z_{0}) = 0.$ Multiply out and gather the constants into one letter $d = - (a x_{0} + b y_{0} + c z_{0})$ : $n_{x} a x + n_{y} b y + n_{z} c z + shift d = 0.$

Step 6 — Costume 1: normal form (make the pencil length exactly 1)

WHAT. Shrink $n$ to a unit-length pencil $\overset{n}{^}$ by dividing by its length $∣ n ∣ = a^{2} + b^{2} + c^{2}$ . Then $\overset{n}{^} = (l, m, n)$ where $l, m, n$ are the direction cosines (see Direction cosines and direction ratios) and satisfy $l^{2} + m^{2} + n^{2} = 1$ .

WHY. With a unit pencil, the equation reads $\overset{n}{^} \cdot r = p$ , and the right side $p$ becomes the honest perpendicular distance from the origin to the sheet — a real, measurable length. Divide $a x + b y + cz = - d$ by $\pm ∣ n ∣$ , choosing the sign so $p \geq 0$ (a distance can't be negative — that is Trap B in the parent note). $\overset{n}{^} \cdot r l x + m y + n z = \geq 0 distance p .$

PICTURE. Drop a perpendicular from $O$ to the sheet; it lands at foot $N$ with $O N = p \overset{n}{^}$ . The segment $O N$ is the distance $p$ .

Step 7 — Costume 2: intercept form (where the sheet pokes the walls)

WHAT. Take general form with $d \neq = 0$ , write it as $A x + B y + C z = k$ with $k \neq = 0$ , and ask where the sheet crosses each axis. On the $x$ -axis, $y = z = 0$ , so $x = k / A$ — call it $a$ . Likewise $b = k / B$ , $c = k / C$ . Substituting and dividing by $k$ to make the right side $1$ : $crosses x-axis at a \frac{x}{a} + at b \frac{y}{b} + at c \frac{z}{c} = 1.$

WHY. Dividing to force RHS $= 1$ makes the three intercepts readable straight off the denominators — no algebra needed at a glance.

Degenerate case. If $d = 0$ the sheet passes through $O$ , so one or more intercepts are $0$ , and $\frac{x}{0}$ is undefined — intercept form does not exist (Trap C). We must not force it.

PICTURE. The triangle where the sheet slices the three axes, with intercepts $a, b, c$ marked; a second faded sheet through $O$ shows the "no-intercept" degenerate case.

Step 8 — Worked example: one plane, all three costumes

The one-picture summary

The single diagram below is the whole page: the nail $r_{0}$ , the roaming point $r$ , the little in-plane arrow $r - r_{0}$ at a right angle to the pencil $n$ — and, branching off, the unit pencil $\overset{n}{^}$ with distance $p$ (normal form) and the axis-crossings $a, b, c$ (intercept form). One right angle spawns all three equations.

Recall Feynman: the whole walkthrough in plain words

A point in space is just an arrow from the corner of the room. A flat sheet is described by one pencil stuck straight out of it — that's the only direction that's unique, because inside the sheet there are too many directions to choose from. To test whether some point is on the sheet, draw the little arrow from a known point on the sheet to your test point; if that little arrow is square to the pencil, the point is on the sheet. "Square to" means the dot product — a machine that multiplies matching shadows and adds — comes out to zero. Write that zero out with coordinates and you get $a x + b y + cz + d = 0$ , where $(a, b, c)$ is literally the pencil. Shrink the pencil to length one and the right-hand number becomes the sheet's distance from the corner (normal form). Instead ask where the sheet pokes the three walls and divide so the right side is one, and the denominators are those three crossing points (intercept form). One right angle, three outfits.

Connections

Dot product and projection — the zero-test engine of Step 3.
Direction cosines and direction ratios — the $(l, m, n)$ of normal form.
Distance of a point from a plane — extends the $p$ of Step 6 to any point.
Cross product as area/normal — how to get $n$ from three points.
Angle between two planes — compares two pencils $n_{1}, n_{2}$ .
Vector and Cartesian equation of a line — the sibling built from a direction inside space.