Equation of a plane — normal form, intercept form, general form: Worked examples

Before we start, one reminder of the whole vocabulary we'll use, so nothing sneaks in undefined:

Recall The four objects we lean on (tap to expand)

General form $a x + b y + cz + d = 0$ : the coefficients $(a, b, c)$ ARE a normal vector (the arrow sticking straight out of the plane).
Normal form $l x + m y + n z = p$ : here $(l, m, n)$ is the unit normal (length 1), and $p \geq 0$ is the straight-line distance from the origin to the plane.
Intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ : $a, b, c$ are where the plane pokes through the three axes.
The bridges: divide by $\pm a^{2} + b^{2} + c^{2}$ to reach normal form; divide by the RHS constant to reach intercept form; use $A B \times A C$ (the cross product) to build a normal from three points.

The scenario matrix

Every plane problem lands in one (or more) of these cells. The last column names the example that nails it.

Cell	What makes it tricky	Covered by
C1 All-positive general form	the "friendly" baseline	Ex 1
C2 Mixed signs in $(a, b, c)$	a negative intercept, negative direction cosine	Ex 2
C3 Wrong-sign constant $\Rightarrow$ fix $p \geq 0$	must flip the whole equation	Ex 3
C4 A zero coefficient	plane parallel to an axis, missing term	Ex 4
C5 Degenerate: plane through origin ( $d = 0$ )	intercept form fails	Ex 5
C6 Two zero coefficients	plane parallel to a coordinate plane	Ex 6
C7 Build from point + normal	reverse direction	Ex 7
C8 Build from three points	cross-product route	Ex 8
C9 Real-world word problem	translate English $\to$ equation	Ex 9
C10 Exam twist: given distance & direction, find the plane(s)	limiting/both-sign case	Ex 10

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

Example 1 — C1: the friendly baseline

Worked example All three forms of

$3 x + 6 y + 2 z = 12$ Convert to normal form and intercept form; state the normal, the unit normal, the distance $p$ , and the three intercepts.

Forecast: the coefficients are $3, 6, 2$ — before reading on, guess $∣ n ∣$ . (Hint: it's a Pythagorean-friendly triple.)

Read the normal off the coefficients: $n = (3, 6, 2)$ . Why this step? In $a x + b y + cz + d = 0$ the coefficients of $x, y, z$ are the components of a normal vector — the parent note's headline fact.
Length of the normal: $∣ n ∣ = 3^{2} + 6^{2} + 2^{2} = 9 + 36 + 4 = 49 = 7$ . Why? To turn $n$ into a unit normal (length 1) we must divide by its own length — that's what "unit" means.
Normal form: divide the whole equation by $7$ : $\frac{3}{7} x + \frac{6}{7} y + \frac{2}{7} z = \frac{12}{7} .$ So $\overset{n}{^} = (\frac{3}{7}, \frac{6}{7}, \frac{2}{7})$ and $p = \frac{12}{7}$ . Why divide by $∣ n ∣$ ? Both sides scale together; the LHS coefficients become direction cosines and the RHS becomes the true perpendicular distance.
Intercept form: divide the original by $12$ : $\frac{x}{4} + \frac{y}{2} + \frac{z}{6} = 1.$ Intercepts: $4, 2, 6$ . Why divide by 12? Intercept form demands RHS $= 1$ ; only then do the denominators read off as intercepts.

Verify: unit normal length $= (3/7)^{2} + (6/7)^{2} + (2/7)^{2} = 49/49 = 1$ ✓. Plug intercept $(4, 0, 0)$ into the original: $3 \cdot 4 = 12$ ✓.

Example 2 — C2: mixed signs

Worked example

$x - 2 y - 2 z = 6$ : expect a negative intercept and a negative direction cosine Give the unit normal, $p$ , and the intercepts.

Forecast: two of the three coefficients are negative. Which intercepts will come out negative?

Normal: $n = (1, - 2, - 2)$ .
Length: $∣ n ∣ = 1 + 4 + 4 = 9 = 3$ .
Normal form: divide by $3$ : $\frac{1}{3} x - \frac{2}{3} y - \frac{2}{3} z = 2, \overset{n}{^} = (\frac{1}{3}, - \frac{2}{3}, - \frac{2}{3}), p = 2.$ Why is $p$ still positive? The RHS $6/3 = 2$ is already $\geq 0$ , so no sign flip is needed — the direction cosines simply carry the minus signs. A negative direction cosine just means the normal tilts toward the negative half of that axis.
Intercept form: divide by $6$ : $\frac{x}{6} + \frac{y}{- 3} + \frac{z}{- 3} = 1.$ Intercepts $6, - 3, - 3$ . Why do the negatives appear? A negative intercept means the plane crosses that axis on its negative side — perfectly legal.

Verify: $(0, - 3, 0)$ : $- 2 \cdot (- 3) = 6$ ✓. $(0, 0, - 3)$ : $- 2 \cdot (- 3) = 6$ ✓.

Example 3 — C3: constant of the wrong sign, force $p \geq 0$

Worked example Normal form of

$2 x - y - 2 z + 9 = 0$ Forecast: move the $9$ across — the RHS goes negative. But distance can't be negative. What must you do?

Rewrite so all variables sit on the left: $2 x - y - 2 z = - 9$ . Why? Normal form compares $r \cdot \overset{n}{^}$ to a distance $p$ , so we need "variables $=$ constant."
Length of normal: $∣ n ∣ = 4 + 1 + 4 = 3$ .
Naive divide by $+ 3$ : $\frac{2}{3} x - \frac{1}{3} y - \frac{2}{3} z = - 3$ . The RHS is $- 3 < 0$ — not allowed, because $p$ is a distance. Why is negative $p$ forbidden? $p$ is the length of the perpendicular from the origin; a length is $\geq 0$ by definition.
Fix — divide by $- 3$ instead (so RHS becomes positive and the normal flips with it): $- \frac{2}{3} x + \frac{1}{3} y + \frac{2}{3} z = 3, \overset{n}{^} = (- \frac{2}{3}, \frac{1}{3}, \frac{2}{3}), p = 3.$ Why flip the whole equation? The line and its opposite point the same plane; flipping every sign keeps the plane identical but makes $p \geq 0$ and keeps $\overset{n}{^}$ consistent with it.

Verify: unit normal length $4/9 + 1/9 + 4/9 = 1$ ✓. The distance also matches the distance formula from the origin: $\frac{∣2 \cdot 0 - 0 - 2 \cdot 0 + 9∣}{3} = \frac{9}{3} = 3 = p$ ✓.

Example 4 — C4: one zero coefficient (parallel to an axis)

Worked example

$4 x + 3 z = 12$ (no $y$ !) Find the normal, describe the geometry, and give whatever intercepts exist.

Forecast: the $y$ term is missing. What does a missing variable say about the plane's tilt?

Normal: $n = (4, 0, 3)$ — its $y$ -component is $0$ .
Geometry: because the normal has no $y$ -part, it is perpendicular to the $y$ -axis; the plane therefore contains the $y$ -direction — it runs parallel to the $y$ -axis and never turns to cross it. Why? A missing variable means changing $y$ (moving parallel to the $y$ -axis) never changes the LHS, so you stay on the plane.
Length & normal form: $∣ n ∣ = 16 + 0 + 9 = 5$ , so $\frac{4}{5} x + \frac{3}{5} z = \frac{12}{5}$ , $p = \frac{12}{5}$ .
Intercepts: $x$ -intercept $= 12/4 = 3$ ; $z$ -intercept $= 12/3 = 4$ ; no $y$ -intercept (set $x = z = 0 \Rightarrow 0 = 12$ , impossible), consistent with "parallel to the $y$ -axis." Why no $y$ -intercept? Parallel lines/planes to an axis never meet it.

Verify: $(3, 0, 0)$ : $4 \cdot 3 = 12$ ✓. $(0, 0, 4)$ : $3 \cdot 4 = 12$ ✓. $(0, 5, 0)$ : $0 \neq = 12$ — indeed misses the $y$ -axis ✓.

Example 5 — C5: degenerate, plane through the origin

Worked example Does

$x + 2 y + 2 z = 0$ have an intercept form? Forecast: the constant is $0$ . Predict yes/no before reading.

Spot $d = 0$ : the RHS is $0$ , so the origin $(0, 0, 0)$ satisfies it — the plane passes through $O$ .
Try intercept form: dividing by the RHS means dividing by $0$ — undefined. Every "intercept" would be $x /0$ , which has no meaning. Why does it fail? Intercept form assumes the plane crosses each axis at a nonzero point; a plane through the origin crosses all three axes at the origin itself, so no finite intercept exists.
What still works: normal form is fine, with $p = 0$ . $∣ n ∣ = 1 + 4 + 4 = 3$ , so $\frac{1}{3} x + \frac{2}{3} y + \frac{2}{3} z = 0$ , $p = 0$ . Why $p = 0$ ? The distance from the origin to a plane through the origin is zero.

Verify: origin on plane: $0 + 0 + 0 = 0$ ✓. Distance from origin $= ∣0∣/3 = 0 = p$ ✓. Intercept form: does not exist.

Example 6 — C6: two zero coefficients (parallel to a coordinate plane)

Example 7 — C7: build a plane from a point and a normal

Example 8 — C8: build a plane from three points

Worked example Plane through

$A (1, 1, 1), B (2, 0, 1), C (1, 2, 3)$ Forecast: we have no normal handed to us. Which tool manufactures one?

Two in-plane arrows: $A B = (1, - 1, 0)$ , $A C = (0, 1, 2)$ . Why? Both arrows lie in the plane; their cross product is perpendicular to both, hence to the whole plane.
Cross product $\Rightarrow$ normal: $n = A B \times A C = \hat{i} 10 \hat{j} - 1 1 \hat{k} 02 = \hat{i} (- 2 - 0) - \hat{j} (2 - 0) + \hat{k} (1 - 0) = (- 2, - 2, 1) .$ Why the cross product and not the dot product? The dot product produces a number (an angle/projection); we need a direction perpendicular to two vectors — that is exactly what the cross product returns.
Point–normal form through $A$ : $- 2 (x - 1) - 2 (y - 1) + 1 (z - 1) = 0$ .
Expand: $- 2 x - 2 y + z + 3 = 0 \Rightarrow 2 x + 2 y - z = 3$ (multiplied by $- 1$ for tidiness).

Verify: $A : 2 + 2 - 1 = 3$ ✓; $B : 4 + 0 - 1 = 3$ ✓; $C : 2 + 4 - 3 = 3$ ✓. All three points satisfy it.

Example 9 — C9: real-world word problem

Worked example A solar panel

An engineer mounts a flat solar panel so that its outward face points along the direction $(2, 3, 6)$ , and the panel plane passes $14$ metres from the plant's corner (the origin), on the side the normal points to. Find the panel's equation, its intercepts on the three walls (axes), and confirm the $14$ m distance.

Forecast: the direction $(2, 3, 6)$ is a Pythagorean-friendly triple. Guess $∣ n ∣$ before computing.

Normal & its length: $n = (2, 3, 6)$ , $∣ n ∣ = 4 + 9 + 36 = 49 = 7$ .
Normal form from distance: we want $\overset{n}{^} \cdot r = p$ with $p = 14$ . Unit normal $\overset{n}{^} = (\frac{2}{7}, \frac{3}{7}, \frac{6}{7})$ , so $\frac{2}{7} x + \frac{3}{7} y + \frac{6}{7} z = 14.$ Why start from normal form? The problem gives us a direction and a distance — precisely the two ingredients of normal form.
Clear denominators to general form: multiply by $7$ : $2 x + 3 y + 6 z = 98$ .
Intercepts: divide by $98$ : $\frac{x}{49} + \frac{y}{98/3} + \frac{z}{49/3} = 1$ , giving intercepts $49, \frac{98}{3}, \frac{49}{3}$ metres.

Verify: distance from origin $= \frac{∣2 \cdot 0 + 3 \cdot 0 + 6 \cdot 0 - 98∣}{7} = \frac{98}{7} = 14$ m ✓. Intercept $(49, 0, 0)$ : $2 \cdot 49 = 98$ ✓.

Example 10 — C10: exam twist — distance & direction give TWO planes

Worked example Find all planes parallel to

$2 x - 2 y + z = 0$ that lie at distance $6$ from the origin Forecast: "parallel" fixes the direction; "distance 6" fixes $∣ p ∣$ . How many answers — one or two?

Same normal: parallel planes share the normal $n = (2, - 2, 1)$ , so each has the form $2 x - 2 y + z = k$ for some constant $k$ . Why? Parallel $\Leftrightarrow$ same normal direction; only the constant term slides the plane along the normal.
Length: $∣ n ∣ = 4 + 4 + 1 = 3$ .
Distance condition: distance from origin $= \frac{∣ k ∣}{3} = 6 \Rightarrow ∣ k ∣ = 18$ . Why the absolute value? Distance is $\geq 0$ , but $k$ can be positive or negative — the plane can sit on either side of the origin.
Both solutions: $2 x - 2 y + z = 18 and 2 x - 2 y + z = - 18.$ In proper normal form (RHS $\geq 0$ ): $\frac{2}{3} x - \frac{2}{3} y + \frac{1}{3} z = 6$ and $- \frac{2}{3} x + \frac{2}{3} y - \frac{1}{3} z = 6$ (sign flipped so $p = 6$ ).

Verify: for $k = 18$ : $∣18∣/3 = 6$ ✓. For $k = - 18$ : $∣ - 18∣/3 = 6$ ✓. Exactly two planes — one each side of the origin.

Connections

Equation of a plane — normal form, intercept form, general form (parent)
Direction cosines and direction ratios
Distance of a point from a plane
Angle between two planes
Vector and Cartesian equation of a line
Dot product and projection
Cross product as area/normal