Mathematics · Analytic Geometry / Vectors

Plane Equation in Normal Form

A plane is fixed by a point and a normal vector: exactly the position vectors x⃗ with (x⃗ − p⃗)·n⃗ = 0 lie in it.

AdvancedExam-relevant

Study as an exam set I already have an account

Free · no credit card · in your study plan in 2 minutes

Formula

E : (x - p) \cdot n = 0

LaTeX: E: (\vec{x} - \vec{p}) \cdot \vec{n} = 0

Coordinates in length units (LU) · dot product dimensionless

Variables & units – Plane Equation in Normal Form

Symbol	Meaning	Unit
x⃗	Position vector of an arbitrary point of the plane	length units
p⃗	Support vector (position vector of a known point)	length units
n⃗	Normal vector (perpendicular to the plane)	dimensionless

Derivation & background – Plane Equation in Normal Form

The core idea: every connection vector x⃗ − p⃗ inside the plane is perpendicular to n⃗, so the dot product is 0. Multiplied out, the coordinate form ax + by + cz = d appears with (a|b|c) = n⃗ and d = n⃗·p⃗. From a parametric form you obtain n⃗ as the cross product of the two direction vectors. n⃗ is determined only up to multiples: every multiple describes the same plane.

Exam blueprint

Validity range

Describes every plane in space; n⃗ must be perpendicular to the plane and must not be the zero vector. Multiples of n⃗ and any other plane point as support point describe the same plane.

Derivation steps

Orthogonality as the defining property: lying in the plane means running perpendicular to n⃗.

1X lies in E exactly when the connection vector x⃗ − p⃗ runs inside the plane, i.e. perpendicular to n⃗: (x⃗ − p⃗)·n⃗ = 0.
2Multiplying out yields n⃗·x⃗ = n⃗·p⃗, i.e. the coordinate form ax + by + cz = d.

Rearrangements

Coordinate form

a x + b y + cz = n \cdot p

(a|b|c) = n⃗; d arises by inserting the support point.

Normal vector from parametric form

n = u \times v

Cross product of the direction vectors; then insert the support point.

Point test

(q - p) \cdot n = ? 0

Result 0 means: Q lies in the plane.

Task variant

State the coordinate form of the plane with n⃗ = (3|2|−1) through P(2|0|0).

d = n⃗·p⃗ = 3·2 + 0 + 0 = 6, so E: 3x + 2y − z = 6.

Determine a coordinate equation of the plane through A(1|1|0), B(3|1|1) and C(1|2|2).

AB⃗ = (2|0|1), AC⃗ = (0|1|2). n⃗ = AB⃗ × AC⃗ = (−1|−4|2). With A: −1 − 4 + 0 = −5, so −x − 4y + 2z = −5 or x + 4y − 2z = 5. Check with B: 3 + 4 − 2 = 5 ✓.

Common mistakes

Confusing normal vector and direction vector.

n⃗ is perpendicular to the plane; direction vectors lie inside it.

Forgetting d or assuming d = 0.

Compute d = n⃗·p⃗ with a plane point; d = 0 holds only for planes through the origin.

Sign errors in the cross product.

Check the result: n⃗·AB⃗ = 0 and n⃗·AC⃗ = 0 must hold.

Exam context

Conversion between plane representations, relative positions, angle and distance tasks as follow-up steps.

These mistakes cost points in real exams. The set drills them until they stick.

Train these mistakes in the set I already have an account

Formula cluster

Planes in space

Normal form, coordinate form and HNF are the same plane in three guises.

d (P; E) = \frac{∣ a p _{1} + b p _{2} + c p _{3} - d ∣}{a ^{2} + b ^{2} + c ^{2}}

Hesse Normal Form (Distance Point-Plane)

a \times b = a_{2} b_{3} - a_{3} b_{2} a_{3} b_{1} - a_{1} b_{3} a_{1} b_{2} - a_{2} b_{1}

Cross Product (Vector Product)

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = ∣ a ∣ \cdot ∣ b ∣ \cdot cos φ

Dot Product

Worked example

n⃗ = (2|−1|3) and P(1|0|2): d = n⃗·p⃗ = 2·1 − 0 + 3·2 = 8, so E: 2x − y + 3z = 8. Check with P: 2 − 0 + 6 = 8 ✓.

Applications

Converting parametric form into coordinate form, relative positions of point, line and plane, angles between planes via normal vectors, precursor of the Hesse normal form

Quanta exam set

Curated exam set for "Plane Equation in Normal Form":

Question (front)

Which formula describes Plane Equation in Normal Form?

Answer in your set

Question (front)

How do you rearrange E: (x⃗ − p⃗)·n⃗ = 0 for Coordinate form?

Answer in your set

Question (front)

Which common mistake happens with Plane Equation in Normal Form?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

These 10 cards are ready. One click and they sit in your deck, FSRS schedules the reviews until exam day.

Take the set, free I already have an account

Scientific sources

[1]Bronstein, I.N. et al. (2020). Taschenbuch der Mathematik. Springer.

Common notations & search queries

(x-p)*n=0ax+by+cz=dNormalenform EbeneKoordinatenform EbeneEbenengleichung aufstellenNormalenvektor bestimmenplane equation normal formParameterform in Koordinatenform

Related formulas

d (P; E) = \frac{∣ a p _{1} + b p _{2} + c p _{3} - d ∣}{a ^{2} + b ^{2} + c ^{2}}

Hesse Normal Form (Distance Point-Plane)

a \times b = a_{2} b_{3} - a_{3} b_{2} a_{3} b_{1} - a_{1} b_{3} a_{1} b_{2} - a_{2} b_{1}

Cross Product (Vector Product)

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = ∣ a ∣ \cdot ∣ b ∣ \cdot cos φ

Dot Product

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

Free in Quanta

Frequently asked questions about Plane Equation in Normal Form

How do you get from the parametric form to the coordinate form of a plane?+

In three steps. First: form the normal vector from the two direction vectors u⃗ and v⃗ as a cross product, n⃗ = u⃗ × v⃗. Second: write down the left-hand side of the coordinate form, ax + by + cz with (a|b|c) = n⃗. Third: insert the support point to determine d. Example: support point A(1|1|0), u⃗ = (2|0|1), v⃗ = (0|1|2): n⃗ = (−1|−4|2), inserting A gives −1 − 4 + 0 = −5, so −x − 4y + 2z = −5, nicer x + 4y − 2z = 5. Check: a second plane point must satisfy the equation too. The way back is easier: guess three plane points or take intercepts.

How do you set up the normal form of a plane?+

You need exactly two ingredients: a point P of the plane (support vector p⃗) and a normal vector n⃗ perpendicular to the plane. Then the normal form is (x⃗ − p⃗)·n⃗ = 0: a point X lies in the plane exactly when its connection vector to P is perpendicular to n⃗. Example: P(1|0|2) and n⃗ = (2|−1|3) give (x⃗ − (1|0|2))·(2|−1|3) = 0; multiplied out 2x − y + 3z = 8. Depending on the task you obtain the normal vector from the cross product of two direction vectors, from the coordinate form (read off coefficients) or from a perpendicularity condition. Every multiple of n⃗ works the same.

What does the number d in the coordinate form ax + by + cz = d tell you?+

d is the dot product n⃗·p⃗ with any plane point and ties the plane to its position in space: changing d shifts the plane parallel to itself. d = 0 means the plane passes through the origin. After normalization d becomes directly interpretable geometrically: |d|/|n⃗| is the distance of the plane from the origin. Example: 2x − y + 3z = 8 has |n⃗| = √14, the distance from the origin is 8/√14 ≈ 2.14 LU. Careful: d alone (without division by |n⃗|) is not a distance, because the equation can be multiplied through by any factor without changing the plane.

How do you check whether a point lies in a plane?+

With the point test: insert the coordinates of the point into the plane equation and see whether a true statement results. In the coordinate form 2x − y + 3z = 8 you test e.g. Q(1|1|1): 2 − 1 + 3 = 4 ≠ 8, Q does not lie in the plane. In the normal form you compute (q⃗ − p⃗)·n⃗ and check for 0. The value of the deviation is not the distance; for that you would still divide by |n⃗| (Hesse normal form). The point test is also the checking tool after every conversion: support and spanning points of the original parametric form must satisfy the new coordinate equation, otherwise something went wrong with the cross product or with d.

How do you calculate the angle between two planes?+

Via their normal vectors: cos φ = |n⃗₁·n⃗₂|/(|n⃗₁|·|n⃗₂|). The absolute value in the numerator ensures you get the acute intersection angle (between 0° and 90°), regardless of the directions the normal vectors point. Example: E₁ with n⃗₁ = (1|0|1) and E₂ with n⃗₂ = (0|1|1): cos φ = |0 + 0 + 1|/(√2·√2) = 1/2, so φ = 60°. Two planes are perpendicular when n⃗₁·n⃗₂ = 0; they are parallel when the normal vectors are multiples of each other. The line-plane angle works analogously, but with the sine: sin φ = |u⃗·n⃗|/(|u⃗|·|n⃗|).

Retain Plane Equation in Normal Form for exams

Create a curated FSRS exam set for E: (x⃗ − p⃗)·n⃗ = 0: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Study as an exam set Already registered? Continue directly

Free · curated formula set · LaTeX · FSRS spaced repetition

How do you calculate with Plane Equation in Normal Form?

Here is how to work through a typical Plane Equation in Normal Form (E: (x⃗ − p⃗)·n⃗ = 0) task step by step:

1
Task
State the coordinate form of the plane with n⃗ = (3|2|−1) through P(2|0|0).
Solution path
d = n⃗·p⃗ = 3·2 + 0 + 0 = 6, so E: 3x + 2y − z = 6.
2
Task
Determine a coordinate equation of the plane through A(1|1|0), B(3|1|1) and C(1|2|2).
Solution path
AB⃗ = (2|0|1), AC⃗ = (0|1|2). n⃗ = AB⃗ × AC⃗ = (−1|−4|2). With A: −1 − 4 + 0 = −5, so −x − 4y + 2z = −5 or x + 4y − 2z = 5. Check with B: 3 + 4 − 2 = 5 ✓.