Mathematics · Analytic Geometry / Vectors

Hesse Normal Form (Distance Point-Plane)

The Hesse normal form gives the distance of a point from a plane: insert the point into the coordinate equation and divide by the length of the normal vector.

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Formula

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

Coordinates and d(P;E) in length units (LU) · normal vector coordinates dimensionless

Variables & units

Variables & units
Symbol	Meaning	Unit
P(p₁\|p₂\|p₃)	Point whose distance is required	length units
a, b, c	Coordinates of the normal vector n⃗ of the plane	dimensionless
$d$	Right-hand side of the plane equation ax + by $+ cz = d$	dimensionless
\|n⃗\|	Length of the normal vector $\sqrt{(a^{2} + b^{2} + c^{2})}$	dimensionless

LaTeX

d(P;E) = \frac{|a p_{1} + b p_{2} + c p_{3} - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}

QUANTA STUDYquanta-study.de

Slanted plane E with normal vector n; from a point P above it a highlighted perpendicular leads down to the plane, marked as d(P;E). — The distance d(P;E) is the length of the perpendicular from P to E; it runs parallel to the normal vector n.

Derivation & background

Named after Otto Hesse (19th century). If the coordinate equation is normalized to |n⃗ $| = 1$ , inserting a point directly yields the signed distance: positive on the side the normal vector points to, negative on the other. The absolute value turns this into the geometric distance; a point in the plane gives 0. The distance of parallel planes also follows in one step.

Exam blueprint

Validity range

Holds for planes in coordinate form ax + by $+ cz = d$ in space; the normal vector (a|b|c) must not be the zero vector. Without the absolute value one gets the signed distance with side information.

Derivation steps

Projection of the connection vector onto the unit normal vector.

1For a plane point A the distance of P is the length of the projection of AP⃗ onto n⃗: $d = |AP$ ⃗·n⃗ $\frac{|}{|}n$ ⃗|.
2Writing it out with n⃗·a⃗ = d gives d(P;E) = |a·p₁ + b·p₂ + c·p₃ − d|/√(a² + b² + c²).

Rearrangements

Normalized plane equation

\frac{a x + b y + c z - d}{\sqrt{a^{2}+b^{2}+c^{2}}} = 0

After normalization, inserting a point directly yields the distance.

Distance of parallel planes

d = \frac{|d_{1} - d_{2}|}{|\vec{n}|}

Assuming the same normal vector; otherwise align first.

Points at a required distance

a x + b y + c z = d \pm k \cdot |\vec{n}|

Both parallel planes at distance k, e.g. for tangent planes.

Task variant

E: $x + 2y + 2z = 6$ . What is the distance of the origin from E?

$d = |0 + 0 + 0 -\frac{ 6|}{\sqrt{(1 + 4 + 4)}} =\frac{ 6}{3} = 2 LU$ .

How far apart are the parallel planes $2x + y + 2z = 9$ and $2x + y + 2z = 3$ ?

|n⃗ $| = \sqrt{(4 + 1 + 4)} = 3$ , so $d = |9 -\frac{ 3|}{3} = 2 LU$ .

Common mistakes

Forgetting to divide by |n⃗|.

Inserting alone gives only a multiple; only /√(a²+b²+c²) turns it into the distance.

Not subtracting the right-hand side d.

The numerator needs $a\cdot{}p_{1} + b\cdot{}p_{2} + c\cdot{}p_{3} - d$ , not just the left-hand side.

Interpreting a negative result as an error.

Without the absolute value the sign only indicates the side of the plane; the distance is the absolute value.

Exam context

Distance tasks point-plane and plane-plane, tangent planes to spheres, distances in applied tasks (flight paths).

Formula cluster

Distances in space

HNF for point-plane, cross product formula for point-line: together they cover the distance tasks.

E: (x⃗ − p⃗)·n⃗ = 0

Plane Equation in Normal Form

d(Q;g) = |(q⃗−p⃗)×u⃗| / |u⃗|

Distance Point-Line

a⃗·b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Dot Product

Worked example

E: $2x + y + 2z = 9$ and P(3|1|5): numerator $|2\cdot{}3 + 1 + 2\cdot{}5 - 9| = |8| = 8$ , denominator $\sqrt{(4 + 1 + 4)} = 3$ . Distance $d(P;E) = \frac{8}{3} \approx 2.67\,\mathrm{LU}$ .

Applications

Distance tasks in final exams (point-plane, parallel planes), tangent planes to spheres, safety distances of flight paths, collision checks in computer graphics

How do you calculate with Hesse Normal Form (Distance Point-Plane)?

Here is how to work through a typical Hesse Normal Form (Distance Point-Plane) (d(P;E) = |n⃗·p⃗ − d| / |n⃗|) task step by step:

1
Task
E: $x + 2y + 2z = 6$ . What is the distance of the origin from E?
Solution path
$d = |0 + 0 + 0 -\frac{ 6|}{\sqrt{(1 + 4 + 4)}} =\frac{ 6}{3} = 2 LU$ .
2
Task
How far apart are the parallel planes $2x + y + 2z = 9$ and $2x + y + 2z = 3$ ?
Solution path
|n⃗ $| = \sqrt{(4 + 1 + 4)} = 3$ , so $d = |9 -\frac{ 3|}{3} = 2 LU$ .

Frequently asked questions about Hesse Normal Form (Distance Point-Plane)

How do you calculate the distance of a point from a plane?+

In three steps with the Hesse normal form. First: bring the plane into coordinate form ax + by $+ cz = d$ . Second: insert the point and subtract d; the absolute value of this is the numerator. Third: divide by the length of the normal vector $\sqrt{(a^{2} + b^{2} + c^{2})}$ . Example: E: $2x + y + 2z = 9$ and P(3|1|5): inserting gives $6 + 1 + 10 = 17$ , minus 9 leaves 8; the denominator is $\sqrt{9} = 3$ , so d(P;E) = 8/3 ≈ 2.67 LU. Result 0 means the point lies in the plane. Forget neither subtracting d nor the division, those are the two classics.

Why must you divide by the magnitude of the normal vector?+

Because the same plane has infinitely many coordinate equations: $2x + y + 2z = 9$ and $4x + 2y + 4z = 18$ describe identical point sets. If you only insert a point, you get different numbers depending on the equation, here 8 or 16; that cannot be a distance yet. Only the division by |n⃗| normalizes the equation and makes the result unique: $\frac{8}{3} =\frac{ 16}{6}$ . Geometrically a projection is behind it: the distance is the length of the projection of the connection vector onto the normal direction, and for that you need the unit normal vector n⃗/|n⃗| of length 1. The division is thus not a convention but the actual core of the formula.

What does the sign mean if you drop the absolute value?+

Without the absolute value the Hesse normal form yields the signed distance: the sign tells on which side of the plane the point lies. Positive means the point is on the side the normal vector points to; negative means the opposite side. With this you can check whether two points lie on the same side of a plane (equal signs) or whether a segment pierces the plane (different signs) without computing the intersection. Example: for E: $2x + y + 2z = 9$ , P(3|1|5) gives $+\frac{8}{3}$ , while the origin gives $-\frac{9}{3} = -3$ ; so they lie on different sides. For the geometric distance you take the absolute value at the end.

How do you calculate the distance of two parallel planes?+

Bring both planes to the same normal vector; then the equations differ only in the right-hand side, and $d = |d_{1} -\frac{ d_{2}|}{|}n$ ⃗| holds. Example: $2x + y + 2z = 9$ and $2x + y + 2z = 3$ have |n⃗ $| = 3$ , so the distance is $|9 -\frac{ 3|}{3} = 2 LU$ . If the second plane comes with a multiple of the normal vector (say 4x + 2y + 4z = 6), divide the equation by the factor first. Alternatively pick any point of one plane and compute its distance to the other with the ordinary HNF formula; both lead to the same result. Non-parallel planes intersect, their distance is 0.

What do you need the Hesse normal form for with spheres?+

For relative positions of sphere and plane the distance of the centre M from the plane compared with the radius r decides. If d(M;E) > r, the plane misses the sphere; if d(M;E) = r, it touches the sphere (tangent plane); if d(M;E) < r, it cuts out a circle. Example: sphere with M(3|1|5), $r = 3$ and E: $2x + y + 2z = 9$ : $d =\frac{ 8}{3} \approx{} 2.67 < 3$ , so the plane cuts the sphere; by Pythagoras the intersection circle has radius $\sqrt{(r^{2} - d^{2})} = \sqrt{(9 -\frac{ 64}{9})} \approx{} 1.37$ . Tangent planes are constructed the same way: all planes at distance r from the centre.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 3, §2.5: Equations of Lines and Planes in Space

More formulas: Analytic geometry and vectors

\vec{a} \times \vec{b} = \begin{pmatrix} a_{2}b_{3} - a_{3}b_{2} \\ a_{3}b_{1} - a_{1}b_{3} \\ a_{1}b_{2} - a_{2}b_{1} \end{pmatrix}

Cross Product (Vector Product)

|\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}

Magnitude of a Vector (Length)

u_k=v_k-\sum_{j<k}\langle v_k,q_j\rangle q_j, \quad q_k=\frac{u_k}{\|u_k\|}

Gram–Schmidt Orthogonalization

|\langle u,v\rangle|\leq\|u\|\,\|v\|

Cauchy–Schwarz Inequality

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Hesse Normal Form (Distance Point-Plane)

The Hesse normal form gives the distance of a point from a plane: insert the point into the coordinate equation and divide by the length of the normal vector.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Variables & units

Symbol

Meaning

Unit

P(p₁|p₂|p₃)

Point whose distance is required

length units

a, b, c

Coordinates of the normal vector n⃗ of the plane

dimensionless

d

Right-hand side of the plane equation ax + by

+ cz = d

dimensionless

|n⃗|

Length of the normal vector

\sqrt{(a^{2} + b^{2} + c^{2})}

dimensionless

Derivation & background

Exam blueprint

Validity range

Holds for planes in coordinate form ax + by $+ cz = d$ in space; the normal vector (a|b|c) must not be the zero vector. Without the absolute value one gets the signed distance with side information.

Derivation steps

Projection of the connection vector onto the unit normal vector.

1For a plane point A the distance of P is the length of the projection of AP⃗ onto n⃗: $d = |AP$ ⃗·n⃗ $\frac{|}{|}n$ ⃗|.
2Writing it out with n⃗·a⃗ = d gives d(P;E) = |a·p₁ + b·p₂ + c·p₃ − d|/√(a² + b² + c²).

How do you calculate with Hesse Normal Form (Distance Point-Plane)?

Here is how to work through a typical Hesse Normal Form (Distance Point-Plane) (d(P;E) = |n⃗·p⃗ − d| / |n⃗|) task step by step:

Task

E: $x + 2y + 2z = 6$ . What is the distance of the origin from E?

Solution path

$d = |0 + 0 + 0 -\frac{ 6|}{\sqrt{(1 + 4 + 4)}} =\frac{ 6}{3} = 2 LU$ .

Task

How far apart are the parallel planes $2x + y + 2z = 9$ and $2x + y + 2z = 3$ ?

Solution path

|n⃗ $| = \sqrt{(4 + 1 + 4)} = 3$ , so $d = |9 -\frac{ 3|}{3} = 2 LU$ .

Frequently asked questions about Hesse Normal Form (Distance Point-Plane)

How do you calculate the distance of a point from a plane?+

Why must you divide by the magnitude of the normal vector?+

What does the sign mean if you drop the absolute value?+

How do you calculate the distance of two parallel planes?+

What do you need the Hesse normal form for with spheres?+

Retain Hesse Normal Form (Distance Point-Plane) for exams

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

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential