Mathematics.

coordinate geometry

Lines and Planes in Space

Analytic Geometry30 minDifficulty5 out of 10

You should know: three dimensional coordinates, vectors

Overview

In three-dimensional space, a line is determined by a point and a direction vector, while a plane is determined by a point and a normal vector — a single vector perpendicular to every line lying in the plane. This is a step up in subtlety from 2D coordinate geometry: two linear equations in x, y, z (like a plane's equation) no longer pin down a point, they pin down a whole plane, and a line in space is usually described parametrically or as the intersection of two planes rather than by a single equation. The plane, in particular, is the fundamental flat object of 3D geometry, playing the role that the line played in the 2D plane.

Intuition

A line in space needs a starting point and a direction to walk in — exactly one direction vector, since a line is one-dimensional. A plane, being two-dimensional, could be described by two direction vectors lying in it, but there's a slicker trick: instead of describing directions *in* the plane, describe the single direction *perpendicular* to it. That normal vector n, together with one known point, pins the plane down completely, because a point Q lies in the plane exactly when the vector from the known point to Q is perpendicular to n — which is precisely what the dot-product equation n · (Q − P₀) = 0 says.

Formal Definition

Definition

A line through point P₀ = (x₀, y₀, z₀) with direction vector v = ⟨a, b, c⟩, and a plane through point P₀ with normal vector n = ⟨A, B, C⟩:

r(t)=r0+tv,t∈R\mathbf{r}(t) = \mathbf{r_0} + t\mathbf{v}, \quad t \in \mathbb{R}

r₀ = ⟨x₀,y₀,z₀⟩ is a point on the line

Vector form of a line
x=x0+at,y=y0+bt,z=z0+ctx = x_0 + at,\quad y = y_0 + bt,\quad z = z_0 + ct
Parametric equations of a line
x−x0a=y−y0b=z−z0c\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}

Valid where a, b, c are all nonzero

Symmetric equations of a line
A(x−x0)+B(y−y0)+C(z−z0)=0A(x-x_0) + B(y-y_0) + C(z-z_0) = 0
Point-normal form of a plane
Ax+By+Cz+D=0Ax + By + Cz + D = 0

n = ⟨A, B, C⟩ is normal to the plane

General (scalar) form of a plane

Derivation

The plane equation follows directly from the definition of perpendicularity via the dot product. If P₀ is a known point in the plane and n = ⟨A,B,C⟩ is normal to it, then any other point P = (x,y,z) lies in the plane exactly when the vector P₀P is orthogonal to n:

n⋅P0P→=0\mathbf{n} \cdot \overrightarrow{P_0P} = 0

Orthogonality condition defining the plane

⟨A,B,C⟩⋅⟨x−x0, y−y0, z−z0⟩=0\langle A,B,C\rangle \cdot \langle x-x_0,\ y-y_0,\ z-z_0\rangle = 0

Expand the dot product component-wise

A(x−x0)+B(y−y0)+C(z−z0)=0  ⟹  Ax+By+Cz+(−Ax0−By0−Cz0)⏟D=0A(x-x_0)+B(y-y_0)+C(z-z_0)=0 \;\Longrightarrow\; Ax+By+Cz+\underbrace{(-Ax_0-By_0-Cz_0)}_{D} = 0

Collecting constants gives the general form Ax+By+Cz+D=0

Properties

Direction vector

v=⟨a,b,c⟩ is parallel to the line; scalar multiples of v describe the same line\mathbf{v} = \langle a,b,c\rangle \text{ is parallel to the line; scalar multiples of } \mathbf{v} \text{ describe the same line}

Normal vector

n=⟨A,B,C⟩ is perpendicular to every vector lying in the plane\mathbf{n} = \langle A,B,C\rangle \text{ is perpendicular to every vector lying in the plane}

Parallel planes

Two planes are parallel iff their normal vectors are scalar multiples of each other\text{Two planes are parallel iff their normal vectors are scalar multiples of each other}

Line of intersection

Two non-parallel planes intersect in a line whose direction is n1×n2\text{Two non-parallel planes intersect in a line whose direction is } \mathbf{n_1}\times\mathbf{n_2}

Distance from point to plane

d=∣Ax1+By1+Cz1+D∣A2+B2+C2 for a point (x1,y1,z1)d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \text{ for a point } (x_1,y_1,z_1)

Applications

3D rendering engines represent every polygon face by its plane equation (using the normal vector) to compute lighting via the angle between the normal and a light source.

Worked Examples

  1. Apply the point-normal form with A=2, B=-1, C=4 and (x₀,y₀,z₀)=(1,2,3).

    2(x−1)−1(y−2)+4(z−3)=02(x-1) - 1(y-2) + 4(z-3) = 0

  2. Expand and collect constants.

    2x−2−y+2+4z−12=0  ⟹  2x−y+4z−12=02x - 2 - y + 2 + 4z - 12 = 0 \implies 2x - y + 4z - 12 = 0

Answer: 2x − y + 4z − 12 = 0

Practice Problems

Difficulty 5/10

Find the equation of the plane through (0, 0, 0) with normal vector n = ⟨1, 2, 2⟩.

Difficulty 5/10

Find the distance from the point (1, 1, 1) to the plane 2x − y + 2z − 4 = 0.

Difficulty 6/10

A flat solar panel lies in the plane 3x + 4y + 12z = 60 (units in meters). Find the foot of the perpendicular dropped from the origin onto the panel — the closest point on the panel to the origin.

Common Mistakes

Common Mistake

Trying to describe a plane using a single linear equation the same way a line in 2D is described by y = mx + b, and expecting one equation in x, y, z to pin down a point.

In 3D, one linear equation Ax+By+Cz+D=0 describes an entire plane, not a point or a line. A line in space needs two such equations (the intersection of two planes) or a parametric/vector description.

Common Mistake

Confusing a plane's normal vector with a direction vector lying in the plane.

The normal vector is perpendicular to the plane, while direction vectors used in a line's equation are parallel to that line. A plane has infinitely many direction vectors lying within it but (up to scalar multiples) only one normal direction.

Quiz

A plane in space is most naturally determined by a point together with:
Two planes with normal vectors n₁ and n₂ are parallel exactly when:
The direction of the line where two non-parallel planes intersect is given by:

Summary

  • A line in space is described by a point plus a direction vector: r(t) = r₀ + tv, or equivalently by parametric or symmetric equations.
  • A plane is described by a point plus a normal vector n = ⟨A,B,C⟩, giving the point-normal form A(x−x₀)+B(y−y₀)+C(z−z₀)=0, which expands to Ax+By+Cz+D=0.
  • Two planes are parallel iff their normal vectors are scalar multiples; otherwise they intersect in a line with direction n₁ × n₂.
  • The distance from a point to a plane is |Ax₁+By₁+Cz₁+D| / √(A²+B²+C²).

References

  1. WebsiteWikipedia — Plane (geometry)
  2. WebsiteWikipedia — Line (geometry)