Line-plane intersection

From Wikipedia, the free encyclopedia

It has been suggested that Line–plane intersection be merged into this article or section. (Discuss)

This article may require cleanup to meet Wikipedia's quality standards.
Please discuss this issue on the talk page or replace this tag with a more specific message.
This article has been tagged since December 2005.

The three possible plane-line intersections:
1. No intersection.
2. Point intersection.
3. Line intersection.

In analytic geometry, the intersection of a line and a plane can be the empty set, a point, or a line. Distinguishing these cases, and determining equations for the point and line in the latter cases have use, for example, in computer graphics, motion planning, and collision detection.

[edit] Equations in 3D Euclidean space

[edit] Parametric form

A line is described by all points that are a given direction from a point. Thus a line can be represented as

$\mathbf{p}_a + (\mathbf{p}_b - \mathbf{p}_a)t, \quad t\in \mathbb{R},$

where $\mathbf{p}_a=(x_a,y_a,z_a)$ and $\mathbf{p}_b=(x_b,y_b,z_b)$ are two distinct points along the line.

Similarly a plane can be represented as

$\mathbf{p}_0 + (\mathbf{p}_1-\mathbf{p}_0)u + (\mathbf{p}_2-\mathbf{p}_0)v, \quad u,v\in\mathbb{R}$

where $\mathbf{p}_k=(x_k,y_k,z_k)$ , $k = 0,1,2$ are three points in the plane.

The point at which the line intersects the plane is therefore described by setting the line equal to the plane in the parametric equation:

$\mathbf{p}_a + (\mathbf{p}_b - \mathbf{p}_a)t = \mathbf{p}_0 + (\mathbf{p}_1-\mathbf{p}_0)u + (\mathbf{p}_2-\mathbf{p}_0)v$

This can be simplified to

$\mathbf{p}_a - \mathbf{p}_0 = (\mathbf{p}_a - \mathbf{p}_b)t + (\mathbf{p}_1-\mathbf{p}_0)u + (\mathbf{p}_2-\mathbf{p}_0)v,$

which can be expressed in matrix form as:

$\begin{bmatrix} x_a - x_0 \\ y_a - y_0 \\ z_a - z_0 \end{bmatrix} = \begin{bmatrix} x_a - x_b & x_1 - x_0 & x_2 - x_0 \\ y_a - y_b & y_1 - y_0 & y_2 - y_0 \\ z_a - z_b & z_1 - z_0 & z_2 - z_0 \end{bmatrix} \begin{bmatrix} t \\ u \\ v \end{bmatrix}$

The point of intersection is then equal to

$\mathbf{p}_a + (\mathbf{p}_b - \mathbf{p}_a)t$

[edit] Usage

If the solution satisfies the condition $t \in [0,1],$ , then the intersection point is on the line between $\mathbf{p}_a$ and $\mathbf{p}_b$ .

If the solution satisfies

$u,v \in [0,1], \;\;\; (u+v) \leq 1,$

then the intersection point is in the plane inside the triangle spanned by the three points $\mathbf{p}_0$ , $\mathbf{p}_1$ and $\mathbf{p}_2$ .

This problem is typically solved by expressing it in matrix form, and inverting it:

$\begin{bmatrix} t \\ u \\ v \end{bmatrix} = \begin{bmatrix} x_a - x_b & x_1 - x_0 & x_2 - x_0 \\ y_a - y_b & y_1 - y_0 & y_2 - y_0 \\ z_a - z_b & z_1 - z_0 & z_2 - z_0 \end{bmatrix}^{-1} \begin{bmatrix} x_a - x_0 \\ y_a - y_0 \\ z_a - z_0 \end{bmatrix}$