Homogeneous coordinates & points at infinity

Points and rays

A 2D point gets a third coordinate 1. Converting back divides by the last coordinate:

(x, y) \to \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} x \\ y \\ w \end{bmatrix} \to \left( \frac{x}{w}, \frac{y}{w} \right)

The same works in 3D, which is how scene points enter the camera's projection matrix:

(x, y, z) \to \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} \to \left( \frac{x}{w}, \frac{y}{w}, \frac{z}{w} \right)

Homogeneous coordinates are scale invariant: for any $k \neq 0$ ,

k \begin{bmatrix} x \\ y \\ w \end{bmatrix} = \begin{bmatrix} kx \\ ky \\ kw \end{bmatrix} \to \left( \frac{kx}{kw}, \frac{ky}{kw} \right) = \left( \frac{x}{w}, \frac{y}{w} \right)

So a point $(x, y)$ is really the whole ray $(wx, wy, w)$ through the origin, and the image plane $w = 1$ is just where we look at it. In the 3D view, every vector on the ray through $p_1$ lands on $p_1$ . This is also why a camera cannot recover depth from a single image: the whole projection ray maps to one pixel.

Lines and intersections

A line $ax + by + c = 0$ is written as the vector $l = (a, b, c)^\top$ . A point $p = (x, y, 1)^\top$ lies on the line exactly when $p \cdot l = 0$ . Like points, lines are only defined up to scale. Two cross products do all the work:

\text{line through two points: } l = p_1 \times p_2, \qquad \text{intersection of two lines: } q = l_1 \times l_2

In the 3D view each line is a plane through the origin (the tinted triangles). Two such planes meet in a ray, and that ray is the intersection point $q$ .

Parallel lines and points at infinity

In Cartesian coordinates parallel lines never meet. In homogeneous coordinates the cross product still gives an answer, $q = (x, y, 0)^\top$ . A point with $w = 0$ cannot be divided out: it is a point at infinity in the direction $(x, y)$ . In the 3D view, the planes of two parallel lines meet in a ray that lies in the plane $w = 0$ , parallel to the image plane, so it never hits it.

This is exactly what happens in perspective images. Parallel lines in the scene have the same direction, and their images meet at a vanishing point: the image of their common point at infinity. Unlike here, the vanishing point is usually at a finite position in the image, because the camera looks at the lines at an angle.

Why bother? In homogeneous coordinates, operations that are not linear in Cartesian coordinates become matrix products. Perspective division moves to the very end of the pinhole camera's projection. A translation becomes a matrix product too:

\begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & t_x \\ 0 & 1 & t_y \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}

The axes follow the image convention:

x

right,

y

down, and

w

points away from the viewer, like the camera's optical axis.

Try this

Move $k$ in “Scale invariance”. The highlighted vector $k\,p$ slides along its ray, but $\left(\frac{kx}{kw}, \frac{ky}{kw}\right)$ stays the same.
Set $k$ to a negative value. The vector moves to the other side of the origin and is still the same point.
Set $k = 0$ . $(0, 0, 0)$ is not a point, which is why $k$ must be non-zero.
Check that $p_1 \cdot l_1 = 0$ in the values panel, then drag $p_1$ and see that it stays 0.
Drag $p_4$ until the lines are almost parallel. $q$ moves far away and its $w$ gets close to 0.
Press “Make l₂ parallel to l₁”. Now $w = 0$ exactly, and in the 3D view the intersection ray lies in the plane $w = 0$ .
Select $q$ as the vector to scale while the lines are parallel. Scaling keeps $w = 0$ : the point stays at infinity.
Drag $p_2$ onto $p_1$ . The cross product of two equal points is $(0, 0, 0)$ , so they do not define a line.