In a previous post, I talked about the image points and image lines. In this post, I will talk about how to interpret the image geometry which is different from the geometry we learned before.

• Point of view I: good, easy

Let an image point with coordinates $(x,y)$. A line passing the point is

$ax+by+c=0$

Hence we denote $m=[x,y,1]^T$ as an image point and $l=[a,b,c]^T$ as an image line. Then we know:
a) a point is on a line iff $l^Tm=0$
b) the line passing two points is $l=m_1\times m_2$
c) the intersection point of two lines is $m\sim l_1 \times l_2$

Remark: This point of view is quite easy. No need to care what happens outside the image plane.

• Point of view II: a 3D interseption

Image point: The formula of the normalized image plane is $z=1$. So the 3D coordinates of a point $(x,y)$ on the image plane is $[x,y,1]^T$.
Image Line: A line in the image plane can be treated as the intersection line between a plane $\pi$ passing through the origin and plane $z=1$. Note any plane $\pi$ passing through the origin is uniquely determined by its norm vector $l$. Hence any vector

$l\Rightarrow \pi \Rightarrow$ intersect with plane $z=1\Rightarrow$ an image line

Any points $m$ on the image line must have $l^Tm=0$ since $l$ is orthogonal to all vectors in the plane $\pi$

Remark: here we interpret them in 3D camera reference frame.