Given a point and a line in homogeneous coordinates, what is the coordinate of the projection of the point on the line?

Question

Knowing $M$ and $L$ in homogeneous coordinates, how can we determine the coordinates of the point $P$?

Answer 1

I assume the standard $z=1$ embedding, i.e. points homogenized as $(x,y)\mapsto[x:y:1]$ and line $[a:b:c]$ represents $ax+by+c=0$.

The coordinates $(a,b)$ represent the direction orthogonal to the line. $[a:b:0]$ is the point at infinity orthogonal to line $L$. You can connect that to $M$ to get the orthogonal line passing through $M$, and intersect that line with $L$ to get the orthogonal projection. Joining points and intersecting lines can both be expressed using the cross product.

$$P = ((F\cdot L)\times M)\times L\qquad\text{with }F=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$

(If you are dealing with Cayley-Klein metrics, $F$ would be the dual fundamental conic of the Euclidean plane, which is crucial for angle measurements, including the right angle used here.)

Answer 2

it still depends on what kind of projection is required; Given the constraints in the original post, it looks like an orthographic parallel projection. Both central projection and parallel projection (oblique or orthographic) are computation-friendly redefined as a common algebraic formulation equation 3.1 in Householder elemenetary matrix form this arXiv.org article: Unified Framework of Elementary Geometric Transformation

Suppose the homogeneous coordinates of the point $(x,y,1)^T$ and the line $(a,b,c)^T$; then an orthographic parallel projection onto the line $(a,b,c)^T$ uniquely determined by $(a,b,c)^T$ and its normal direction per definition 3.6, definition 3.7 and equation 3.1 (line No. 2 of table 1 in page 9) for 2D projective transformations in Unified Framework of Elementary Geometric Transformation has the following form:

$$\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]-\dfrac{\left[ \begin{array}{c} a \\ b \\ 0 \\ \end{array} \right]\cdot \left[a,b,c\right]}{\left[a,b,0\right]\cdot \left[ \begin{array}{c} a \\ b \\ c \\ \end{array} \right]} =\dfrac{1}{a^2+b^2}\left[ \begin{array}{ccc} b^2 & -a b & -a c \\ -a b & a^2 & -b c \\ 0 & 0 & a^2+b^2 \\ \end{array} \right]$$

Then the orthographic parallel projection of $(x,y,1)^T$ onto the line $(a,b,c)^T$ is: $$\dfrac{1}{a^2+b^2}\left[ \begin{array}{ccc} b^2 & -a b & -a c \\ -a b & a^2 & -b c \\ 0 & 0 & a^2+b^2 \\ \end{array} \right]\cdot \left[ \begin{array}{c} x \\ y \\ 1 \\ \end{array} \right]= \dfrac{1}{a^2+b^2}\left[ \begin{array}{c} b^2 x-a b y-a c \\ a^2 y-a b x-b c \\ a^2+b^2 \\ \end{array} \right]$$