0

We consider the projection of three-dimensional projective space from a center $Z$ onto image plane $\pi$: \begin{align} X \longmapsto \alpha(X) = (Z \lor X) \cap \pi \end{align}

since $\pi$ is a plane and given by its homogeneous coordinates $u$, I'm assuming that $u$ is a normal vector of that plane - but is that true? If so: Does the length of that vector matter or is it sufficient that it is orthogonal on that plane and why does it make sense to intersect this vector with the line $(X \lor Z)$ in order to get the projected point on plane $\pi$?

I hope my questions are clear. If not please let me know.

Notes:

Here is the description from my lecture notes. Please let me know if you need more information:

We consider the projection of three-dimensional projective space from a center $Z$ onto image plane $\pi$:

\begin{align} X \longmapsto \alpha(X) = (Z \lor X) \cap \pi \end{align}

We assume that $Z$ is given by its homogeneous coordinate vector $z$ and that the plane is given by its homogeneous coordinates vector $u$. For an arbitrary point $X$ with coordinates $x$ the line $X \lor Z$ is decscribed by points $\lambda x + \mu z$. We determine the intersection of that line with the image plane, so we have to solve for $\lambda$,$\mu$ such that

\begin{align} u^T(\lambda x + \mu z) &= 0 \\ \lambda u^Tx + \mu u^Tz &= 0 \end{align} \begin{align} \Rightarrow \lambda &= -u^Tz \\ \Rightarrow \mu &= u^Tx \end{align}

Thus the image point is given by coordinates $(-u^Tz)x + (u^Tx)z$

\begin{align} \alpha (X) = (-(u^Tz) E + zu^T)x \end{align}

Here $-(u^Tz) E + zu^T$ is the matrix describing the projection - and the projection is thereby recognized as a projective mapping ($E$ is here the identity matrix).

Stefan Falk
  • 1,217
  • 1
  • 11
  • 24

1 Answers1

0

In projective three-space, a point $X$ described by a homogeneous vector $x$ lies on a plane $\pi$ described by a homogeneous coordinate vector $u$ if the dot product $\langle u,x\rangle$ is zero (or in the notation from your lecture notes, if $u^Tx=0$).

So if you think of $X$ as the vector $x=(x_1,x_2,x_3,1)^T$ then that plane $u=(u_1,u_2,u_3,u_4)^T$ corresponds to

$$\{(x_1,x_2,x_3)^T\in\mathbb R^3\mid u_1x_1+u_2x_2+u_3x_3=-u_4\}$$

for inhomogeneous (affine) points. Generalized to generic homogeneous point coordinates $x=(x_1,x_2,x_3,x_4)^T$, this becomes

$$\{(x_1,x_2,x_3,x_4)^T\in\mathbb{RP}^3\mid u_1x_1+u_2x_2+u_3x_3+u_4x_4=0\}$$

The first three coordinates of $u$ are the normal vector of the plane, not neccessarily of unit length. The forth coordinate encodes the distance of that plane from the origin. Since the whole vector is homogeneous, (non-zero) multiples of that vector represent the same plane. That corresponds to a multiplication of the equation above by a scalar factor, which doesn't affect its set of solutions. Just be sure to scale $u_4$ along with the other components.

In the ambient space $\mathbb R^4$ you can consider the whole vector $u$ as being orthogonal to the vectors $x$ of all incident points. So you might consider it a normal vector in $\mathbb R^4$ as well. The fact that the length does not matter does remain.

Your lecture notes already tell you that any point on $X\vee Z$ can be written as a linear combination $\lambda x+\mu z$ for suitable $\lambda,\mu\in\mathbb R$. So let's combine this, like your notes do, but in slightly different notation. You want to find $\lambda,\mu$ such that

$$0=\langle u,\lambda x+\mu z\rangle=\lambda\langle u,x\rangle + \mu\langle u,z\rangle$$

The choice of $\lambda,\mu$ is not unique, since scalar multiples of a solution are again a solution. The equation is homogeneous. One way to solve the equation without having to divide is by choosing

$$\lambda=-\langle u,z\rangle \qquad \mu=\langle u,x\rangle$$

Simply plug this into the equation above and you see that it works. So your projected point has homogeneous coordinates

$$\lambda x+\mu z = -\langle u,z\rangle x+\langle u,x\rangle z$$

MvG
  • 42,596