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Projective geometry - intersections

Impo

New member
Apr 2, 2013
17
Hi,

Sorry if my post would be in the wrong section, but I didn't know where to put it as there's no topic called 'projective geometry'.

I'm working on the following problem. Suppose we're working in the projective space [tex]\mathbb{P}^4[/tex] where we have given the Plucker coordinates of a plane [tex] \alpha: (a_0:a_1:a_2:a_3)[/tex] and the Plucker coordinates of a line [tex]l = (d:m)=(d_1:d_2:d_3:m_1:m_2:m_3)[/tex]. The question is, what are the coordinates of the intersection point?

I can represent the plane [tex]\alpha[/tex] analytically as [tex]a_0x_0+a_1x_1+a_2x_2+a_3x_3=0[/tex] which can be useful to work with, but I also want an analytically expression of the line, which I cannot find. I did some research and I found on wikipedia that the coordinates of the intersection point are given by: [tex](x_0:x)=(a\cdot d: a \times m - a_0d)[/tex] where [tex]x=(x_1:x_2:x_3)[/tex] and [tex]a=(a_1:a_2:a_3)[/tex]. Honestly, I don't know how they came to that answer as I have no analytical expression for the line [tex]l[/tex].

How do I have to approach this problem?

Many thanks,
Impo
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Hi,

Sorry if my post would be in the wrong section, but I didn't know where to put it as there's no topic called 'projective geometry'.

I'm working on the following problem. Suppose we're working in the projective space [tex]\mathbb{P}^4[/tex] where we have given the Plucker coordinates of a plane [tex] \alpha: (a_0:a_1:a_2:a_3)[/tex] and the Plucker coordinates of a line [tex]l = (d:m)=(d_1:d_2:d_3:m_1:m_2:m_3)[/tex]. The question is, what are the coordinates of the intersection point?

I can represent the plane [tex]\alpha[/tex] analytically as [tex]a_0x_0+a_1x_1+a_2x_2+a_3x_3=0[/tex] which can be useful to work with, but I also want an analytically expression of the line, which I cannot find. I did some research and I found on wikipedia that the coordinates of the intersection point are given by: [tex](x_0:x)=(a\cdot d: a \times m - a_0d)[/tex] where [tex]x=(x_1:x_2:x_3)[/tex] and [tex]a=(a_1:a_2:a_3)[/tex]. Honestly, I don't know how they came to that answer as I have no analytical expression for the line [tex]l[/tex].

How do I have to approach this problem?

Many thanks,
Impo
Welcome to MHB, Impo! :)

A line in $\mathbb R^3$ is of the form $\mathbf x = \mathbf s + \lambda \mathbf d$.
We can write this in $\mathbb{P}^3$ as $(1:\dfrac {\mathbf s} {s_0} + \lambda \dfrac {\mathbf d} {d_0} )$.
Generalizing to projective geometry, we get $(x_0:\mathbf x) = (s_0d_0:d_0 \mathbf s + \lambda s_0 \mathbf d)$.


In addition we have the relationship $\mathbf m = \mathbf s \times \mathbf d$.


Furthermore we have the cross product identity $\mathbf a \times (\mathbf b \times \mathbf c) = \mathbf b (\mathbf a \cdot \mathbf c) - \mathbf c (\mathbf a \cdot \mathbf b)$.


Combine those?
 

Impo

New member
Apr 2, 2013
17
Hi I like Serena,

thanks for the answer! Unfortunately, I don't see the relationship between the equation of a line you gave and the one with the Plucker coordinates I gave. Also, where do we use the equation of the plane?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Hi I like Serena,

thanks for the answer! Unfortunately, I don't see the relationship between the equation of a line you gave and the one with the Plucker coordinates I gave. Also, where do we use the equation of the plane?
Substitute the equation for the line that I gave into the equation of the plane.
Then try to rewrite to a form using $\mathbf m$.

Note that the Plücker form of the line is $\mathbf m \cdot \mathbf d = (\mathbf s \times \mathbf d) \cdot \mathbf d = 0$
 

Impo

New member
Apr 2, 2013
17
If I substitute the equation of the line into the equation of the plane then I get:
$a_0s_0+(a\cdot s)d_0+(a\cdot d)\lambda s_0$
$\Rightarrow a_0s_0+(a \cdot s)d_0-(a\cdot d)(-\lambda)s_0$

with the given identity I get
$a_0s_0+a \times (d_0 \times s_0(-\lambda))$

I don't know how to go further ...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
If I substitute the equation of the line into the equation of the plane then I get:
$a_0s_0+(a\cdot s)d_0+(a\cdot d)\lambda s_0$
$\Rightarrow a_0s_0+(a \cdot s)d_0-(a\cdot d)(-\lambda)s_0$

with the given identity I get
$a_0s_0+a \times (d_0 \times s_0(-\lambda))$

I don't know how to go further ...
You seem to be mixing up vectors and scalars.
In particular in your last equation you are using scalars in a cross product.
But a cross product is only defined for vectors.

Let's write what you did a little cleaner.

We have the plane $\alpha$:
$$a_0x_0 + \mathbf a \cdot \mathbf x = 0 \qquad (1)$$
And the line $l$:
$$(x_0:\mathbf x) = (s_0d_0 : d_0\mathbf s + \lambda s_0 \mathbf d) \qquad (2)$$

Substituting (2) in (1) gives:
$$a_0(s_0d_0) + \mathbf a \cdot (d_0\mathbf s + \lambda s_0 \mathbf d) = 0 \qquad$$
$$a_0s_0d_0 + d_0 (\mathbf a \cdot \mathbf s) + \lambda s_0 (\mathbf a \cdot \mathbf d) = 0$$
$$\lambda = - \frac {a_0s_0d_0 + d_0 (\mathbf a \cdot \mathbf s)} {s_0 (\mathbf a \cdot \mathbf d)} \qquad (3)$$
Perhaps you can continue from there.
 

Impo

New member
Apr 2, 2013
17
Yeah, I have to substitute the $\lambda$ into the equation of the projective line.

We have $(x_0:x)=(s_0d_0:d_0s+\lambda s_0d)$
$x = d_0s + \displaystyle \left[\frac{-a_0s_0d_0-d_0(a\cdot s)}{a\cdot d}\right]d$
$=\displaystyle \frac{d_0\left[s(a\cdot d) - a_0s_0 d - (a\cdot s)d\right]}{a\cdot d}$
$=\displaystyle \frac{d_0\left[a \times (s \times d)-a_0s_0d\right]}{a\cdot d}$
$=\displaystyle \frac{d_0 \left[a \times m - a_0s_0d \right]}{a\cdot d}$

Hence
$(x_0:x) = \left(s_0d_0: \displaystyle \frac{d_0 \left(a \times m - a_0s_0d \right)}{a\cdot d} \right) = \left(a\cdot d: \displaystyle \frac{a \times m - a_0s_0d}{s_0}\right)$

It seems there's an $s_0$ missing in the first term of the numerator, I guess that is a mistake of me.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Yeah, I have to substitute the $\lambda$ into the equation of the projective line.

We have $(x_0:x)=(s_0d_0:d_0s+\lambda s_0d)$
$x = d_0s + \displaystyle \left[\frac{-a_0s_0d_0-d_0(a\cdot s)}{a\cdot d}\right]d$
$=\displaystyle \frac{d_0\left[s(a\cdot d) - a_0s_0 d - (a\cdot s)d\right]}{a\cdot d}$
$=\displaystyle \frac{d_0\left[a \times (s \times d)-a_0s_0d\right]}{a\cdot d}$
$=\displaystyle \frac{d_0 \left[a \times m - a_0s_0d \right]}{a\cdot d}$

Hence
$(x_0:x) = \left(s_0d_0: \displaystyle \frac{d_0 \left(a \times m - a_0s_0d \right)}{a\cdot d} \right) = \left(a\cdot d: \displaystyle \frac{a \times m - a_0s_0d}{s_0}\right)$

It seems there's an $s_0$ missing in the first term of the numerator, I guess that is a mistake of me.
Good! You found it. ;)

And yes, you dropped an $s_0$ in the first step.
 

Impo

New member
Apr 2, 2013
17
Okay, thanks for all the help!
 

Impo

New member
Apr 2, 2013
17
I have another question about this subject.

Suppose now we're working with homogenous coordinates. How can I find the intersection point of the plane $\alpha$ and a line $l$ by using the exterior product $\wedge$?