# Projective geometry - intersections

#### Impo

##### New member
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 $$\mathbb{P}^4$$ where we have given the Plucker coordinates of a plane $$\alpha: (a_0:a_1:a_2:a_3)$$ and the Plucker coordinates of a line $$l = (d:m)=(d_1:d_2:d_3:m_1:m_2:m_3)$$. The question is, what are the coordinates of the intersection point?

I can represent the plane $$\alpha$$ analytically as $$a_0x_0+a_1x_1+a_2x_2+a_3x_3=0$$ 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: $$(x_0:x)=(a\cdot d: a \times m - a_0d)$$ where $$x=(x_1:x_2:x_3)$$ and $$a=(a_1:a_2:a_3)$$. Honestly, I don't know how they came to that answer as I have no analytical expression for the line $$l$$.

How do I have to approach this problem?

Many thanks,
Impo

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $$\mathbb{P}^4$$ where we have given the Plucker coordinates of a plane $$\alpha: (a_0:a_1:a_2:a_3)$$ and the Plucker coordinates of a line $$l = (d:m)=(d_1:d_2:d_3:m_1:m_2:m_3)$$. The question is, what are the coordinates of the intersection point?

I can represent the plane $$\alpha$$ analytically as $$a_0x_0+a_1x_1+a_2x_2+a_3x_3=0$$ 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: $$(x_0:x)=(a\cdot d: a \times m - a_0d)$$ where $$x=(x_1:x_2:x_3)$$ and $$a=(a_1:a_2:a_3)$$. Honestly, I don't know how they came to that answer as I have no analytical expression for the line $$l$$.

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
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
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
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
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
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
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
Okay, thanks for all the help!

#### Impo

##### New member
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 $\wedge$?