6
$\begingroup$

For calculations in 2D space, there exist a few useful equations to compute general geometry with the vector dot product . and the vector cross product x when working with homogeneous coordinates (remember even though we are working in 2D space the homogeneous representation of lines and point are 3D vectors):

1) The point where two lines cross x = l1 x l2.

2) The line between two points l = x1 x x2

3) A point lies on a line if x . l = 0

In 3D space a plane can be described by a normal vector n = [n1, n2, n3] and a point on the plane x = [x1, x2, x3] both in Euclidean coordinates.

In homogeneous coordinated the plane can be defined as p = [n1, n2, n3, -(n1*x1 + n2*x2 + n3*x3)].

Is there a short equation for finding the point where a line passes through a plane? Intuitively it feels as it should be x = l x p in homogeneous coordinates, but this computation does not exist, since there is no cross product in 4 dimensions.

At the moment I am only able to compute the intersection by defining the line with the equation l(t) = a + b(t), where a is a point on the line and b is the direction of the line in Euclidean coordinates.

For a = [a1, a2, a3], b = [b1, b2, b3] and the plane in question p = [p1, p2, p3, p4], the point of intersection x = [x1, x2, x3] can be obtained by substituting t in the line equation with t = -(p . [a1, a2, a3, 1])/(p . [b1, b2, b3, 0])

In summary, is there a elegant equation to find a point x where a line l crosses through a plane p preferably in homogeneous coordinates?

$\endgroup$
7
$\begingroup$

Definition

You are right, 3D points and planes are described with 4 homogeneous coordinates. A point at $\vec{r}$ is $P=(\vec{r};1)$ and a plane $W=(\vec{n};-d)=(\vec{n};-\vec{r}\cdot\vec{n})$ with normal $\vec{n}$ through point $\vec{r}$, or with minimum distance to origin $d$.

A line needs 6 coordinates (plücker coordinates) describing the direction and moment about the axis. A line along $\vec{e}$ through a point $\vec{r}$ has coordinates $L=[\vec{e};\vec{r}\times\vec{e}]$. Given a line $L=[\vec{l};\vec{m}]$, the direction is recovered by $\vec{e}=\frac{\vec{l}}{|\vec{l}|}$ and the position by $\vec{r} = \frac{\vec{l}\times\vec{m}}{|\vec{l}|^2}$

Now derive the point $P=(\vec{r};1)$ where line $L=[\vec{l};\vec{m}]$ meets plane $W=(\vec{w};\epsilon)$ as follows:

  • See that for the point to be on the plane you must have $\epsilon = - \vec{r}\cdot \vec{w}$
  • For the point to be on the line you must have $\vec{m} = \vec{r} \times \vec{l}$
  • Use the vector triple product to get $$ \vec{w} \times \vec{m} = \vec{w} \times \left( \vec{r} \times \vec{l} \right) = \vec{r} (\vec{w}\cdot\vec{l})-\vec{l}(\vec{w}\cdot\vec{r}) $$

$$ \vec{w} \times \vec{m} = \vec{r} (\vec{w}\cdot\vec{l}) - \vec{l}(-\epsilon) $$

$$ \vec{r} = \frac{\vec{w}\times\vec{m}-\epsilon \vec{l}}{\vec{w}\cdot\vec{l}} $$

  • Define the line-plane meet operator as

$$ \begin{aligned} P & = [W\times] L \\ \begin{pmatrix} \vec{p} \\ \delta \end{pmatrix} & = \begin{bmatrix} -\epsilon {\bf 1} & \vec{w}\times \\ \vec{w}^\top & 0 \end{bmatrix} \begin{pmatrix} \vec{l} \\ \vec{m} \end{pmatrix} \end{aligned}$$

where $\vec{w}\times = \begin{pmatrix}x\\y\\z\end{pmatrix} \times = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$ is the cross product matrix operator in 3×3 form.

  • The meet operator $[W\times]$ has dimensions 4×6 to work between lines and points.

Example

  • A plane normal to the $x$ axis located at $x=3$ has coordinates $W=(1,0,0;-3)$
  • A line through $y=2$ directed towards $\hat{i}+\hat{k}$ has coordinates $L=[1,0,1;2,0,-2]$
  • The meet operator is $$ [W\times] = \left[ \begin{array}{ccc|ccc} 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 & -1 \\ 0 & 0 & 3 & 0 & 1 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right]$$
  • The point where the line meets the plane is $P=[W\times]L$ $$P=\left[ \begin{array}{ccc|ccc} 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 & -1 \\ 0 & 0 & 3 & 0 & 1 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \begin{bmatrix}1\\0\\1\\ \hline 2 \\ 0 \\ -2 \end{bmatrix} =\begin{pmatrix}3\\2\\3\\ \hline 1 \end{pmatrix}$$
  • The point is located at $\vec{r} = (3,2,3)$
$\endgroup$
  • $\begingroup$ Thanks, this is a very easy to code answer, exactly what I was looking for. $\endgroup$ – Simon Streicher Sep 8 '14 at 12:49
  • $\begingroup$ Note that a line between points $P_1=(\vec{p}_1;\delta_1)$ and $P_2=(\vec{p}_2;\delta_2)$ is defined by $$L=\begin{pmatrix} \delta_1 \vec{p}_2 - \delta_2 \vec{p}_1 \\ \vec{p}_1 \times \vec{p}_2 \end{pmatrix}$$ $\endgroup$ – ja72 Sep 8 '14 at 12:55
  • $\begingroup$ Also in 2D lines are dual to points, but in 3D lines are dual to lines, and planes are dual to points. In 2D a line can be viewed as a 3D plane normal to the paper, or a line lying on the paper. A 2D point is either a 3D line coming out of the paper or a 3D point lying on the paper. It depends on the context. $\endgroup$ – ja72 Sep 10 '14 at 19:01
1
$\begingroup$

To answer your question about the line and plane in $\mathbb R^3$, think of it this way: You want to find the line of intersection of a $2$-dimensional subspace $L$ and a $3$-dimensional subspace $P$ in $\mathbb R^4$. Let $\mathbf n$ be the normal vector of $P$; you want the line in $L$ orthogonal to $\mathbf n$. This is really the same as your approach.

If you want something more like your approach one dimension lower, you need two vectors orthogonal to $L$, i.e., a basis for $L^\perp$, say $\mathbf v_1$ and $\mathbf v_2$. Then compute the cross product $\mathbf n\times \mathbf v_1\times \mathbf v_2$, and, voilà! You compute the cross product of $n$ vectors in $\mathbb R^{n+1}$ exactly as in the case of $n=2$.

$\endgroup$
  • $\begingroup$ Thanks for the quick reply :). So $L$ is decomposed into two orthogonal vectors $\mathbf a$ and $\mathbf b$ (would this translate to a point on the line and the direction of the line?). So when you apply $\mathbf v\times \mathbf a\times \mathbf b$ what does the answer represent? I know it is not the point of intersection on the surface. How do I manipulate this value to get to the point on the surface? $\endgroup$ – Simon Streicher May 29 '13 at 1:03
  • $\begingroup$ If you're talking about my second paragraph, you want two vectors perpendicular to the plane $L\subset\mathbb R^4$, so they need to be vectors in $\mathbb R^4$ perpendicular to your $\mathbf a$ and $\mathbf b$ (you can find them, for example, by finding the nullspace of the matrix with rows $\mathbf a$ and $\mathbf b$). Yes, when you compute the cross product $\mathbf n\times\mathbf v_1\times\mathbf v_2$, this is the direction vector of a line in $\mathbb R^4$ that gives the homogeneous coordinates of the point you seek. $\endgroup$ – Ted Shifrin May 29 '13 at 1:24
0
$\begingroup$

In a similar vein to ja72’s answer, you can use the Plücker matrix of a line to compute its intersection with a plane: The Plücker matrix of a line through points $\mathbf p$ and $\mathbf q$ is $L=\mathbf p\mathbf q^T-\mathbf q\mathbf p^T$, and its intersection with the plane $\mathbf\pi$ is simply $$L\pi = (\mathbf q^T\mathbf\pi)\mathbf p - (\mathbf p^T\mathbf\pi)\mathbf q.$$ If this product vanishes, then the line lies on $\mathbf\pi$. The Plücker matrix $L$ is in a certain sense the three-dimensional analog of the formula $\mathbf l = \mathbf p\times\mathbf q$ in 2-D.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.