# Plucker's line definition in PGA

How does a Plucker’s coordinate for line works in PGA? Is it e01,e02,e03 for direction and e23,e31,e12 to specify a point located on that line, or is it the wedge product (Point ^ line) ? Does the direction have to be normalized?

My goal was in fact to draw a line, aiming in a specific direction , passing
through a specific point (knowing the Plucker’s line was a way to do that)…Here is the solution, I was looking for:

var dir = (x,y,z)=>!(x1e1 + y1e2 + z1e3);
var point = (x,y,z)=>!(1e0 + x
1e1 + y1e2 + z1e3);
var Pt = point(-0.5,0.5,1)
var u = dir(1,1,1)
var L = (Pt&u)

That’s certainly a valid solution. More generally the PGA form of the Plucker condition is simply L \wedge L = 0. There are plenty of ways to generate valid Plucker lines:

• The wedge of any two planes a,b produces a Plucker line at their intersection : L = a \wedge b
• The join of any two Euclidean points p_1,p_2 produces a Plucker line through them : L = p_1 \vee p_2
• The join of a Euclidean point p_1 and an infinite point (direction) p_\infty also does : L = p_1 \vee p_\infty
• The inner product between a point p and a plane a produces a line through the point, \perp to the plane : L = a \cdot p

And probably some more I’m forgetting

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Thank you.
In fact, I am still looking for a proper understanding of the Plucker line coordinate. I was trying to trace a line from the formula itself(see below), by entering the direction components and the moment components. Even though the direction components seem obvious, I can’t grasp the geometric meaning of the moment components!!
/*
Pt = (a,b,c) // any Point on the line
u = (u,v,w) // line direction
L = (u1e23 - v1e13 + w*1e12) + ( (av-bu)*1e03 -(aw-cu)*1e02 +(bw-cv)*1e01) // tested ok
// L = (“direction”) + ( " moment" ) // derived from !(Pt & u) = (!Pt ^ !u)
*/

@charpelletier

Here’s a bit of geometry that may help to understand what the Pluecker coordinates mean. The main idea is to write a given line as the sum of a line through the origin and an ideal line. Each of these parts contains important information from which you can completely describe the original line.

Given a line L := a_{01}e_{01} + a_{02}e_{02}+a_{03}e_{03}+ a_{12}e_{12}+a_{31}e_{31}+ a_{23}e_{23}.
Write L as the sum of two lines L = L_o + L_\infty:
L_o := a_{12}e_{12}+a_{31}e_{31}+ a_{23}e_{23} and
L_\infty = a_{01}e_{01} + a_{02}e_{02}+a_{03}e_{03}

Set-up

• The “Pluecker condition” that a bivector L is a line is a_{01}a_{23} + a_{02} a_{31} + a_{03}a_{12} = 0.
• For what follows it’s useful to define \{E_1,E_2,E_3\} := \{e_{032},e_{013},e_{021}\}, the x-, y-, and z- coordinate directions (ideal points).
• Also recall that the orthogonal complement \mathbf{X}^\perp := \mathbf{XI} where \mathbf{I} is the unit pseudoscalar.

Exercise:

1. L_o is a line through the origin.
2. The ideal point of L_o (and L) is \mathbf{V} := a_{23}E_1 + a_{31}E_2+ a_{12}E_3. [Notice the reversal of order in the triple of Pluecker coordinates!]
3. L_\infty is an ideal line that passes through \mathbf{V} hence meets L. This is equivalent to the Pluecker condition on the coordinates as can be seen by computing L_\infty \vee \mathbf{V}.
4. Since L_o and L_\infty have a common point V, any linear combination aL_o + bL_\infty will also pass through V. The set of such lines is called the line pencil spanned by L_o and L_\infty. In particular L = L_o + L_\infty belongs to this pencil (for a=1,b=1).
5. L is obtained from L_o by translating it parallel to itself in the plane L\wedge L_\infty, a distance equal to \|L\|_\infty. (As we learn in euclidean PGA, this translation can be thought of as “rotation” in the plane L\wedge L_\infty around the ideal point \mathbf{V}.)
6. The ideal point \mathbf{T} := a_{01}E_1 + a_{02}E_2 + a_{03}E_3 is the “pole” of the ideal line L_\infty (i. e., is at right angles to all the points of L_\infty).
7. \mathbf{T} = (L_\infty \vee \mathbf{P})^\perp for any euclidean normalized euclidean point \mathbf{P}.
8. The Pluecker condition, in older texts, is expressed by the condition that \mathbf{T} and \mathbf{V} are perpendicular vectors, i. e., a_{01}a_{23} + a_{02} a_{31} + a_{03}a_{12} = 0.
9. The direction of the translation L_o+L_\infty is (L \vee \mathbf{T})^\perp, the direction normal to the plane spanned by L and \mathbf{T}.
10. Carry out the above analysis for the line L = e_{03} - d E_{31}.

Remark: This leads us to a nice generalization of a well-known fact from ordinary vector algebra:
For a euclidean point P and an ideal point V , P \rightarrow P + V translates P by V. The distance moved equals \|V\|_\infty.
Our new result is:
For a euclidean line L and an ideal line M (satisfying L \wedge M= 0), L \rightarrow L + M translates L by M. The distance moved equals \|M\|_\infty.

Exercise: deduce a similar result for translating euclidean planes.

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The following sketch attempts to picture the decomposition L = L_o + L_\infty described in my previous post.

The right view is the more traditional one, but in perspective so \mathbf{V} is visible.

The left view focus on the plane at infinity, picturing it as a very big sphere. It allowed me to include the vector \mathbf{T}, the “pole” of L_\infty. The reader/viewer should imagine that the sphere continues to grow until it is flat.

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Thank you, that is very helpful.
I am right to say that when L is obtained from L=a L_o + b L_\infty, then all L’s(pink lines) are part of the same line pencil (located on plane: L_o ^ L_\infty) , but if we translate L_o to L, the way shown below (red lines), then L becomes part of another line pencil? I still struggle on the way to translate L_o to the exact same L with a rotation (L2 green).

//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
var a23=1, //u direction through the origine
a31=1, //v
a12=0 //w
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
var V = a231e032+a311e013+a12*1e021 // The ideal point of Lo
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
var Lo = (1e123 & V) // Line: Pt & dir
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
var sqrt = motor => (Math.sign(motor.s) + motor).Normalized;
var motor = (line,angle_or_distance)=>Math.E**(angle_or_distance/2 * line);
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
var here = point(1,0,1)
var Trans = ()=>sqrt(1e123 * here)
var Rotat = motor(Lo,here/2) //----------<<<<<--------------
var L = ~Trans * Lo * Trans
var L2 = ~Rotat * Lo * Rotat
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++