Geometric constraint and Line: Difference between pages

From Rigid Geometric Algebra
(Difference between pages)
Jump to navigation Jump to search
No edit summary
 
No edit summary
 
Line 1: Line 1:
An element $$\mathbf x$$ of a geometric algebra possesses the ''geometric property'' if and only if the [[geometric product]] between $$\mathbf x$$ and its own reverse is a scalar, which is given by the [[dot product]], and the [[geometric antiproduct]] between $$\mathbf x$$ and its own antireverse is an antiscalar, which is given by the [[antidot product]]. That is,
[[Image:line.svg|400px|thumb|right|'''Figure 1.''' A line is the intersection of a 4D bivector with the 3D subspace where $$w = 1$$.]]
In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, a ''line'' $$\boldsymbol l$$ is a bivector having the general form


:$$\mathbf x \mathbin{\unicode{x27D1}} \mathbf{\tilde x} = \mathbf x \mathbin{\unicode{x25CF}} \mathbf x$$
:$$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$ .


and
The components $$(l_{vx}, l_{vy}, l_{vz})$$ correspond to the line's direction, and the components $$(l_{mx}, l_{my}, l_{mz})$$ correspond to the line's moment. (These are equivalent to the six Plücker coordinates of a line.) To satisfy the [[geometric constraint]], the components of $$\boldsymbol l$$ must satisfy the equation


:$$\mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{x}}} = \mathbf x \mathbin{\unicode{x25CB}} \mathbf x$$ .
:$$l_{vx} l_{mx} + l_{vy} l_{my} + l_{vz} l_{mz} = 0$$ ,


The set of all elements possessing the geometric property is closed under both the [[geometric product]] and [[geometric antiproduct]].
which means that, when regarded as vectors, the direction and moment of a line are perpendicular.


The following table lists the requirements that must be satisfied for the main types in the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$ to possess the geometric property. Points and planes do not have any requirements—they all possess the geometric property.
The [[bulk]] of a line is given by its $$mx$$, $$my$$, and $$mz$$ coordinates, and the [[weight]] of a line is given by its $$vx$$, $$vy$$, and $$vz$$ coordinates. A line is [[unitized]] when $$l_{vx}^2 + l_{vy}^2 + l_{vz}^2 = 1$$. The [[attitude]] of a line is the vector $$l_{vx} \mathbf e_1 + l_{vy} \mathbf e_2 + l_{vz} \mathbf e_3$$ corresponding to its direction.


{| class="wikitable"
When used as an operator in the sandwich with the [[geometric antiproduct]], a line is a specific kind of [[motor]] that performs a 180-degree rotation about itself.
! Type !! Definition !! Requirement
 
|-
<br clear="right" />
| style="padding: 12px;" | [[Magnitude]]
== Lines at Infinity ==
| style="padding: 12px;" | $$\mathbf z = x \mathbf 1 + y {\large\unicode{x1d7d9}}$$
 
| style="padding: 12px;" | $$xy = 0$$
[[Image:line_infinity.svg|400px|thumb|right|'''Figure 2.''' A line at infinity consists of all points at infinity in directions perpendicular to the moment $$\mathbf m$$.]]
|-
If the weight of a line is zero (i.e., its $$vx$$, $$vy$$, and $$vz$$ coordinates are all zero), then the line is contained in the horizon infinitely far away in all directions perpendicular to its moment $$\mathbf m = (l_{mx}, l_{my}, l_{mz})$$, regarded as a vector, as shown in Figure 2. Such a line cannot be unitized, but it can be normalized by dividing by its [[bulk norm]].
| style="padding: 12px;" | [[Point]]
 
| style="padding: 12px;" | $$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$
When the moment $$\mathbf m$$ is regarded as a bivector, a line at infinity can be thought of as all directions $$\mathbf v$$ parallel to the moment, which satisfy $$\mathbf m \wedge \mathbf v = 0$$.
| style="padding: 12px;" | &mdash;
 
|-
<br clear="right" />
| style="padding: 12px;" | [[Line]]
== Skew Lines ==
| style="padding: 12px;" | $$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$
 
| style="padding: 12px;" | $$l_{vx} l_{mx} + l_{vy} l_{my} + l_{vz} l_{mz} = 0$$
[[Image:skew_lines.svg|400px|thumb|right|'''Figure 3.''' The line $$\mathbf j$$ connecting skew lines.]]
|-
Given two skew lines $$\boldsymbol l$$ and $$\mathbf k$$, as shown in Figure 3, a third line $$\mathbf j$$ that contains a point on each of the lines $$\boldsymbol l$$ and $$\mathbf k$$ is given by the axis of the [[motor]] $$\boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{k}}}$$. The line $$\mathbf j$$ can be found by first calculating the line
| style="padding: 12px;" | [[Plane]]
 
| style="padding: 12px;" | $$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$
:$$\mathbf i = [\boldsymbol l, \mathbf k]^{\Large\unicode{x27C7}}_- = (l_{vy} k_{vz} - l_{vz} k_{vy})\mathbf e_{41} + (l_{vz} k_{vx} - l_{vx} k_{vz})\mathbf e_{42} + (l_{vx} k_{vy} - l_{vy} k_{vx})\mathbf e_{43} + (l_{vy} k_{mz} - l_{vz} k_{my} + l_{my} k_{vz} - l_{mz} k_{vy})\mathbf e_{23} + (l_{vz} k_{mx} - l_{vx} k_{mz} + l_{mz} k_{vx} - l_{mx} k_{vz})\mathbf e_{31} + (l_{vx} k_{my} - l_{vy} k_{mx} + l_{mx} k_{vy} - l_{my} k_{vx})\mathbf e_{12}$$
| style="padding: 12px;" | &mdash;
 
|-
and then orthogonalizing its direction and moment to obtain
| style="padding: 12px;" | [[Motor]]
 
| style="padding: 12px;" | $$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$
:$$\mathbf j = i_{vx} \mathbf e_{41} + i_{vy} \mathbf e_{42} + i_{vz} \mathbf e_{43} + (i_{mx} - s i_{vx})\mathbf e_{23} + (i_{my} - s i_{vy})\mathbf e_{31} + (i_{mz} - s i_{vz})\mathbf e_{12}$$ ,
| style="padding: 12px;" | $$Q_{vx} Q_{mx} + Q_{vy} Q_{my} + Q_{vz} Q_{mz} + Q_{vw} Q_{mw} = 0$$
 
|-
where
| style="padding: 12px;" | [[Flector]]
 
| style="padding: 12px;" | $$\mathbf F = F_{px} \mathbf e_1 + F_{py} \mathbf e_2 + F_{pz} \mathbf e_3 + F_{pw} \mathbf e_4 + F_{gx} \mathbf e_{423} + F_{gy} \mathbf e_{431} + F_{gz} \mathbf e_{412} + F_{gw} \mathbf e_{321}$$
:$$s = \dfrac{i_{vx}i_{mx} + i_{vy}i_{my} + i_{vz}i_{mz}}{i_{vx}^2 + i_{vy}^2 + i_{vz}^2}$$ .
| style="padding: 12px;" | $$F_{px} F_{gx} + F_{py} F_{gy} + F_{pz} F_{gz} + F_{pw} F_{gw} = 0$$
 
|}
If $$l_{vx}k_{vx} + l_{vy}k_{vy} + l_{vz}k_{vz} = 0$$, meaning that the directions of the two lines are perpendicular, then $$\mathbf j = \mathbf i$$.
 
The direction of $$\mathbf j$$ is perpendicular to the directions of $$\boldsymbol l$$ and $$\mathbf k$$, and it contains the closest points of approach between $$\boldsymbol l$$ and $$\mathbf k$$. The points themselves can then be found by calculating $$(\mathbf j \wedge \operatorname{att}(\boldsymbol l)) \vee \mathbf k$$ and $$(\mathbf j \wedge \operatorname{att}(\mathbf k)) \vee \boldsymbol l$$, where $$\operatorname{att}$$ is the [[attitude]] function.
 
<br clear="right" />


== See Also ==
== See Also ==


* [[Geometric norm]]
* [[Point]]
* [[Plane]]

Latest revision as of 01:01, 9 February 2024

Figure 1. A line is the intersection of a 4D bivector with the 3D subspace where $$w = 1$$.

In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, a line $$\boldsymbol l$$ is a bivector having the general form

$$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$ .

The components $$(l_{vx}, l_{vy}, l_{vz})$$ correspond to the line's direction, and the components $$(l_{mx}, l_{my}, l_{mz})$$ correspond to the line's moment. (These are equivalent to the six Plücker coordinates of a line.) To satisfy the geometric constraint, the components of $$\boldsymbol l$$ must satisfy the equation

$$l_{vx} l_{mx} + l_{vy} l_{my} + l_{vz} l_{mz} = 0$$ ,

which means that, when regarded as vectors, the direction and moment of a line are perpendicular.

The bulk of a line is given by its $$mx$$, $$my$$, and $$mz$$ coordinates, and the weight of a line is given by its $$vx$$, $$vy$$, and $$vz$$ coordinates. A line is unitized when $$l_{vx}^2 + l_{vy}^2 + l_{vz}^2 = 1$$. The attitude of a line is the vector $$l_{vx} \mathbf e_1 + l_{vy} \mathbf e_2 + l_{vz} \mathbf e_3$$ corresponding to its direction.

When used as an operator in the sandwich with the geometric antiproduct, a line is a specific kind of motor that performs a 180-degree rotation about itself.


Lines at Infinity

Figure 2. A line at infinity consists of all points at infinity in directions perpendicular to the moment $$\mathbf m$$.

If the weight of a line is zero (i.e., its $$vx$$, $$vy$$, and $$vz$$ coordinates are all zero), then the line is contained in the horizon infinitely far away in all directions perpendicular to its moment $$\mathbf m = (l_{mx}, l_{my}, l_{mz})$$, regarded as a vector, as shown in Figure 2. Such a line cannot be unitized, but it can be normalized by dividing by its bulk norm.

When the moment $$\mathbf m$$ is regarded as a bivector, a line at infinity can be thought of as all directions $$\mathbf v$$ parallel to the moment, which satisfy $$\mathbf m \wedge \mathbf v = 0$$.


Skew Lines

Figure 3. The line $$\mathbf j$$ connecting skew lines.

Given two skew lines $$\boldsymbol l$$ and $$\mathbf k$$, as shown in Figure 3, a third line $$\mathbf j$$ that contains a point on each of the lines $$\boldsymbol l$$ and $$\mathbf k$$ is given by the axis of the motor $$\boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{k}}}$$. The line $$\mathbf j$$ can be found by first calculating the line

$$\mathbf i = [\boldsymbol l, \mathbf k]^{\Large\unicode{x27C7}}_- = (l_{vy} k_{vz} - l_{vz} k_{vy})\mathbf e_{41} + (l_{vz} k_{vx} - l_{vx} k_{vz})\mathbf e_{42} + (l_{vx} k_{vy} - l_{vy} k_{vx})\mathbf e_{43} + (l_{vy} k_{mz} - l_{vz} k_{my} + l_{my} k_{vz} - l_{mz} k_{vy})\mathbf e_{23} + (l_{vz} k_{mx} - l_{vx} k_{mz} + l_{mz} k_{vx} - l_{mx} k_{vz})\mathbf e_{31} + (l_{vx} k_{my} - l_{vy} k_{mx} + l_{mx} k_{vy} - l_{my} k_{vx})\mathbf e_{12}$$

and then orthogonalizing its direction and moment to obtain

$$\mathbf j = i_{vx} \mathbf e_{41} + i_{vy} \mathbf e_{42} + i_{vz} \mathbf e_{43} + (i_{mx} - s i_{vx})\mathbf e_{23} + (i_{my} - s i_{vy})\mathbf e_{31} + (i_{mz} - s i_{vz})\mathbf e_{12}$$ ,

where

$$s = \dfrac{i_{vx}i_{mx} + i_{vy}i_{my} + i_{vz}i_{mz}}{i_{vx}^2 + i_{vy}^2 + i_{vz}^2}$$ .

If $$l_{vx}k_{vx} + l_{vy}k_{vy} + l_{vz}k_{vz} = 0$$, meaning that the directions of the two lines are perpendicular, then $$\mathbf j = \mathbf i$$.

The direction of $$\mathbf j$$ is perpendicular to the directions of $$\boldsymbol l$$ and $$\mathbf k$$, and it contains the closest points of approach between $$\boldsymbol l$$ and $$\mathbf k$$. The points themselves can then be found by calculating $$(\mathbf j \wedge \operatorname{att}(\boldsymbol l)) \vee \mathbf k$$ and $$(\mathbf j \wedge \operatorname{att}(\mathbf k)) \vee \boldsymbol l$$, where $$\operatorname{att}$$ is the attitude function.


See Also