A Very Brief Introduction to Reflections in 2D Geometric Algebra, and their Use in Solving "Construction" Problems
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This document introduces reflections in 2D geometric algebra and their application to solving tangency problems. It provides formulas for inner and outer products, explains the reflection of vectors and geometric products, and demonstrates the use of reflections to construct circles tangent to given circles with a specific point. The methods discussed build upon previously established techniques in geometric algebra and are illustrated with various geometric interpretations.
![A Very Brief Introduction to Reflections in 2D
Geometric Algebra, and their Use in Solving
“Construction” Problems
Jim Smith
QueLaMateNoTeMate.webs.com
email: nitac14b@yahoo.com
June 24, 2016
Abstract
This document is intended to be a convenient collection of explana-
tions and techniques given elsewhere ([1]-[3]) in the course of solving tan-
gency problems via Geometric Algebra.
1](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-1-320.jpg)
![Geometric-Algebra Formulas
for Plane (2D) Geometry
The Geometric Product, and Relations Derived from It
For any two vectors a and b,
a · b = b · a
b ∧ a = −a ∧ b
ab = a · b + a ∧ b
ba = b · a + b ∧ a = a · b − a ∧ b
ab + ba = 2a · b
ab − ba = 2a ∧ b
ab = 2a · b + ba
ab = 2a ∧ b − ba
Definitions of Inner and Outer Products (Macdonald A. 2010 p. 101.)
The inner product
The inner product of a j-vector A and a k-vector B is
A · B = AB k−j. Note that if j>k, then the inner product doesn’t exist.
However, in such a case B · A = BA j−k does exist.
The outer product
The outer product of a j-vector A and a k-vector B is
A ∧ B = AB k+j.
Relations Involving the Outer Product and the Unit Bivector, i.
For any two vectors a and b,
ia = −ai
a ∧ b = [(ai) · b] i = − [a · (bi)] i = −b ∧ a
Equality of Multivectors
For any two multivectors M and N,
M = N if and only if for all k, M k = N k.
Formulas Derived from Projections of Vectors
and Equality of Multivectors
Any two vectors a and b can be written in the form of “Fourier expansions”
with respect to a third vector, v:
a = (a · ˆv) ˆv + [a · (ˆvi)] ˆvi and b = (b · ˆv) ˆv + [b · (ˆvi)] ˆvi.
Using these expansions,
ab = {(a · ˆv) ˆv + [a · (ˆvi)] ˆvi} {(b · ˆv) ˆv + [b · (ˆvi)] ˆvi}
Equating the scalar parts of both sides of that equation,
2](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-2-320.jpg)
![a · b = [a · ˆv] [b · ˆv] + [a · (ˆvi)] [b · (ˆvi)], and
a ∧ b = {[a · ˆv] [b · (ˆvi)] − [a · (ˆvi)] [b · (ˆvi)]} i.
Also, a2
= [a · ˆv]
2
+ [a · (ˆvi)]
2
, and b2
= [b · ˆv]
2
+ [b · (ˆvi)]
2
.
Reflections of Vectors, Geometric Products, and Rotation operators
For any vector a, the product ˆvaˆv is the reflection of a with respect to the
direction ˆv.
For any two vectors a and b, ˆvabˆv = ba, and vabv = v2
ba. Therefore,
ˆveθiˆv = e−θi
, and veθi
v = v2
e−θi
.
A useful relationship that is valid only in plane geometry: abc = cba.
Here is a brief proof:
abc = {a · b + a ∧ b} c
= {a · b + [(ai) · b] i} c
= (a · b) c + [(ai) · b] ic
= c (a · b) − c [(ai) · b] i
= c (a · b) + c [a · (bi)] i
= c (b · a) + c [(bi) · a] i
= c {b · a + [(bi) · a] i}
= c {b · a + b ∧ a}
= cba.
3](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-3-320.jpg)
![1 Introduction
This document discusses reflections of vectors and of geometrical products of two
vectors, in two-dimensional Geometric Algebra (GA). It then uses reflections to
solve a simple tangency problem.
2 Reflections in 2D GA
2.1 Reflections of a single vector
For any two vectors ˆu and v, the product ˆuvˆu is
ˆuvˆu = {2ˆu ∧ v + vˆu} ˆu (2.1)
= v + 2 [(ˆui) · v] iˆu (2.2)
= v − 2 [v · (ˆui)] ˆui, (2.3)
which evaluates to the reflection of the reflection of v with respect to ˆu (Fig.
2.1).
Figure 2.1: Geometric interpretation of ˆuvˆu, showing why it evaluates to the
reflection of v with respect to ˆu.
We also note that because u = |u| ˆu,
uvu = u2
(ˆuvˆu) = u2
v − 2 [v · (ui)] ui. (2.4)](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-4-320.jpg)
![2.2 Reflections of a bivector, and of a geometric product
of two vectors
The product ˆuvwˆu is
ˆuvwˆu = ˆu (v · w + v ∧ w) ˆu
= ˆu (v · w) ˆu + ˆu (v ∧ w) ˆu
= ˆu2
(v · w) + ˆu [(vi) · w] iˆu
= v · w + ˆu [−v · (wi)] (−ˆui)
= v · w + ˆu2
[(wi) · v] i
= w · v + w ∧ v
= wv.
In other words, the reflection of the geometric product vw is wv, and does
not depend on the direction of the vector with respect to which it is reflected.
We saw that the scalar part of vw was unaffected by the reflection, but the
bivector part was reversed.
Further to that point, the reflection of geometric product of v and w is
equal to the geometric product of the two vectors’ reflections:
ˆuvwˆu = ˆuv (ˆuˆu) wˆu
= (ˆuvˆu) (ˆuwˆu) .
That observation provides a geometric interpretation (Fig. 2.2) of why reflecting
a bivector changes its sign: the direction of the turn from v to w reverses.
Figure 2.2: Geometric interpretation of ˆuvwˆu, showing why it evaluates to
the reflection of v with respect to ˆu. Note that ˆuvwˆu = ˆuv (ˆuˆu) wˆu =
(ˆuvˆu) (ˆuwˆu).
5](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-5-320.jpg)
![3 Use of reflections to solve a simple tangency
problem
The problem that we will solve is
“Given two coplanar circles, with a point Q on one of them, con-
struct the circles that are tangent to both of the given circles, with
point Q as one of the points of tangency” (Fig. 3.1).
Figure 3.1: Diagram for our problem: “Given two coplanar circles, with a point
Q on one of them, construct the circles that are tangent to both of the given
circles, with point Q as one of the points of tangency.”
Several solutions that use rotations are given by [1], but here we will use
reflections. The triangle TQC3 is isosceles, so ˆt is the reflection of ˆw with
respect to the mediatrix of segment QT. In order to make use of that fact,
we need to express the direction of that mediatrix as a vector written in terms
of known quantities. We can do so by constructing another isosceles triangle
(C1SC3) that has the same mediatrix (Fig. 3.2).
The vector from C1 to S is c2+(r2 − r1) ˆw, so the direction of the mediatrix
of QT is the vector [c2 + (r2 − r1) ˆw] i. The unit vector with that direction is
[c2 + (r2 − r1) ˆw] i
c2 + (r2 − r1) ˆw
. Therefore, to express ˆt as the reflection of ˆw with respect
6](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-6-320.jpg)
![Figure 3.2: Adding segment C1S to Fig. 3.1 to produce a new isosceles triangle
with the same mediatrix as QT.
to the mediatrix, we write
ˆt =
[c2 + (r2 − r1) ˆw] i
c2 + (r2 − r1) ˆw
[ ˆw]
[c2 + (r2 − r1) ˆw] i
c2 + (r2 − r1) ˆw
=
{[c2 + (r2 − r1) ˆw] i} [ ˆw] {[c2 + (r2 − r1) ˆw] i}
[c2 + (r2 − r1) ˆw]
2
=
[c2 + (r2 − r1) ˆw] [ ˆw] [c2 + (r2 − r1) ˆw] ii
[c2 + (r2 − r1) ˆw]
2 ,
from which
t = r1
ˆt = −r1
[c2 + (r2 − r1) ˆw] [ ˆw] [c2 + (r2 − r1) ˆw]
[c2 + (r2 − r1) ˆw]
2 . (3.1)
Interestingly, the geometric interpretation of that result is that ˆt and − ˆw
are reflections of each other with respect to the vector c2 + (r2 − r1) ˆw. After
expanding and rearranging the numerator and denominator of (3.1), then using
w = r2 ˆw, we obtain
t = r1
c2
2
− (r2 − r1)
2
w − 2 [c2 · w + r2 (r2 − r1)] c2
r2c2
2 + 2 (r2 − r1) c2 · w + r2 (r2 − r1)
2
. (3.2)
References
[1] J. Smith, “Rotations of Vectors Via Geometric Algebra: Explanation, and
Usage in Solving Classic Geometric ”Construction” Problems” (Version
of 11 February 2016). Available at http://vixra.org/abs/1605.0232 .
7](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-7-320.jpg)
![[2] “Solution of the Special Case ”CLP” of the Problem of Apollo-
nius via Vector Rotations using Geometric Algebra”. Available at
http://vixra.org/abs/1605.0314.
[3] “The Problem of Apollonius as an Opportunity for Teaching Students to
Use Reflections and Rotations to Solve Geometry Problems via Geometric
(Clifford) Algebra”. Available at http://vixra.org/abs/1605.0233.
8](https://image.slidesharecdn.com/simplereflectiondocument-160624114535/85/A-Very-Brief-Introduction-to-Reflections-in-2D-Geometric-Algebra-and-their-Use-in-Solving-Construction-Problems-8-320.jpg)
A Very Brief Introduction to Reflections in 2D Geometric Algebra, and their Use in Solving "Construction" Problems
- 1. A Very Brief Introduction to Reflections in 2D Geometric Algebra, and their Use in Solving “Construction” Problems Jim Smith QueLaMateNoTeMate.webs.com email: nitac14b@yahoo.com June 24, 2016 Abstract This document is intended to be a convenient collection of explana- tions and techniques given elsewhere ([1]-[3]) in the course of solving tan- gency problems via Geometric Algebra. 1
- 2. Geometric-Algebra Formulas for Plane (2D) Geometry The Geometric Product, and Relations Derived from It For any two vectors a and b, a · b = b · a b ∧ a = −a ∧ b ab = a · b + a ∧ b ba = b · a + b ∧ a = a · b − a ∧ b ab + ba = 2a · b ab − ba = 2a ∧ b ab = 2a · b + ba ab = 2a ∧ b − ba Definitions of Inner and Outer Products (Macdonald A. 2010 p. 101.) The inner product The inner product of a j-vector A and a k-vector B is A · B = AB k−j. Note that if j>k, then the inner product doesn’t exist. However, in such a case B · A = BA j−k does exist. The outer product The outer product of a j-vector A and a k-vector B is A ∧ B = AB k+j. Relations Involving the Outer Product and the Unit Bivector, i. For any two vectors a and b, ia = −ai a ∧ b = [(ai) · b] i = − [a · (bi)] i = −b ∧ a Equality of Multivectors For any two multivectors M and N, M = N if and only if for all k, M k = N k. Formulas Derived from Projections of Vectors and Equality of Multivectors Any two vectors a and b can be written in the form of “Fourier expansions” with respect to a third vector, v: a = (a · ˆv) ˆv + [a · (ˆvi)] ˆvi and b = (b · ˆv) ˆv + [b · (ˆvi)] ˆvi. Using these expansions, ab = {(a · ˆv) ˆv + [a · (ˆvi)] ˆvi} {(b · ˆv) ˆv + [b · (ˆvi)] ˆvi} Equating the scalar parts of both sides of that equation, 2
- 3. a · b = [a · ˆv] [b · ˆv] + [a · (ˆvi)] [b · (ˆvi)], and a ∧ b = {[a · ˆv] [b · (ˆvi)] − [a · (ˆvi)] [b · (ˆvi)]} i. Also, a2 = [a · ˆv] 2 + [a · (ˆvi)] 2 , and b2 = [b · ˆv] 2 + [b · (ˆvi)] 2 . Reflections of Vectors, Geometric Products, and Rotation operators For any vector a, the product ˆvaˆv is the reflection of a with respect to the direction ˆv. For any two vectors a and b, ˆvabˆv = ba, and vabv = v2 ba. Therefore, ˆveθiˆv = e−θi , and veθi v = v2 e−θi . A useful relationship that is valid only in plane geometry: abc = cba. Here is a brief proof: abc = {a · b + a ∧ b} c = {a · b + [(ai) · b] i} c = (a · b) c + [(ai) · b] ic = c (a · b) − c [(ai) · b] i = c (a · b) + c [a · (bi)] i = c (b · a) + c [(bi) · a] i = c {b · a + [(bi) · a] i} = c {b · a + b ∧ a} = cba. 3
- 4. 1 Introduction This document discusses reflections of vectors and of geometrical products of two vectors, in two-dimensional Geometric Algebra (GA). It then uses reflections to solve a simple tangency problem. 2 Reflections in 2D GA 2.1 Reflections of a single vector For any two vectors ˆu and v, the product ˆuvˆu is ˆuvˆu = {2ˆu ∧ v + vˆu} ˆu (2.1) = v + 2 [(ˆui) · v] iˆu (2.2) = v − 2 [v · (ˆui)] ˆui, (2.3) which evaluates to the reflection of the reflection of v with respect to ˆu (Fig. 2.1). Figure 2.1: Geometric interpretation of ˆuvˆu, showing why it evaluates to the reflection of v with respect to ˆu. We also note that because u = |u| ˆu, uvu = u2 (ˆuvˆu) = u2 v − 2 [v · (ui)] ui. (2.4)
- 5. 2.2 Reflections of a bivector, and of a geometric product of two vectors The product ˆuvwˆu is ˆuvwˆu = ˆu (v · w + v ∧ w) ˆu = ˆu (v · w) ˆu + ˆu (v ∧ w) ˆu = ˆu2 (v · w) + ˆu [(vi) · w] iˆu = v · w + ˆu [−v · (wi)] (−ˆui) = v · w + ˆu2 [(wi) · v] i = w · v + w ∧ v = wv. In other words, the reflection of the geometric product vw is wv, and does not depend on the direction of the vector with respect to which it is reflected. We saw that the scalar part of vw was unaffected by the reflection, but the bivector part was reversed. Further to that point, the reflection of geometric product of v and w is equal to the geometric product of the two vectors’ reflections: ˆuvwˆu = ˆuv (ˆuˆu) wˆu = (ˆuvˆu) (ˆuwˆu) . That observation provides a geometric interpretation (Fig. 2.2) of why reflecting a bivector changes its sign: the direction of the turn from v to w reverses. Figure 2.2: Geometric interpretation of ˆuvwˆu, showing why it evaluates to the reflection of v with respect to ˆu. Note that ˆuvwˆu = ˆuv (ˆuˆu) wˆu = (ˆuvˆu) (ˆuwˆu). 5
- 6. 3 Use of reflections to solve a simple tangency problem The problem that we will solve is “Given two coplanar circles, with a point Q on one of them, con- struct the circles that are tangent to both of the given circles, with point Q as one of the points of tangency” (Fig. 3.1). Figure 3.1: Diagram for our problem: “Given two coplanar circles, with a point Q on one of them, construct the circles that are tangent to both of the given circles, with point Q as one of the points of tangency.” Several solutions that use rotations are given by [1], but here we will use reflections. The triangle TQC3 is isosceles, so ˆt is the reflection of ˆw with respect to the mediatrix of segment QT. In order to make use of that fact, we need to express the direction of that mediatrix as a vector written in terms of known quantities. We can do so by constructing another isosceles triangle (C1SC3) that has the same mediatrix (Fig. 3.2). The vector from C1 to S is c2+(r2 − r1) ˆw, so the direction of the mediatrix of QT is the vector [c2 + (r2 − r1) ˆw] i. The unit vector with that direction is [c2 + (r2 − r1) ˆw] i c2 + (r2 − r1) ˆw . Therefore, to express ˆt as the reflection of ˆw with respect 6
- 7. Figure 3.2: Adding segment C1S to Fig. 3.1 to produce a new isosceles triangle with the same mediatrix as QT. to the mediatrix, we write ˆt = [c2 + (r2 − r1) ˆw] i c2 + (r2 − r1) ˆw [ ˆw] [c2 + (r2 − r1) ˆw] i c2 + (r2 − r1) ˆw = {[c2 + (r2 − r1) ˆw] i} [ ˆw] {[c2 + (r2 − r1) ˆw] i} [c2 + (r2 − r1) ˆw] 2 = [c2 + (r2 − r1) ˆw] [ ˆw] [c2 + (r2 − r1) ˆw] ii [c2 + (r2 − r1) ˆw] 2 , from which t = r1 ˆt = −r1 [c2 + (r2 − r1) ˆw] [ ˆw] [c2 + (r2 − r1) ˆw] [c2 + (r2 − r1) ˆw] 2 . (3.1) Interestingly, the geometric interpretation of that result is that ˆt and − ˆw are reflections of each other with respect to the vector c2 + (r2 − r1) ˆw. After expanding and rearranging the numerator and denominator of (3.1), then using w = r2 ˆw, we obtain t = r1 c2 2 − (r2 − r1) 2 w − 2 [c2 · w + r2 (r2 − r1)] c2 r2c2 2 + 2 (r2 − r1) c2 · w + r2 (r2 − r1) 2 . (3.2) References [1] J. Smith, “Rotations of Vectors Via Geometric Algebra: Explanation, and Usage in Solving Classic Geometric ”Construction” Problems” (Version of 11 February 2016). Available at http://vixra.org/abs/1605.0232 . 7
- 8. [2] “Solution of the Special Case ”CLP” of the Problem of Apollo- nius via Vector Rotations using Geometric Algebra”. Available at http://vixra.org/abs/1605.0314. [3] “The Problem of Apollonius as an Opportunity for Teaching Students to Use Reflections and Rotations to Solve Geometry Problems via Geometric (Clifford) Algebra”. Available at http://vixra.org/abs/1605.0233. 8