Three Solutions of the LLP Limiting Case of the Problem of Apollonius via Geometric Algebra, Using Reflections and Rotations
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This document presents three geometric algebra solutions to the limiting case of the problem of Apollonius, which involves constructing circles tangent to two intersecting lines passing through a given point. It reviews the mathematical techniques of reflections and rotations in 2D geometry and applies them to derive the location of the solution circle's center. By exploring multiple methods, the document aims to enhance understanding of geometric algebra applications in solving geometric problems.
![Three Solutions of the LLP Limiting Case of the
Problem of Apollonius via Geometric Algebra,
Using Reflections and Rotations
Jim Smith
QueLaMateNoTeMate.webs.com
email: nitac14b@yahoo.com
July 26, 2016
Abstract
This document adds to the collection of solved problems presented
in [1]-[4]. After reviewing, briefly, how reflections and rotations can be
expressed and manipulated via GA, it solves the LLP limiting case of the
Problem of Apollonius in three ways.
“Given a point P between two intersecting lines, construct the circles
that are tangent to both of the lines, and pass through P”.
1](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-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/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-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/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-3-320.jpg)
![1 Introduction
This document adds to the collection of solved problems presented in [1]-[4].
After reviewing, briefly, how reflections and rotations can be expressed and ma-
nipulated via GA, it solves the LLP limiting case of the Problem of Apollonius
in three ways.
2 A brief review of reflections and rotations in
2D GA
2.1 Review of reflections
2.1.1 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/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-4-320.jpg)
![2.1.2 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/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-5-320.jpg)
![2.2 Review of rotations
One of the most-important rotations—for our purposes—is the one that is pro-
duced when a vector is multiplied by the unit bivector, i, for the plane: vi
is v’s 90-degree counter-clockwise rotation, while iv is v’s 90-degree clockwise
rotation.
Every geometric product bc is a rotation operator, whether we use it as
such or not:
bc = b c eθi
.
where θ is the angle of rotation from b to c. From that equation, we obtain the
identity
eθi
=
bc
b c
=
b
ˆb
c
ˆc
= ˆbˆc .
Note that a [ˆuˆv] evaluates to the rotation of a by the angle from ˆu to ˆv, while
[ˆuˆv] a evaluates to the rotation of a by the angle from ˆv to ˆu.
A useful corollary of the foregoing is that any product of an odd number
of vectors evaluates to a vector, while the product of an odd number of vectors
evaluates to the sum of a scalar and a bivector.
An interesting example of a rotation is the product ˆuˆvabˆvˆu. Writing that
product as (ˆuˆva) (bˆvˆu), and recalling that any geometric product of two vectors
acts as a rotation operator in 2D (2.2), we can see why ˆuˆvabˆvˆu = ab (Fig.
2.3).
Figure 2.3: Left-multiplying a by the geometric product ˆuˆv rotates a by an
angle equal (in sign as well as magnitude) to that from ˆv to ˆu. Right-multiplying
b by the geometric product ˆvˆu rotates b by that same angle. The orientations
of a and b with respect to each other are the same after the rotation, and the
magnitudes of a and b are unaffected. Therefore, the geometric products ab
and (ˆuˆva) (bˆvˆu) are equal.
We can also write ˆuˆvabˆvˆu as ˆuˆv [ab] ˆvˆu, giving it the form of a rotation of
the geometric product ab. Considered in this way, the result ˆuˆvabˆvˆu = ab is
a special case of an important theorem: rotations preserve geometric products
[5].
6](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-6-320.jpg)
![3 Use of reflections and rotations to solve the
LLP limiting case of the Problem of Apollo-
nius
The Problem of Apollonius is described in detail in [3]. Its LLP limiting case,
which we will solve here, is
“Given a point P between two intersecting lines, construct the circles
that are tangent to both of the lines, and pass through P” (Fig. 3.1).
Figure 3.1: Statement of the problem: “Given a point P between two intersect-
ing lines, construct the circles that are tangent to both of the lines, and pass
through P.”
Because [1]-[4] deal at length with the “translation” of such problems into
GA terms, we will go directly to the solution process here. We’ll present three
solution methods, in each of which we will identify the location of the center of
the solution circle.
The three methods’ algebraic manipulations have much in common, but
their differences are instructive. Please note that the third method makes,
perhaps, the best use of GA.
3.1 First solution method
We’ll begin by identifying key elements that can be expressed in GA terms (Fig.
3.2):
The directions of the given lines are captured as ˆa and ˆb. As our origin,
we’ve chosen the point of intersection of those lines. With respect to that
origin, the center (C) of the solution circle lies along the direction ˆa + ˆb. For
our convenience, we’ve defined a unit vector (ˆu) in that direction. Therefore,
we can write the location of C as a scalar multiple of ˆu:
c = λˆu.
7](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-7-320.jpg)
![Figure 3.2: One of several possible ways to express key elements of the problem
in terms of reflections and rotations, via GA. See the text for explanation.
We then use that result to express the location of the point of tangency T:
t = λ (ˆu · ˆa) ˆa.
The point P is P’s reflection with respect to ˆu. The vector from the origin to
P is ˆupˆu (Section 2.1.1). Because P is P’s reflection with respect to a line
through C, P belongs to the solution circle.
As discussed in [1], the angles θ are equal. Therefore, we can write (see
2.2)
p − t
p − t
p − t
p − t
=
p − c
p − c
(−ˆu) . (3.1)
Using ideas presented in [1]-[4], we can transform that result into
ˆu [p − c] [p − t] [p − t] = − p − t p − t p − c ,
the right side of which is a scalar. Therefore, the according to the postulate for
the Equality of Multivectors,
ˆu [p − c] [p − t] [p − t] 2 = − p − t p − t p − c 2 = 0.
Now, we make the substitutions c = λˆu, p = ˆupˆu, and t = λ (ˆu · ˆa) ˆa to
obtain
ˆu [p − λˆu] [p − λ (ˆu · ˆa) ˆa] [ˆupˆu − λ (ˆu · ˆa) ˆa] 2 = 0. (3.2)
Expanding,
ˆup2
ˆupˆu − ˆup2
λ (ˆu · ˆa) ˆa − ˆupλ (ˆu · ˆa) ˆaˆupˆu
+ˆupλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa − ˆuλˆupˆupˆu + ˆuλˆupλ (ˆu · ˆa) ˆa
+ˆuλˆuλ (ˆu · ˆa) ˆaˆupˆu − ˆuλˆuλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa 2 = 0.
8](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-8-320.jpg)
![To simplify that equation, we note (see 2.3 )that
ˆupλ (ˆu · ˆa) ˆaˆupˆu = ˆup [λ (ˆu · ˆa) ˆaˆu] pˆu
= λp2
(ˆu · ˆa) ˆaˆu.
We also note that
ˆuλˆuλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa = λ3
(ˆu · ˆa)
2
,
which is a scalar. Therefore, its bivector part is zero.
Making use of such observations to find the bivector part of the left-hand
side, we obtain (after simplifying) the following quadratic equation for λ:
λ2
(ˆu · ˆa)
2
(ˆu ∧ p) − λ[2 (ˆu · p) (ˆu ∧ p)] + p2
(ˆu ∧ p) = 0.
The product ˆu ∧ p is zero only if p is a scalar multiple of ˆu; that is, if the given
point P lies along the centerline of the angle formed by the two given lines.
We’ll omit that possibility from consideration, and multiply both sides of the
preceding equation by (ˆu ∧ p)
−1
to arrive at
λ2
(ˆu · ˆa)
2
− λ[2ˆu · p] + p2
= 0. (3.3)
The two solutions to that quadratic are
λ =
ˆu · p ± (ˆu · p)
2
− p2 (ˆu · ˆa)
2
(ˆu · ˆa)
2 . (3.4)
To write that solution in terms of ˆa, ˆb, and p, we use the relations
ˆu =
ˆa + ˆb
ˆa + ˆb
=
ˆa + ˆb
2 1 + ˆa · ˆb
1
2
.
to transform (3.3) into
λ2 1 + ˆa · ˆb
2
− λ
√
2 ˆa + ˆb · p
1 + ˆa · ˆb
1
2
+ p2
= 0, (3.5)
the solutions of which are
λ =
√
2
ˆa + ˆb · p ± ˆa + ˆb · p
2
− p2
1 + ˆa · ˆb
2
1
2
1 + ˆa · ˆb
3
2
. (3.6)
The two solution circles are shown in Fig. 3.3:
9](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-9-320.jpg)
![Figure 3.3: The two solution circles.
3.2 Second solution method
This second solution uses only rotations, instead of a combination of rotations
and reflections. It is similar to many solutions given in [1]. Again, we’ll begin
by identifying key elements that can be expressed in GA terms (Fig. 3.4):
Figure 3.4: A second way to express key elements of the problem in terms of
GA. See the text for explanation.
We’ve made use of the fact that the two tangents drawn two a circle from
any external point are equal in length, to write that t = λ (ˆu · ˆa) ˆb. Now, we
use theorems about angles and circles to write
φ =
1
2
[ψ − (2π − ψ)] ; ∴
ψ
2
−
φ
2
=
π
2
.
Next, we use that relationship between ψ and φ to obtain an equation for λ.
We’ll keep in mind that ˆu · ˆb = ˆu · ˆa = 1 + ˆa · ˆb, ˆu 2
= 2 1 + ˆa · ˆb , and
10](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-10-320.jpg)
![ˆap + pˆa = 2ˆa · p:
e
ψ
2
−
φ
2
i
= e
π
2
i
e
ψ
2
i
e
−
φ
2
i
=ˆuˆa
= i
e
ψ
2
i
ˆuˆa 0 = i 0
t − p
t − p
t − p
t − p
ˆuˆa 0 = 0
[t − p] [t − p] ˆuˆa 0 = 0
[λ (ˆu · ˆa) ˆa − p] λ (ˆu · ˆa) ˆb − p ˆuˆa 0 = 0
λ
1 + ˆbˆa
ˆa + ˆb
− p λ
1 + ˆaˆb
ˆa + ˆb
− p
1 + ˆbˆa
ˆa + ˆb
0 = 0.
From here, we would expand the left-hand side, thereby arriving at (3.5).
3.3 Third solution method
Of the three solutions presented in this document, this one probably makes
the best use of GA. Again, we’ll begin by identifying key elements that can be
expressed in GA terms (Fig. 3.5). Note that the angle of rotation from iˆa to
ˆu: is half the angle ψ that was shown in Fig. 3.4:
Figure 3.5: A third way, based upon Fig. 3.4, to express key elements of the
problem in terms of GA. See the text for explanation.
As noted in 2.2, the geometric product of two unit vectors is a rotation
operator. In this solution, we’ll use the equality of the angles ψ/2 to write that
11](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-11-320.jpg)
![the product
t − p
t − p
t − p
t − p
is an operator that rotates iˆa into ˆu:
[iˆa]
t − p
t − p
t − p
t − p
= ˆu.
Now, we transform that equation into a product of four vectors that is equal to
a bivector:
[iˆa]
t − p
t − p
t − p
t − p
= ˆu
−i [iˆa]
t − p
t − p
t − p
t − p
= −iˆu
−ii [ˆa]
t − p
t − p
t − p
t − p
= −iˆu
[ˆa]
t − p
t − p
t − p
t − p
= −iˆu
ˆa
t − p
t − p
t − p
t − p
= ˆui
ˆuˆa
t − p
t − p
t − p
t − p
= i
ˆuˆa [t − p] [t − p] = t − p t − p i,
from which
ˆuˆa [t − p] [t − p] 0 = 0. (3.7)
The simplification of ˆuˆapˆb 0
uses the fact that ˆbˆapˆb is a
reflection of ˆap:
ˆuˆapˆb 0 =
ˆa + ˆb
ˆa + ˆb
ˆapˆb 0
=
ˆaˆapˆb + ˆbˆapˆb
ˆa + ˆb
0
=
pˆb + pˆa
ˆa + ˆb
0
=
ˆa + ˆb
ˆa + ˆb
· p
= ˆu · p.
From 3.7, we could obtain a quadratic in λ in many ways, one of which is
the following, which uses ˆu = ˆa + ˆb / ˆa + ˆb ; ˆu · ˆa = ˆu · ˆb; t = λ (ˆu · ˆa) ˆa:
t = λ (ˆu · ˆa) ˆb; and (ˆu · ˆa)
2
= 1 + ˆa · ˆb /2 :
ˆuˆa [t − p] [t − p] 0 = 0
ˆuˆa [λ (ˆu · ˆa) ˆa − p] λ (ˆu · ˆa) ˆb − p 0 = 0.
Expanding and simplifying,
λ2
(ˆu · ˆa)
2
ˆuˆb − λ (ˆu · ˆa) ˆup + ˆuˆapˆb + ˆuˆap = 0
λ2
(ˆu · ˆa)
2
ˆu · ˆa
=ˆu·ˆb
−λ {(ˆu · ˆa) [2ˆu · p]} + (ˆu · ˆa) p = 0.
Eliminating the factor ˆu · ˆa,
λ2
(ˆu · ˆa)
2
− λ [2ˆu · p] + (ˆu · ˆa) p = 0,
which is identical to 3.3.
12](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-12-320.jpg)
![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 .
[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.
[4] “A Very Brief Introduction to Reflections in 2D Geometric Alge-
bra, and their Use in Solving ‘Construction’ Problems”. Available at
http://vixra.org/abs/1606.0253.
[5] A. Macdonald, Linear and Geometric Algebra (First Edition) p. 126, Cre-
ateSpace Independent Publishing Platform (Lexington, 2012).
13](https://image.slidesharecdn.com/apolloniusllpviareflections-160714031424/85/Three-Solutions-of-the-LLP-Limiting-Case-of-the-Problem-of-Apollonius-via-Geometric-Algebra-Using-Reflections-and-Rotations-13-320.jpg)
Three Solutions of the LLP Limiting Case of the Problem of Apollonius via Geometric Algebra, Using Reflections and Rotations
- 1. Three Solutions of the LLP Limiting Case of the Problem of Apollonius via Geometric Algebra, Using Reflections and Rotations Jim Smith QueLaMateNoTeMate.webs.com email: nitac14b@yahoo.com July 26, 2016 Abstract This document adds to the collection of solved problems presented in [1]-[4]. After reviewing, briefly, how reflections and rotations can be expressed and manipulated via GA, it solves the LLP limiting case of the Problem of Apollonius in three ways. “Given a point P between two intersecting lines, construct the circles that are tangent to both of the lines, and pass through P”. 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 adds to the collection of solved problems presented in [1]-[4]. After reviewing, briefly, how reflections and rotations can be expressed and ma- nipulated via GA, it solves the LLP limiting case of the Problem of Apollonius in three ways. 2 A brief review of reflections and rotations in 2D GA 2.1 Review of reflections 2.1.1 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. 2.1.2 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. 2.2 Review of rotations One of the most-important rotations—for our purposes—is the one that is pro- duced when a vector is multiplied by the unit bivector, i, for the plane: vi is v’s 90-degree counter-clockwise rotation, while iv is v’s 90-degree clockwise rotation. Every geometric product bc is a rotation operator, whether we use it as such or not: bc = b c eθi . where θ is the angle of rotation from b to c. From that equation, we obtain the identity eθi = bc b c = b ˆb c ˆc = ˆbˆc . Note that a [ˆuˆv] evaluates to the rotation of a by the angle from ˆu to ˆv, while [ˆuˆv] a evaluates to the rotation of a by the angle from ˆv to ˆu. A useful corollary of the foregoing is that any product of an odd number of vectors evaluates to a vector, while the product of an odd number of vectors evaluates to the sum of a scalar and a bivector. An interesting example of a rotation is the product ˆuˆvabˆvˆu. Writing that product as (ˆuˆva) (bˆvˆu), and recalling that any geometric product of two vectors acts as a rotation operator in 2D (2.2), we can see why ˆuˆvabˆvˆu = ab (Fig. 2.3). Figure 2.3: Left-multiplying a by the geometric product ˆuˆv rotates a by an angle equal (in sign as well as magnitude) to that from ˆv to ˆu. Right-multiplying b by the geometric product ˆvˆu rotates b by that same angle. The orientations of a and b with respect to each other are the same after the rotation, and the magnitudes of a and b are unaffected. Therefore, the geometric products ab and (ˆuˆva) (bˆvˆu) are equal. We can also write ˆuˆvabˆvˆu as ˆuˆv [ab] ˆvˆu, giving it the form of a rotation of the geometric product ab. Considered in this way, the result ˆuˆvabˆvˆu = ab is a special case of an important theorem: rotations preserve geometric products [5]. 6
- 7. 3 Use of reflections and rotations to solve the LLP limiting case of the Problem of Apollo- nius The Problem of Apollonius is described in detail in [3]. Its LLP limiting case, which we will solve here, is “Given a point P between two intersecting lines, construct the circles that are tangent to both of the lines, and pass through P” (Fig. 3.1). Figure 3.1: Statement of the problem: “Given a point P between two intersect- ing lines, construct the circles that are tangent to both of the lines, and pass through P.” Because [1]-[4] deal at length with the “translation” of such problems into GA terms, we will go directly to the solution process here. We’ll present three solution methods, in each of which we will identify the location of the center of the solution circle. The three methods’ algebraic manipulations have much in common, but their differences are instructive. Please note that the third method makes, perhaps, the best use of GA. 3.1 First solution method We’ll begin by identifying key elements that can be expressed in GA terms (Fig. 3.2): The directions of the given lines are captured as ˆa and ˆb. As our origin, we’ve chosen the point of intersection of those lines. With respect to that origin, the center (C) of the solution circle lies along the direction ˆa + ˆb. For our convenience, we’ve defined a unit vector (ˆu) in that direction. Therefore, we can write the location of C as a scalar multiple of ˆu: c = λˆu. 7
- 8. Figure 3.2: One of several possible ways to express key elements of the problem in terms of reflections and rotations, via GA. See the text for explanation. We then use that result to express the location of the point of tangency T: t = λ (ˆu · ˆa) ˆa. The point P is P’s reflection with respect to ˆu. The vector from the origin to P is ˆupˆu (Section 2.1.1). Because P is P’s reflection with respect to a line through C, P belongs to the solution circle. As discussed in [1], the angles θ are equal. Therefore, we can write (see 2.2) p − t p − t p − t p − t = p − c p − c (−ˆu) . (3.1) Using ideas presented in [1]-[4], we can transform that result into ˆu [p − c] [p − t] [p − t] = − p − t p − t p − c , the right side of which is a scalar. Therefore, the according to the postulate for the Equality of Multivectors, ˆu [p − c] [p − t] [p − t] 2 = − p − t p − t p − c 2 = 0. Now, we make the substitutions c = λˆu, p = ˆupˆu, and t = λ (ˆu · ˆa) ˆa to obtain ˆu [p − λˆu] [p − λ (ˆu · ˆa) ˆa] [ˆupˆu − λ (ˆu · ˆa) ˆa] 2 = 0. (3.2) Expanding, ˆup2 ˆupˆu − ˆup2 λ (ˆu · ˆa) ˆa − ˆupλ (ˆu · ˆa) ˆaˆupˆu +ˆupλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa − ˆuλˆupˆupˆu + ˆuλˆupλ (ˆu · ˆa) ˆa +ˆuλˆuλ (ˆu · ˆa) ˆaˆupˆu − ˆuλˆuλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa 2 = 0. 8
- 9. To simplify that equation, we note (see 2.3 )that ˆupλ (ˆu · ˆa) ˆaˆupˆu = ˆup [λ (ˆu · ˆa) ˆaˆu] pˆu = λp2 (ˆu · ˆa) ˆaˆu. We also note that ˆuλˆuλ (ˆu · ˆa) ˆaλ (ˆu · ˆa) ˆa = λ3 (ˆu · ˆa) 2 , which is a scalar. Therefore, its bivector part is zero. Making use of such observations to find the bivector part of the left-hand side, we obtain (after simplifying) the following quadratic equation for λ: λ2 (ˆu · ˆa) 2 (ˆu ∧ p) − λ[2 (ˆu · p) (ˆu ∧ p)] + p2 (ˆu ∧ p) = 0. The product ˆu ∧ p is zero only if p is a scalar multiple of ˆu; that is, if the given point P lies along the centerline of the angle formed by the two given lines. We’ll omit that possibility from consideration, and multiply both sides of the preceding equation by (ˆu ∧ p) −1 to arrive at λ2 (ˆu · ˆa) 2 − λ[2ˆu · p] + p2 = 0. (3.3) The two solutions to that quadratic are λ = ˆu · p ± (ˆu · p) 2 − p2 (ˆu · ˆa) 2 (ˆu · ˆa) 2 . (3.4) To write that solution in terms of ˆa, ˆb, and p, we use the relations ˆu = ˆa + ˆb ˆa + ˆb = ˆa + ˆb 2 1 + ˆa · ˆb 1 2 . to transform (3.3) into λ2 1 + ˆa · ˆb 2 − λ √ 2 ˆa + ˆb · p 1 + ˆa · ˆb 1 2 + p2 = 0, (3.5) the solutions of which are λ = √ 2 ˆa + ˆb · p ± ˆa + ˆb · p 2 − p2 1 + ˆa · ˆb 2 1 2 1 + ˆa · ˆb 3 2 . (3.6) The two solution circles are shown in Fig. 3.3: 9
- 10. Figure 3.3: The two solution circles. 3.2 Second solution method This second solution uses only rotations, instead of a combination of rotations and reflections. It is similar to many solutions given in [1]. Again, we’ll begin by identifying key elements that can be expressed in GA terms (Fig. 3.4): Figure 3.4: A second way to express key elements of the problem in terms of GA. See the text for explanation. We’ve made use of the fact that the two tangents drawn two a circle from any external point are equal in length, to write that t = λ (ˆu · ˆa) ˆb. Now, we use theorems about angles and circles to write φ = 1 2 [ψ − (2π − ψ)] ; ∴ ψ 2 − φ 2 = π 2 . Next, we use that relationship between ψ and φ to obtain an equation for λ. We’ll keep in mind that ˆu · ˆb = ˆu · ˆa = 1 + ˆa · ˆb, ˆu 2 = 2 1 + ˆa · ˆb , and 10
- 11. ˆap + pˆa = 2ˆa · p: e ψ 2 − φ 2 i = e π 2 i e ψ 2 i e − φ 2 i =ˆuˆa = i e ψ 2 i ˆuˆa 0 = i 0 t − p t − p t − p t − p ˆuˆa 0 = 0 [t − p] [t − p] ˆuˆa 0 = 0 [λ (ˆu · ˆa) ˆa − p] λ (ˆu · ˆa) ˆb − p ˆuˆa 0 = 0 λ 1 + ˆbˆa ˆa + ˆb − p λ 1 + ˆaˆb ˆa + ˆb − p 1 + ˆbˆa ˆa + ˆb 0 = 0. From here, we would expand the left-hand side, thereby arriving at (3.5). 3.3 Third solution method Of the three solutions presented in this document, this one probably makes the best use of GA. Again, we’ll begin by identifying key elements that can be expressed in GA terms (Fig. 3.5). Note that the angle of rotation from iˆa to ˆu: is half the angle ψ that was shown in Fig. 3.4: Figure 3.5: A third way, based upon Fig. 3.4, to express key elements of the problem in terms of GA. See the text for explanation. As noted in 2.2, the geometric product of two unit vectors is a rotation operator. In this solution, we’ll use the equality of the angles ψ/2 to write that 11
- 12. the product t − p t − p t − p t − p is an operator that rotates iˆa into ˆu: [iˆa] t − p t − p t − p t − p = ˆu. Now, we transform that equation into a product of four vectors that is equal to a bivector: [iˆa] t − p t − p t − p t − p = ˆu −i [iˆa] t − p t − p t − p t − p = −iˆu −ii [ˆa] t − p t − p t − p t − p = −iˆu [ˆa] t − p t − p t − p t − p = −iˆu ˆa t − p t − p t − p t − p = ˆui ˆuˆa t − p t − p t − p t − p = i ˆuˆa [t − p] [t − p] = t − p t − p i, from which ˆuˆa [t − p] [t − p] 0 = 0. (3.7) The simplification of ˆuˆapˆb 0 uses the fact that ˆbˆapˆb is a reflection of ˆap: ˆuˆapˆb 0 = ˆa + ˆb ˆa + ˆb ˆapˆb 0 = ˆaˆapˆb + ˆbˆapˆb ˆa + ˆb 0 = pˆb + pˆa ˆa + ˆb 0 = ˆa + ˆb ˆa + ˆb · p = ˆu · p. From 3.7, we could obtain a quadratic in λ in many ways, one of which is the following, which uses ˆu = ˆa + ˆb / ˆa + ˆb ; ˆu · ˆa = ˆu · ˆb; t = λ (ˆu · ˆa) ˆa: t = λ (ˆu · ˆa) ˆb; and (ˆu · ˆa) 2 = 1 + ˆa · ˆb /2 : ˆuˆa [t − p] [t − p] 0 = 0 ˆuˆa [λ (ˆu · ˆa) ˆa − p] λ (ˆu · ˆa) ˆb − p 0 = 0. Expanding and simplifying, λ2 (ˆu · ˆa) 2 ˆuˆb − λ (ˆu · ˆa) ˆup + ˆuˆapˆb + ˆuˆap = 0 λ2 (ˆu · ˆa) 2 ˆu · ˆa =ˆu·ˆb −λ {(ˆu · ˆa) [2ˆu · p]} + (ˆu · ˆa) p = 0. Eliminating the factor ˆu · ˆa, λ2 (ˆu · ˆa) 2 − λ [2ˆu · p] + (ˆu · ˆa) p = 0, which is identical to 3.3. 12
- 13. 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 . [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. [4] “A Very Brief Introduction to Reflections in 2D Geometric Alge- bra, and their Use in Solving ‘Construction’ Problems”. Available at http://vixra.org/abs/1606.0253. [5] A. Macdonald, Linear and Geometric Algebra (First Edition) p. 126, Cre- ateSpace Independent Publishing Platform (Lexington, 2012). 13