This document uses geometric algebra to solve the limiting case of the Problem of Apollonius known as the Circle-Line-Point problem. It presents three solutions: one using only rotations, one using a combination of reflections and rotations, and one in the appendix using only rotations. The solutions identify either the points of tangency between the solution circles and the given circle, or the points of tangency between the solution circles and the given line. The document reviews reflections and rotations in geometric algebra to establish the necessary foundations before presenting the solutions.
A Very Brief Introduction to Reflections in 2D Geometric Algebra, and their U...James Smith
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.
As a demonstration of the coherence of Geometric Algebra's (GA's) geometric and algebraic concepts of bivectors, we add three geometric bivectors according to the procedure described by Hestenes and Macdonald, then use bivector identities to determine, from the result, two vectors whose outer product is equal to the initial sum. In this way, we show that the procedure that GA's inventors defined for adding geometric bivectors is precisely that which is needed to give results that coincide with those obtained by calculating outer products of vectors that are expressed in terms of a 3D basis. We explain that that accomplishment is no coincidence: it is a consequence of the attributes that GA's designers assigned (or didn't) to bivectors.
How to Effect a Composite Rotation of a Vector via Geometric (Clifford) AlgebraJames Smith
We show how to express the representation of a composite rotation in terms that allow the rotation of a vector to be calculated conveniently via a spreadsheet that uses formulas developed, previously, for a single rotation. The work presented here (which includes a sample calculation) also shows how to determine the bivector angle that produces, in a single operation, the same rotation that is effected by the composite of two rotations.
Via Geometric (Clifford) Algebra: Equation for Line of Intersection of Two Pl...James Smith
As a high-school-level example of solving a problem via Geometric Algebra (GA), we show how to derive an equation for the line of intersection between two given planes. The solution method that we use emphasizes GA's capabilities for expressing and manipulating projections and rotations of vectors.
Solution of a High-School Algebra Problem to Illustrate the Use of Elementary...James Smith
This document is the first in what is intended to be a collection of solutions of high-school-level problems via Geometric Algebra (GA). GA is very much "overpowered" for such problems, but students at that level who plan to go into more-advanced math and science courses will benefit from seeing how to "see" basic problems in GA terms, and to then solve those problems using GA identities and common techniques.
E-Cordial Labeling of Some Mirror GraphsWaqas Tariq
Let G be a bipartite graph with a partite sets V1 and V2 and G\' be the copy of G with corresponding partite sets V1\' and V2\' . The mirror graph M(G) of G is obtained from G and G\' by joining each vertex of V2 to its corresponding vertex in V2\' by an edge. Here we investigate E-cordial labeling of some mirror graphs. We prove that the mirror graphs of even cycle Cn, even path Pn and hypercube Qk are E-cordial graphs.
Impact of facts devices on zonal congestion management in deregulated power s...Alexander Decker
This document summarizes a research paper that investigates the impact of Flexible AC Transmission System (FACTS) devices on zonal congestion management in deregulated power systems. The paper proposes a method using Transmission Congestion Distribution Factors (TCDFs) to identify congestion clusters based on line sensitivities. It then evaluates the effectiveness of Thyristor Controlled Phase Angle Regulators (TCPARs) and Thyristor Controlled Series Capacitors (TCSCs) at reducing transmission congestion costs when optimally located based on a proposed performance index. The method is tested on the IEEE 118-bus and 62-bus Indian utility systems. Key mathematical models of transmission lines with TCS
Some Soliton Solutions of Non Linear Partial Differential Equations by Tan-Co...IOSR Journals
Tan-cot method is applied to get exact soliton solutions of non-linear partial differential equations notably generalized Benjamin-Bona-Mahony, Zakharov-Kuznetsov Benjamin-Bona-Mahony, Kadomtsov-Petviashvilli Benjamin-Bona-Mahony and Korteweg-de Vries equations, which are important evolution equations with wide variety of physical applications. Elastic behavior and soliton fusion/fission is shown graphically and discussed physically as far as possible
A Very Brief Introduction to Reflections in 2D Geometric Algebra, and their U...James Smith
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.
As a demonstration of the coherence of Geometric Algebra's (GA's) geometric and algebraic concepts of bivectors, we add three geometric bivectors according to the procedure described by Hestenes and Macdonald, then use bivector identities to determine, from the result, two vectors whose outer product is equal to the initial sum. In this way, we show that the procedure that GA's inventors defined for adding geometric bivectors is precisely that which is needed to give results that coincide with those obtained by calculating outer products of vectors that are expressed in terms of a 3D basis. We explain that that accomplishment is no coincidence: it is a consequence of the attributes that GA's designers assigned (or didn't) to bivectors.
How to Effect a Composite Rotation of a Vector via Geometric (Clifford) AlgebraJames Smith
We show how to express the representation of a composite rotation in terms that allow the rotation of a vector to be calculated conveniently via a spreadsheet that uses formulas developed, previously, for a single rotation. The work presented here (which includes a sample calculation) also shows how to determine the bivector angle that produces, in a single operation, the same rotation that is effected by the composite of two rotations.
Via Geometric (Clifford) Algebra: Equation for Line of Intersection of Two Pl...James Smith
As a high-school-level example of solving a problem via Geometric Algebra (GA), we show how to derive an equation for the line of intersection between two given planes. The solution method that we use emphasizes GA's capabilities for expressing and manipulating projections and rotations of vectors.
Solution of a High-School Algebra Problem to Illustrate the Use of Elementary...James Smith
This document is the first in what is intended to be a collection of solutions of high-school-level problems via Geometric Algebra (GA). GA is very much "overpowered" for such problems, but students at that level who plan to go into more-advanced math and science courses will benefit from seeing how to "see" basic problems in GA terms, and to then solve those problems using GA identities and common techniques.
E-Cordial Labeling of Some Mirror GraphsWaqas Tariq
Let G be a bipartite graph with a partite sets V1 and V2 and G\' be the copy of G with corresponding partite sets V1\' and V2\' . The mirror graph M(G) of G is obtained from G and G\' by joining each vertex of V2 to its corresponding vertex in V2\' by an edge. Here we investigate E-cordial labeling of some mirror graphs. We prove that the mirror graphs of even cycle Cn, even path Pn and hypercube Qk are E-cordial graphs.
Impact of facts devices on zonal congestion management in deregulated power s...Alexander Decker
This document summarizes a research paper that investigates the impact of Flexible AC Transmission System (FACTS) devices on zonal congestion management in deregulated power systems. The paper proposes a method using Transmission Congestion Distribution Factors (TCDFs) to identify congestion clusters based on line sensitivities. It then evaluates the effectiveness of Thyristor Controlled Phase Angle Regulators (TCPARs) and Thyristor Controlled Series Capacitors (TCSCs) at reducing transmission congestion costs when optimally located based on a proposed performance index. The method is tested on the IEEE 118-bus and 62-bus Indian utility systems. Key mathematical models of transmission lines with TCS
Some Soliton Solutions of Non Linear Partial Differential Equations by Tan-Co...IOSR Journals
Tan-cot method is applied to get exact soliton solutions of non-linear partial differential equations notably generalized Benjamin-Bona-Mahony, Zakharov-Kuznetsov Benjamin-Bona-Mahony, Kadomtsov-Petviashvilli Benjamin-Bona-Mahony and Korteweg-de Vries equations, which are important evolution equations with wide variety of physical applications. Elastic behavior and soliton fusion/fission is shown graphically and discussed physically as far as possible
A new kind of quantum gates, higher braiding gates, as matrix solutions of the polyadic braid equations (different from the generalized Yang–Baxter equations) is introduced. Such gates lead to another special multiqubit entanglement that can speed up key distribution and accelerate algorithms. Ternary braiding gates acting on three qubit states are studied in detail. We also consider exotic non-invertible gates, which can be related with qubit loss, and define partial identities (which can be orthogonal), partial unitarity, and partially bounded operators (which can be non-invertible). We define two classes of matrices, star and circle ones, such that the magic matrices (connected with the Cartan decomposition) belong to the star class. The general algebraic structure of the introduced classes is described in terms of semigroups, ternary and 5-ary groups and modules. The higher braid group and its representation by the higher braid operators are given. Finally, we show, that for each multiqubit state, there exist higher braiding gates that are not entangling, and the concrete conditions to be non-entangling are given for the obtained binary and ternary gates.
METRIC DIMENSION AND UNCERTAINTY OF TRAVERSING ROBOTS IN A NETWORKgraphhoc
Metric dimension in graph theory has many applications in the real world. It has been applied to the
optimization problems in complex networks, analyzing electrical networks; show the business relations,
robotics, control of production processes etc. This paper studies the metric dimension of graphs with
respect to contraction and its bijection between them. Also an algorithm to avoid the overlapping between
the robots in a network is introduced.
A new method of constructing srgd & resolvable srgd designsAlexander Decker
This academic article describes new methods for constructing semi-regular group divisible (SRGD) designs and resolvable SRGD designs. It begins with background on block designs, group divisible designs, and properties of resolvable designs. It then presents two theorems. The first theorem shows that the incidence matrix of a self-complementary balanced incomplete block design can be used to construct an SRGD design. The second theorem demonstrates that a series of resolvable SRGD designs can be constructed when the parameter s is a prime or prime power, by repeating and modifying the partial parallel classes of an existing design. Examples of new designs constructed by these methods are provided.
The document summarizes several physics concepts and problems:
1. It describes an ammeter and its readings, including calculating current and avoiding parallax error.
2. It discusses forces acting on a car, including calculating net force and acceleration from changes in engine thrust.
3. It covers pressure in liquids and buoyant force, such as calculating tension in a cable lowering a block into water.
4. It explains heat transfer and specific heat capacity, like calculating heat removed by a cooling pad.
5. It compares free fall, impulse forces during collisions, and changes in momentum for coconuts striking different surfaces.
6. It analyzes floating and submerged lengths of drinking straws
The idea of metric dimension in graph theory was introduced by P J Slater in [2]. It has been found
applications in optimization, navigation, network theory, image processing, pattern recognition etc.
Several other authors have studied metric dimension of various standard graphs. In this paper we
introduce a real valued function called generalized metric G X × X × X ® R+ d : where X = r(v /W) =
{(d(v,v1),d(v,v2 ),...,d(v,v ) / v V (G))} k Î , denoted d G and is used to study metric dimension of graphs. It
has been proved that metric dimension of any connected finite simple graph remains constant if d G
numbers of pendant edges are added to the non-basis vertices.
Solution of a Vector-Triangle Problem Via Geometric (Clifford) AlgebraJames Smith
As a high-school-level application of Geometric Algebra (GA), we show how to solve a simple vector-triangle problem. Our method highlights the use of outer products and inverses of bivectors.
This document summarizes an article from the International Journal of Computer Engineering and Technology. The article discusses decomposing complete graphs into circulant graphs. It presents two theorems: 1) A complete graph K2m+1 can be decomposed into m edge-disjoint circulant graphs C2m+1(j) for m ≥ 2. 2) K2m+1 can be decomposed into edge-disjoint circulant graphs when 2m+1 is prime, equals 3⋅3n for n ≥ 1, or equals 3s for s ≥ 5 prime. The article also discusses an algorithm for decomposing the complete graph and applying it to the traveling salesman problem.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
This document discusses operations on interval-valued fuzzy graphs. It begins with an introduction to fuzzy graphs and definitions related to fuzzy graph theory. It then presents the main results which prove properties of the union and join operations on interval-valued fuzzy graphs. Specifically, it proves that the union of two interval-valued fuzzy graphs G1 and G2 is isomorphic to their join, and vice versa. It also discusses subgraphs, complements, and neighbors in fuzzy graph theory.
Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
The document provides instructions for an online aerospace engineering examination. It states that the exam has 65 multiple-choice questions worth a total of 100 marks. Questions are either worth 1 or 2 marks depending on the question number. There is no negative marking for numerical answer questions but negative marking for multiple choice questions. Calculators are allowed but no other materials. The exam is timed for 3 hours.
Solution of a Sangaku ``Tangency" Problem via Geometric AlgebraJames Smith
Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to solve one of the beautiful \emph{sangaku} problems from 19th-Century Japan. Among the GA operations that prove useful is the rotation of vectors via the unit bivector
This document discusses balanced incomplete block designs (BIBD) which are incomplete block designs where:
- Each block contains the same number (k) of plots
- Each treatment is replicated the same number (r) of times
- Every pair of treatments occurs together in the same number (λ) of blocks
The key properties of a BIBD are:
I) The total number of plots (bk) equals the total number of treatment replications (vr)
II) The number of treatment pairs (λ) is equal to the number of block-treatment pairs minus the number of treatment pairs
III) The number of blocks (b) must be greater than the number of treatments (v)
function f is said to be an even harmonious labeling of a graph G with q edges if f is an injection from the vertices of G to the integers from 0 to 2q and the induced function f* from the edges ofG to {0, 2…….….2(q-1)} defined by f* (uv) = f (u) +f (v) (mod 2q) is bijective. The graph G is said to have an even harmonious labeling. In this paper the even harmonious labeling of a class of graph namely H (2n, 2t+1) is established.
Commutativity in prime gamma rings with jordan leftAlexander Decker
This document discusses commutativity in prime gamma rings with Jordan left derivations. It begins with introducing key concepts such as gamma rings, prime gamma rings, derivations, and Jordan derivations. It then presents two lemmas regarding properties of Jordan left derivations in 2-torsion free gamma rings satisfying a certain condition. The main theorem proves that if a 2-torsion free prime gamma ring has a non-zero Jordan left derivation, then the gamma ring must be commutative, and the Jordan left derivation is actually a left derivation. The document provides context, definitions, lemmas, and the main result regarding commutativity induced by Jordan left derivations in certain types of gamma rings.
The document discusses the displacement or stiffness method for analyzing structures. It begins by introducing the stiffness method, which allows the same approach to analyze both statically determinate and indeterminate structures. The key concepts are that nodal displacements are the unknowns, and equilibrium equations relating the unknown displacements to known stiffness coefficients are written and solved.
It then defines stiffness, stiffness coefficients, and stiffness matrices. The stiffness of a member is the force required to produce a unit displacement. Stiffness coefficients relate the forces and displacements at nodes. Element stiffness matrices are developed and combined to form the overall structure stiffness matrix, which relates the total nodal forces and displacements.
In this paper, we study semi symmetric and pseudo symmetric conditions in S -manifolds, those are RR = 0 , RC = 0 , C R = 0 , C C = 0 , = ( , ) R R L1Q g R , = ( , ) RC L2Q g C , = ( , ) CR L3Q g R , and = ( , ) C C L4Q g C , where C is the Concircular curvature tensor and 1 2 3 4 L ,L ,L ,L are the smooth functions on M , further we discuss about Ricci soliton
https://arxiv.org/abs/2006.07865
A general mechanism for "breaking" commutativity in algebras is proposed: if the underlying set is taken to be not a crisp set, but rather an obscure/fuzzy set, the membership function, reflecting the degree of truth that an element belongs to the set, can be incorporated into the commutation relations. The special "deformations" of commutativity and e-commutativity are introduced in such a way that equal degrees of truth result in the "nondeformed" case. We also sketch how to "deform" e-Lie algebras and Weyl algebras. Further, the above constructions are extended to n-ary algebras for which the projective representations and e-commutativity are studied.
Cómo entender el uso de escalas logarítmicasJames Smith
Hojas con escalas logarítmicas son herramientas visuales muy poderosas para el análisis de datos, pero pueden ser una fuente de confusión también. Como es el caso para muchas herramientas, la confusión se minimiza o elimina cuando, al aprender su uso, tomamos la siguiente postura:
“Esta herramienta fue inventada para facilitar la resolución de
alg ún tipo problema. Entonces, ¿cuál fue?”
El documento contiene 19 problemas de geometría que involucran figuras como rectángulos, triángulos, círculos y trapecios. Los problemas piden calcular perímetros, áreas y longitudes de lados y diagonales de las figuras dadas, y determinar cuál figura tiene el mayor área.
El documento resume las reglas básicas del cálculo diferencial y integral. Explica que la derivada de una suma o resta de funciones es la suma o resta de sus derivadas respectivas, y que la derivada del producto de una constante por una función es igual al producto de la constante por la derivada de la función. También cubre las reglas para derivar potencias de x y tomar integrales de sumas, diferencias, productos constantes y potencias de x. Proporciona ejemplos ilustrativos de cada regla.
Why Does the Atmosphere Rotate? Trajectory of a desorbed moleculeJames Smith
As a step toward understanding why the Earth's atmosphere "rotates" with the Earth, we use using Geometric (Clifford) Algebra to investigate the trajectory of a single molecule that desorbs vertically upward from the Equator, then falls back to Earth without colliding with any other molecules. Sample calculations are presented for a molecule whose vertical velocity is equal to the surface velocity of the Earth at the Equator (463 m/s) and for one with a vertical velocity three times as high. The latter velocity is sufficient for the molecule to reach the Kármán Line (100,000 m). We find that both molecules fall to Earth behind the point from which they desorbed: by 0.25 degrees of latitude for the higher vertical velocity, but by only 0.001 degrees for the lower.
A new kind of quantum gates, higher braiding gates, as matrix solutions of the polyadic braid equations (different from the generalized Yang–Baxter equations) is introduced. Such gates lead to another special multiqubit entanglement that can speed up key distribution and accelerate algorithms. Ternary braiding gates acting on three qubit states are studied in detail. We also consider exotic non-invertible gates, which can be related with qubit loss, and define partial identities (which can be orthogonal), partial unitarity, and partially bounded operators (which can be non-invertible). We define two classes of matrices, star and circle ones, such that the magic matrices (connected with the Cartan decomposition) belong to the star class. The general algebraic structure of the introduced classes is described in terms of semigroups, ternary and 5-ary groups and modules. The higher braid group and its representation by the higher braid operators are given. Finally, we show, that for each multiqubit state, there exist higher braiding gates that are not entangling, and the concrete conditions to be non-entangling are given for the obtained binary and ternary gates.
METRIC DIMENSION AND UNCERTAINTY OF TRAVERSING ROBOTS IN A NETWORKgraphhoc
Metric dimension in graph theory has many applications in the real world. It has been applied to the
optimization problems in complex networks, analyzing electrical networks; show the business relations,
robotics, control of production processes etc. This paper studies the metric dimension of graphs with
respect to contraction and its bijection between them. Also an algorithm to avoid the overlapping between
the robots in a network is introduced.
A new method of constructing srgd & resolvable srgd designsAlexander Decker
This academic article describes new methods for constructing semi-regular group divisible (SRGD) designs and resolvable SRGD designs. It begins with background on block designs, group divisible designs, and properties of resolvable designs. It then presents two theorems. The first theorem shows that the incidence matrix of a self-complementary balanced incomplete block design can be used to construct an SRGD design. The second theorem demonstrates that a series of resolvable SRGD designs can be constructed when the parameter s is a prime or prime power, by repeating and modifying the partial parallel classes of an existing design. Examples of new designs constructed by these methods are provided.
The document summarizes several physics concepts and problems:
1. It describes an ammeter and its readings, including calculating current and avoiding parallax error.
2. It discusses forces acting on a car, including calculating net force and acceleration from changes in engine thrust.
3. It covers pressure in liquids and buoyant force, such as calculating tension in a cable lowering a block into water.
4. It explains heat transfer and specific heat capacity, like calculating heat removed by a cooling pad.
5. It compares free fall, impulse forces during collisions, and changes in momentum for coconuts striking different surfaces.
6. It analyzes floating and submerged lengths of drinking straws
The idea of metric dimension in graph theory was introduced by P J Slater in [2]. It has been found
applications in optimization, navigation, network theory, image processing, pattern recognition etc.
Several other authors have studied metric dimension of various standard graphs. In this paper we
introduce a real valued function called generalized metric G X × X × X ® R+ d : where X = r(v /W) =
{(d(v,v1),d(v,v2 ),...,d(v,v ) / v V (G))} k Î , denoted d G and is used to study metric dimension of graphs. It
has been proved that metric dimension of any connected finite simple graph remains constant if d G
numbers of pendant edges are added to the non-basis vertices.
Solution of a Vector-Triangle Problem Via Geometric (Clifford) AlgebraJames Smith
As a high-school-level application of Geometric Algebra (GA), we show how to solve a simple vector-triangle problem. Our method highlights the use of outer products and inverses of bivectors.
This document summarizes an article from the International Journal of Computer Engineering and Technology. The article discusses decomposing complete graphs into circulant graphs. It presents two theorems: 1) A complete graph K2m+1 can be decomposed into m edge-disjoint circulant graphs C2m+1(j) for m ≥ 2. 2) K2m+1 can be decomposed into edge-disjoint circulant graphs when 2m+1 is prime, equals 3⋅3n for n ≥ 1, or equals 3s for s ≥ 5 prime. The article also discusses an algorithm for decomposing the complete graph and applying it to the traveling salesman problem.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
This document discusses operations on interval-valued fuzzy graphs. It begins with an introduction to fuzzy graphs and definitions related to fuzzy graph theory. It then presents the main results which prove properties of the union and join operations on interval-valued fuzzy graphs. Specifically, it proves that the union of two interval-valued fuzzy graphs G1 and G2 is isomorphic to their join, and vice versa. It also discusses subgraphs, complements, and neighbors in fuzzy graph theory.
Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
The document provides instructions for an online aerospace engineering examination. It states that the exam has 65 multiple-choice questions worth a total of 100 marks. Questions are either worth 1 or 2 marks depending on the question number. There is no negative marking for numerical answer questions but negative marking for multiple choice questions. Calculators are allowed but no other materials. The exam is timed for 3 hours.
Solution of a Sangaku ``Tangency" Problem via Geometric AlgebraJames Smith
Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to solve one of the beautiful \emph{sangaku} problems from 19th-Century Japan. Among the GA operations that prove useful is the rotation of vectors via the unit bivector
This document discusses balanced incomplete block designs (BIBD) which are incomplete block designs where:
- Each block contains the same number (k) of plots
- Each treatment is replicated the same number (r) of times
- Every pair of treatments occurs together in the same number (λ) of blocks
The key properties of a BIBD are:
I) The total number of plots (bk) equals the total number of treatment replications (vr)
II) The number of treatment pairs (λ) is equal to the number of block-treatment pairs minus the number of treatment pairs
III) The number of blocks (b) must be greater than the number of treatments (v)
function f is said to be an even harmonious labeling of a graph G with q edges if f is an injection from the vertices of G to the integers from 0 to 2q and the induced function f* from the edges ofG to {0, 2…….….2(q-1)} defined by f* (uv) = f (u) +f (v) (mod 2q) is bijective. The graph G is said to have an even harmonious labeling. In this paper the even harmonious labeling of a class of graph namely H (2n, 2t+1) is established.
Commutativity in prime gamma rings with jordan leftAlexander Decker
This document discusses commutativity in prime gamma rings with Jordan left derivations. It begins with introducing key concepts such as gamma rings, prime gamma rings, derivations, and Jordan derivations. It then presents two lemmas regarding properties of Jordan left derivations in 2-torsion free gamma rings satisfying a certain condition. The main theorem proves that if a 2-torsion free prime gamma ring has a non-zero Jordan left derivation, then the gamma ring must be commutative, and the Jordan left derivation is actually a left derivation. The document provides context, definitions, lemmas, and the main result regarding commutativity induced by Jordan left derivations in certain types of gamma rings.
The document discusses the displacement or stiffness method for analyzing structures. It begins by introducing the stiffness method, which allows the same approach to analyze both statically determinate and indeterminate structures. The key concepts are that nodal displacements are the unknowns, and equilibrium equations relating the unknown displacements to known stiffness coefficients are written and solved.
It then defines stiffness, stiffness coefficients, and stiffness matrices. The stiffness of a member is the force required to produce a unit displacement. Stiffness coefficients relate the forces and displacements at nodes. Element stiffness matrices are developed and combined to form the overall structure stiffness matrix, which relates the total nodal forces and displacements.
In this paper, we study semi symmetric and pseudo symmetric conditions in S -manifolds, those are RR = 0 , RC = 0 , C R = 0 , C C = 0 , = ( , ) R R L1Q g R , = ( , ) RC L2Q g C , = ( , ) CR L3Q g R , and = ( , ) C C L4Q g C , where C is the Concircular curvature tensor and 1 2 3 4 L ,L ,L ,L are the smooth functions on M , further we discuss about Ricci soliton
https://arxiv.org/abs/2006.07865
A general mechanism for "breaking" commutativity in algebras is proposed: if the underlying set is taken to be not a crisp set, but rather an obscure/fuzzy set, the membership function, reflecting the degree of truth that an element belongs to the set, can be incorporated into the commutation relations. The special "deformations" of commutativity and e-commutativity are introduced in such a way that equal degrees of truth result in the "nondeformed" case. We also sketch how to "deform" e-Lie algebras and Weyl algebras. Further, the above constructions are extended to n-ary algebras for which the projective representations and e-commutativity are studied.
Cómo entender el uso de escalas logarítmicasJames Smith
Hojas con escalas logarítmicas son herramientas visuales muy poderosas para el análisis de datos, pero pueden ser una fuente de confusión también. Como es el caso para muchas herramientas, la confusión se minimiza o elimina cuando, al aprender su uso, tomamos la siguiente postura:
“Esta herramienta fue inventada para facilitar la resolución de
alg ún tipo problema. Entonces, ¿cuál fue?”
El documento contiene 19 problemas de geometría que involucran figuras como rectángulos, triángulos, círculos y trapecios. Los problemas piden calcular perímetros, áreas y longitudes de lados y diagonales de las figuras dadas, y determinar cuál figura tiene el mayor área.
El documento resume las reglas básicas del cálculo diferencial y integral. Explica que la derivada de una suma o resta de funciones es la suma o resta de sus derivadas respectivas, y que la derivada del producto de una constante por una función es igual al producto de la constante por la derivada de la función. También cubre las reglas para derivar potencias de x y tomar integrales de sumas, diferencias, productos constantes y potencias de x. Proporciona ejemplos ilustrativos de cada regla.
Why Does the Atmosphere Rotate? Trajectory of a desorbed moleculeJames Smith
As a step toward understanding why the Earth's atmosphere "rotates" with the Earth, we use using Geometric (Clifford) Algebra to investigate the trajectory of a single molecule that desorbs vertically upward from the Equator, then falls back to Earth without colliding with any other molecules. Sample calculations are presented for a molecule whose vertical velocity is equal to the surface velocity of the Earth at the Equator (463 m/s) and for one with a vertical velocity three times as high. The latter velocity is sufficient for the molecule to reach the Kármán Line (100,000 m). We find that both molecules fall to Earth behind the point from which they desorbed: by 0.25 degrees of latitude for the higher vertical velocity, but by only 0.001 degrees for the lower.
Trampas comunes en los exámenes de se selección sobre matemáticasJames Smith
El documento describe dos ejemplos comunes de trampas en exámenes de selección. El primero involucra encontrar el valor de x a partir de una ecuación y elegir la respuesta incorrecta que coincide con el valor de x. El segundo involucra calcular las medidas de ángulos en un diagrama y elegir una respuesta que no corresponde con la diferencia requerida por la pregunta. El documento concluye aconsejando volver a leer la pregunta antes de responder.
An additional brief solution of the CPP limiting case of the Problem of Apoll...James Smith
This document adds to the collection of GA solutions to plane-geometry problems, most of them dealing with tangency, that are presented in References 1-7. Reference 1 presented several ways of solving the CPP limiting case of the Problem of Apollonius. Here, we use ideas from Reference 6 to solve that case in yet another way.
Para cualquier practicante de matemáticas nacido en los últimos 400 años es difícil, si no imposible, concebir cuánto el invento de gráficos y del álgebra simplificó la tarea de sus predecesores, por posibilitar que “se hiciera visible” las relaciones entre cantidades que intervinieron en los problemas que trabajaron. Estas herramientas nos ayudarán en este capítulo.
(Véase también http://www.slideshare.net/JamesSmith245/modelando-matemticamente-el-slinky-en-cada-libre .)
El Slinky es un divertido juguete infantil en forma de un resorte de baja resistencia. En este documento, doy énfasis a las bellezas matemáticas de su comportamiento. Claro que mucha de sus matemáticas son avanzadas, pero lo importante es que sí, hasta en las cosas más sencillas se encuentra mucha belleza, que se manifiesta en su matemática. Así que explico detalladamente algunas ideas sencillas, mientras en otras ocasiones presento temas matemáticas avanzadas con poca explicación.
Empiezo por analizar la extensión provocada por su propio peso de un Slinky colgado. Después, se analiza el movimiento vibratorio de un Slinky colgado. Por medio de eses análi-sis, derivo ecuaciones que predicen (1) cuánto se estirará un Slinky bajo su propio peso; (2) el periodo de la vibración de un Slinky colgado; y (3) cómo se relacionan (1) y (2).
Por fin, se examinan los datos empíricos obtenidos de experimentos con Slinkies de acero y de plástico, comparando eses datos con el comportamiento predicho por medio de los aná-lisis matemáticos. Como parte de la comparación, se calcula la aceleración gravitatoria a partir de los datos empíricos.
Cambios de óptica en el curso de un despejeJames Smith
El documento describe los cambios en la perspectiva del autor al resolver una ecuación algebraica. Inicialmente, el autor voltea la ecuación para poner la incógnita en el lado izquierdo. Luego, el autor concibe el término "2v" como un solo número y más tarde como un número negativo "-2v". Finalmente, el autor resuelve la ecuación para encontrar el valor de la incógnita v.
Proporciones de los radios y distancias en una "cadena de Steiner" de 6 circu...James Smith
Si una circunferencia de radio r1 encierre a otra circunferencia menor, el espacio entre el exterior de la menor y el interior de la mayor se puede llenar de una cadena de circunferencias tangentes a ambas, y también entre sí. Una cadena tal se llama un "Cadena de Steiner". Los puntos de de tangencia entre las circunferencias pertenecientes a la cadena yacen sobre una circunferencia de algún radio "r*", cuyo centro queda a alguna distancia "q" del centro de la circunferencia mayor.
En este documento, se identifica la relación entre r*, q , y r tal que haya exactamente seis circunferencias en la cadena.
Simplified Solutions of the CLP and CCP Limiting Cases of the Problem of Apo...James Smith
The new solutions presented herein for the CLP and CCP limiting cases of the Problem of Apollonius are much shorter and more easily understood than those provided by the same author in References [1] and [2]. These improvements result from (1) a better selection of angle relationships as a starting point for the solution process; and (2) better use of GA identities to avoid forming troublesome combinations of terms within the resulting equations.
Modelación matemática de una rueda y eje con momento de incercia variable par...James Smith
Se describe un aparato que consiste en un eje, más una rueda que cuenta con seis masas cuya distancia puede ajustarse para variar el momento de inercia polar del aparato. Se analiza un experimento en el que dicho aparato se cuelga de un soporte, mediante cuerdas fijadas en los dos extremos del eje. Rotando el aparato de manera que cada cuerda se enrede alrededor del extremo al que está fijado, se sube el aparato. Acto seguido, se permite que el aparato descienda, girando libremente. Para un rango de distancias entre las masas y el eje, se mide cuanto tiempo el aparato tarda en descender una distancia predeterminada.
Tras análisis basados en la dinámica y en la conservación de energía, identificamos que el experimento arriba descrito no es suficiente para identificar la masa total de las seis masas. En cambio, es necesario contrastar los resultados de ese experimento, con (por ejemplo) uno que emplee una masa auxiliar conocida.
Construcciones para encontrar la raíz cuadrada y resolver ecuaciones cuadráticasJames Smith
Se presentan y explican construcciones geométricas (realizadas con regla y compás) para encontrar la raíz cuadrada de un número, y resolver ecuaciones cuadráticas.
Cómo resolver problemas con "triángulos rectángulos simultáneos"James Smith
Este documento presenta la resolución de un problema de triángulos rectángulos simultáneos para encontrar las longitudes desconocidas h y c. Se establecen dos ecuaciones con las relaciones trigonométricas y se combinan para obtener una ecuación con una sola incógnita, d. Esto permite calcular d = 10.9 m. Luego, se usa este valor para calcular c = 15.5 m. El documento concluye analizando las fórmulas obtenidas y sugiriendo repasar la solución.
Modelando matemáticamente, el "Slinky'' en caída libreJames Smith
(Véase también http://www.slideshare.net/JamesSmith245/las-ellezas-matemticas-del-slinky, y Véase también, el video https://www.youtube.com/watch?v=2dU5R0ISWcw.)
En este documento, se intentó obtener soluciones analíticas para el "Slinky en caída libre". Los primeros dos, que usaron la ecuación de la onda, fracasaron por razones distintas. El primer intento usó las series Fourier, pero no pudo cumplir las condiciones de frontera. El segundo usó trasformadas Laplace; de esa forma sí, cumplió las condiciones iniciales y de frontera, y predijo, con corrección, que la aceleración del baricentro del Slinky debe ser igual a g, (la aceleración gravitatoria ). Sin embargo, se equivicó en cuanto predijo que el extremo superior del Slinky caería a través de la parte del Slinky que está en reposo. Esta predicción no fue un defecto de las trasformadas Laplace; sino un artefacto del uso de la ecuación de la onda para un resorte de tensión en este caso específico, que trata de una onda de choque.
El tercer intento usó el teorema impulso-momento. Hizo predicciones coherentes entre sí, y que concuerdan con observaciones empíricas.
Cabe señalar que la modelación del Slinky en caída libre presentada aquí trata solamente su movimiento en la dirección vertical, haciendo caso omiso a su rotación.
A Modification of the Lifshitz-Slyozov-Wagner Equation for Predicting Coarsen...James Smith
The story behind this article is instructive, and even a bit troubling. I wrote it in 1991 as a continuation of part of my Doctoral thesis, which I’d completed a few years earlier. During that research, I’d found that scientists who’d done very fine laboratory work on Ostwald ripening during the 1960s had made a curious error in simple mass balances when deriving a rate equation for Ostwald ripening starting from the minimum-entropy-production-rate (MEPR) principle.
That error led the 1960s scientists to reject (with commendable honesty) their hypothesis that the MEPR principle is applicable to Ostwald ripening. Like all the rest of us metallurgists back then, I didn’t catch that error, until I examined the derivation of the MEPR-based rate equation in detail during my thesis work. However, I didn’t manage to re-derive the rate equation fully until I took up the subject again in the early 1990s. The scientists who did such fine lab work in the 1960s would no doubt have been pleased to learn that their empirical results agreed quite well with predictions made by the corrected equation. Thus, those scientists were correct in their hypothesis about the MEPR principle’s applicability.
I continue to wonder how we metallurgists overlooked, for more than two decades, the simple error that led those scientists to conclude, mistakenly but honestly, that they’d been wrong.
I never did manage to publish this article, but the same derivations and analyses were published by other researchers within a few years. Some of the reviewers’ comments on the article are addressed in the second article in this document, “Comments on ‘Ostwald Ripening Growth Rate for Nonideal Systems with Significant Mutual Solubility’”.
Tú sí, puedes, con las ecuaciones simultáneas linealesJames Smith
Este documento presenta un resumen de las ecuaciones simultáneas lineales y varias técnicas para resolverlas, incluyendo gráfica, igualación, sustitución y reducción. Explica qué son las ecuaciones simultáneas y cómo se generan a partir de problemas que involucran múltiples condiciones. También provee ejemplos para ilustrar cada técnica de resolución.
A Solution to the Problem of Apollonius Using Vector Dot ProductsJames Smith
The document presents a solution to the Problem of Apollonius, which is to construct circles tangent to three given coplanar circles, using only vector dot products. It defines variables to represent the circles and derives equations that lead to an expression for the radius of the solution circle that encloses none of the three given circles. The solution method finds the points of tangency of two tangent circles, one enclosing and one not enclosing the three initial circles.
Three Solutions of the LLP Limiting Case of the Problem of Apollonius via Geo...James Smith
This document adds to the collection of solved problems presented in http://www.slideshare.net/JamesSmith245/rotations-of-vectors-via-geometric-algebra-explanation-and-usage-in-solving-classic-geometric-construction-problems-version-of-11-february-2016,http://www.slideshare.net/JamesSmith245/solution-of-the-ccp-case-of-the-problem-of-apollonius-via-geometric-clifford-algebra, http://www.slideshare.net/JamesSmith245/solution-of-the-special-case-clp-of-the-problem-of-apollonius-via-vector-rotations-using-geometric-algebra, and http://www.slideshare.net/JamesSmith245/a-very-brief-introduction-to-reflections-in-2d-geometric-algebra-and-their-use-in-solving-construction-problems. 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 two ways.
Formulas and Spreadsheets for Simple, Composite, and Complex Rotations of Vec...James Smith
We show how to express the representations of single, composite, and "rotated" rotations in GA terms that allow rotations to be calculated conveniently via spreadsheets. Worked examples include rotation of a single vector by a bivector angle; rotation of a vector about an axis; composite rotation of a vector; rotation of a bivector; and the "rotation of a rotation". Spreadsheets for doing the calculations are made available via live links.
Solution of the Special Case "CLP" of the Problem of Apollonius via Vector Ro...James Smith
Using ideas developed in detail in http://www.slideshare.net/JamesSmith245/rotations-of-vectors-via-geometric-algebra-explanation-and-usage-in-solving-classic-geometric-construction-problems-version-of-11-february-2016, this document solves one of the special cases of the famous Problem of Apollonius. A new Appendix presents alternative solutions.
See also:
http://www.slideshare.net/JamesSmith245/solution-of-the-ccp-case-of-the-problem-of-apollonius-via-geometric-clifford-algebra
http://www.slideshare.net/JamesSmith245/rotations-of-vectors-via-geometric-algebra-explanation-and-usage-in-solving-classic-geometric-construction-problems-version-of-11-february-2016
http://www.slideshare.net/JamesSmith245/resoluciones-de-problemas-de-construccin-geomtricos-por-medio-de-la-geometra-clsica-y-el-lgebra-geomtrica-vectorial
Rotations of Vectors via Geometric Algebra: Explanation, and Usage in Solving...James Smith
This document discusses using geometric algebra to solve geometry problems involving rotations of vectors. It begins by reviewing important relationships involving tangents, chords, and angles that are useful for formulating problems. It then examines the relationship between the geometric product and rotations of vectors, showing how a vector can be expressed as the sum of its projections. The document goes on to solve several classic geometry construction problems by representing the problems in terms of vector rotations and using properties of the geometric product to find solutions in a straightforward manner. It includes multiple solutions for some problems to illustrate different approaches.
Using a Common Theme to Find Intersections of Spheres with Lines and Planes v...James Smith
After reviewing the sorts of calculations for which Geometric Algebra (GA) is especially convenient, we identify a common theme through which those types of calculations can be used to find the intersections of spheres with lines, planes, and other spheres.
Solution Strategies for Equations that Arise in Geometric (Clifford) AlgebraJames Smith
Drawing mainly upon exercises from Hestenes's New Foundations for Classical Mechanics, this document presents, explains, and discusses common solution strategies. Included are a list of formulas and a guide to nomenclature.
See also:
http://www.slideshare.net/JamesSmith245/rotations-of-vectors-via-geometric-algebra-explanation-and-usage-in-solving-classic-geometric-construction-problems-version-of-11-february-2016 ;
http://www.slideshare.net/JamesSmith245/resoluciones-de-problemas-de-construccin-geomtricos-por-medio-de-la-geometra-clsica-y-el-lgebra-geomtrica-vectorial ;
http://www.slideshare.net/JamesSmith245/solution-of-the-special-case-clp-of-the-problem-of-apollonius-via-vector-rotations-using-geometric-algebra ;
http://www.slideshare.net/JamesSmith245/solution-of-the-ccp-case-of-the-problem-of-apollonius-via-geometric-clifford-algebra ;
http://www.slideshare.net/JamesSmith245/a-very-brief-introduction-to-reflections-in-2d-geometric-algebra-and-their-use-in-solving-construction-problems ;
http://www.slideshare.net/JamesSmith245/solution-of-the-llp-limiting-case-of-the-problem-of-apollonius-via-geometric-algebra-using-reflections-and-rotations ;
http://www.slideshare.net/JamesSmith245/simplied-solutions-of-the-clp-and-ccp-limiting-cases-of-the-problem-of-apollonius-via-vector-rotations-using-geometric-algebra ;
http://www.slideshare.net/JamesSmith245/additional-solutions-of-the-limiting-case-clp-of-the-problem-of-apollonius-via-vector-rotations-using-geometric-algebra ;
http://www.slideshare.net/JamesSmith245/an-additional-brief-solution-of-the-cpp-limiting-case-of-the-problem-of-apollonius-via-geometric-algebra-ga .
The document contains a list of 6 group members with their names and student identification numbers. The group members are:
1. Ridwan bin shamsudin, student ID: D20101037472
2. Mohd. Hafiz bin Salleh, student ID: D20101037433
3. Muhammad Shamim Bin Zulkefli, student ID: D20101037460
4. Jasman bin Ronie, student ID: D20101037474
5. Hairieyl Azieyman Bin Azmi, student ID: D20101037426
6. Mustaqim Bin Musa, student ID:
This document discusses the dot product of vectors. It defines the dot product as the sum of the products of the corresponding components of two vectors. The dot product is a scalar quantity that can be used to determine the angle between vectors and whether vectors are orthogonal. It also discusses the relationship between the dot product and the projections of one vector onto another vector.
Vector Product of Two Vectors
The vector product, or cross product, of two vectors A and B is a vector C defined as C=A ́B. The magnitude of C is equal to AxBy, where Ax and By are the components of A and B. The direction of C is perpendicular to both A and B and can be determined using the right-hand rule. The vector product is not commutative, so A ́B is not equal to B ́A. The vector product is used to calculate many physical quantities and results in zero if the vectors are parallel or anti-parallel.
This document contains 17 multi-part physics problems involving vectors and dimensional analysis. The problems cover topics like calculating thickness from volume and area, rates of filling tanks, expressing quantities in different units, using dimensional analysis to determine exponents, finding components of vectors, vector addition and subtraction, the cross product of vectors, and determining the resultant of multiple vectors. Detailed step-by-step solutions are provided for each problem.
The document defines a vector as having both magnitude and direction, represented geometrically by an arrow. It discusses representing vectors algebraically using coordinates, and defines operations like addition, subtraction, and scaling of vectors. Key vector concepts covered include the dot product, which yields a scalar when combining two vectors, and unit vectors, which have a magnitude of 1. Examples are provided of using vectors to solve problems and prove geometric properties.
The document defines vectors and discusses their geometric and algebraic representations. Geometrically, a vector has a magnitude and direction represented by an arrow. Algebraically, a vector in a plane can be represented by its coordinates (a1, a2) and in 3D space by coordinates (a1, a2, a3). Vectors can be added by placing them head to tail, subtracted by reversing one and adding, and scaled by a scalar number. The dot product of two vectors A and B yields a scalar value that geometrically equals the magnitudes of A and B multiplied by the cosine of the angle between them.
The document defines vectors and discusses their geometric and algebraic representations. Geometrically, a vector has a magnitude and direction represented by an arrow. Algebraically, a vector in a plane can be represented by its coordinates (a1, a2) and in 3D space by coordinates (a1, a2, a3). Vectors can be added by placing them head to tail, subtracted by reversing one and adding, and scaled by a scalar number. The dot product of two vectors A and B yields a scalar equal to |A||B|cosθ, where θ is the angle between the vectors.
The document discusses vectors and trigonometry. It begins by defining vectors as quantities with both magnitude and direction, such as displacement and velocity. It then covers geometric descriptions of vectors using arrows to represent direction. It also discusses vector operations like addition and scalar multiplication analytically and geometrically. Finally, it shows how vectors can model velocity and how to calculate the true velocity of an object when it experiences additional velocities like wind.
Vectors have both magnitude and direction. They can be represented geometrically as directed line segments or algebraically using ordered pairs and notation like arrows. Fundamental vector operations include addition, subtraction, and scalar multiplication. The dot product or inner product of two vectors produces a scalar value that depends on their magnitudes and the angle between them. It has various applications including finding the angle between vectors and calculating work done by a force.
Higher braiding gates, a new kind of quantum gate, are introduced. These are matrix solutions of the polyadic braid equations (which differ from the generalized Yang-Baxter equations). Such gates support a special kind of multi-qubit entanglement which can speed up key distribution and accelerate the execution of algorithms. Ternary braiding gates acting on three qubit states are studied in detail. We also consider exotic non-invertible gates which can be related to qubit loss, and define partial identities (which can be orthogonal), partial unitarity, and partially bounded operators (which can be non-invertible). We define two classes of matrices, the star and circle types, and find that the magic matrices (connected with the Cartan decomposition) belong to the star class. The general algebraic structure of the classes introduced here is described in terms of semigroups, ternary and 5-ary groups and modules. The higher braid group and its representation by higher braid operators are given. Finally, we show that for each multi-qubit state there exist higher braiding gates which are not entangling, and the concrete conditions to be non-entangling are given for the binary and ternary gates discussed.
This document provides an overview of vectors and their applications in physics. It defines vectors and differentiates them from scalars, discusses vector notation and representation, and covers key concepts like addition, subtraction, and multiplication of vectors. Examples are given of vector quantities like displacement, velocity and force. The document also explains vector operators like gradient, divergence and curl, which allow converting between scalar and vector quantities, and outlines how calculus is important in physics for studying change.
The document provides an overview of the objectives and activities for a lesson on relative motion analysis. It includes sample problems and questions on determining relative position, velocity, and acceleration between two moving frames of reference using vector methods and trigonometric relationships like the laws of sines and cosines. Sample problems demonstrate how to set up and solve for unknown relative motion variables graphically or through vector equations.
The document provides an overview of the objectives and activities for a lesson on relative motion analysis. It includes sample problems and questions on determining relative position, velocity, and acceleration between two moving frames of reference using vector methods and trigonometric relationships like the laws of sines and cosines. Sample problems demonstrate how to set up and solve for unknown relative motion variables graphically or through vector equations.
Similar to Additional Solutions of the Limiting Case "CLP" of the Problem of Apollonius via Vector Rotations using Geometric Algebra (20)
Via Geometric Algebra: Direction and Distance between Two Points on a Spheric...James Smith
As a high-school-level example of solving a problem via Geometric (Clifford) Algebra, we show how to calculate the distance and direction between two points on Earth, given the locations' latitudes and longitudes. We validate the results by comparing them to those obtained from online calculators. This example invites a discussion of the benefits of teaching spherical trigonometry (the usual way of solving such problems) at the high-school level versus teaching how to use Geometric Algebra for the same purpose.
Un acercamiento a los determinantes e inversos de matricesJames Smith
Este documento presenta un resumen de tres oraciones sobre los determinantes e inversos de matrices. Introduce los conceptos de matrices y sistemas de ecuaciones lineales, y explica cómo la resolución de sistemas lleva a las ideas de determinantes de matrices y la inversa de una matriz. Finalmente, compara las versiones matricial y no matricial de resolver sistemas lineales.
Understanding the "Chain Rule" for Derivatives by Deriving Your Own VersionJames Smith
Because the Chain Rule can confuse students as much as it helps them solve real problems, we put ourselves in the shoes of the mathematicians who derived it, so that students may understand the motivation for the rule; its limitations; and why textbooks present it in its customary form. We begin by finding the derivative of sin2x without using the Chain Rule. That exercise, having shown that even a comparatively simple compound function can be bothersome to differentiate using the definition of the derivative as a limit, provides the motivation for developing our own formula for the derivative of the general compound function g[f(x)]. In the course of that development, we see why the function f must be continuous at any value of x to which the formula is applied. We finish by comparing our formula to that which is commonly given.
Learning Geometric Algebra by Modeling Motions of the Earth and Shadows of Gn...James Smith
Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to calculate Solar azimuths and altitudes as a function of time via the heliocentric model. We begin by representing the Earth's motions in GA terms. Our representation incorporates an estimate of the time at which the Earth would have reached perihelion in 2017 if not affected by the Moon's gravity. Using the geometry of the December 2016 solstice as a starting point, we then employ GA's capacities for handling rotations to determine the orientation of a gnomon at any given latitude and longitude during the period between the December solstices of 2016 and 2017. Subsequently, we derive equations for two angles: that between the Sun's rays and the gnomon's shaft, and that between the gnomon's shadow and the direction ``north" as traced on the ground at the gnomon's location. To validate our equations, we convert those angles to Solar azimuths and altitudes for comparison with simulations made by the program Stellarium. As further validation, we analyze our equations algebraically to predict (for example) the precise timings and locations of sunrises, sunsets, and Solar zeniths on the solstices and equinoxes. We emphasize that the accuracy of the results is only to be expected, given the high accuracy of the heliocentric model itself, and that the relevance of this work is the efficiency with which that model can be implemented via GA for teaching at the introductory level. On that point, comments and debate are encouraged and welcome.
Nuevo Manual de la UNESCO para la Enseñanza de CienciasJames Smith
Este documento presenta el prefacio de una nueva edición del Manual de la Unesco para la Enseñanza de las Ciencias. Explica que la nueva edición actualiza el contenido y proporciona más material científico para cursos introductorios de ciencias. Detalla el proceso de revisión llevado a cabo por expertos de varios países bajo la coordinación de la Universidad de Maryland. El objetivo del manual es proveer ideas y recursos para que los maestros puedan enseñar ciencias de manera práctica utilizando materiales disponibles local
Calculating the Angle between Projections of Vectors via Geometric (Clifford)...James Smith
We express a problem from visual astronomy in terms of Geometric (Clifford) Algebra, then solve the problem by deriving expressions for the sine and cosine of the angle between projections of two vectors upon a plane. Geometric Algebra enables us to do so without deriving expressions for the projections themselves.
Estimation of the Earth's "Unperturbed" Perihelion from Times of Solstices an...James Smith
Published times of the Earth's perihelions do not refer to the perihelions of the orbit that the Earth would follow if unaffected by other bodies such as the Moon. To estimate the timing of that ``unperturbed" perihelion, we fit an unperturbed Kepler orbit to the timings of the year 2017's equinoxes and solstices. We find that the unperturbed 2017 perihelion, defined in that way, would occur 12.93 days after the December 2016 solstice. Using that result, calculated times of the year 2017's solstices and equinoxes differ from published values by less than five minutes. That degree of accuracy is sufficient for the intended use of the result.
Projection of a Vector upon a Plane from an Arbitrary Angle, via Geometric (C...James Smith
We show how to calculate the projection of a vector, from an arbitrary direction, upon a given plane whose orientation is characterized by its normal vector, and by a bivector to which the plane is parallel. The resulting solutions are tested by means of an interactive GeoGebra construction.
Kepler and Newton vs. Geocentrism, Flat Earth, and the "Vortex"James Smith
This presentation compares predictions of Kepler's 3rd Law to data from almost all of the orbital systems in the Solar System (including orbits of Pluto's moons), then predicts the height of a geostationary orbit.
Hablando francamente, la suma y resta de fracciones algebraicas suele ser muy molesto. Eso porque la realización de una sola suma puede requerir muchos pasos. Sin embargo, la lógica en la que cada paso se fundamenta, se entiende fácilmente relacionándola con los pasos análogos que se efectúan al sumar fracciones “normales”.
The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
Assessment and Planning in Educational technology.pptxKavitha Krishnan
In an education system, it is understood that assessment is only for the students, but on the other hand, the Assessment of teachers is also an important aspect of the education system that ensures teachers are providing high-quality instruction to students. The assessment process can be used to provide feedback and support for professional development, to inform decisions about teacher retention or promotion, or to evaluate teacher effectiveness for accountability purposes.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...PECB
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
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Additional Solutions of the Limiting Case "CLP" of the Problem of Apollonius via Vector Rotations using Geometric Algebra
1. Additional Solutions of the Limiting Case
“CLP” of the Problem of Apollonius
via Vector Rotations using 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 · ˆv]} 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
.
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. Additional Solutions of the Special Case “CLP”
of the Problem of Apollonius via Vector
Rotations using Geometric Algebra
Jim Smith
QueLaMateNoTeMate.webs.com
email: nitac14b@yahoo.com
August 27, 2016
Contents
1 Introduction 6
2 A brief review of reflections and rotations in 2D GA 7
2.1 Review of reflections . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1.1 Reflections of a single vector . . . . . . . . . . . . . . . . 7
2.1.2 Reflections of a bivector, and of a geometric product of
two vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Review of rotations . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 The Problem of Apollonius, and Its CLP Limiting Case 10
3.1 Observations, and Potentially Useful Elements of the Problem . . 10
3.1.1 Formulating a Strategy . . . . . . . . . . . . . . . . . . . 13
4 Solving the Problem 14
4.1 Solution via Rotations Alone . . . . . . . . . . . . . . . . . . . . 14
4.1.1 Solving for the Solution Circles that Do Not Enclose the
Given Circle . . . . . . . . . . . . . . . . . . . . . . . . . 14
4
5. 4.1.2 Solving for the Solution Circles that Do Enclose the Given
Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Solution via a Combination of Reflections and Rotations . . . . . 24
5 Literature Cited 28
6 Appendix: The Rotations-Only Method Used in the Original
Version of this Document 29
6.1 Identifying the Solution Circles that Don’t Enclose C . . . . . . 29
6.1.1 Formulating a Strategy . . . . . . . . . . . . . . . . . . . 29
6.1.2 Transforming and Solving the Equations that Resulted
from Our Observations and Strategizing . . . . . . . . . . 30
6.2 Identifying the Solution Circles that Enclose C . . . . . . . . . . 36
6.3 The Complete Solution, and Comments . . . . . . . . . . . . . . 38
5
6. ABSTRACT
This document adds to the collection of solved problems presented in [1]-[6].
The solutions presented herein are not as efficient as those in [6], but they give
additional insight into ways in which GA can be used to solve this problem.
After reviewing, briefly, how reflections and rotations can be expressed and ma-
nipulated via GA, it solves the CLP limiting case of the Problem of Apollonius
in three ways, some of which identify the the solution circles’ points of tangency
with the given circle, and others of which identify the solution circles’ points of
tangency with the given line. For comparison, the solutions that were developed
in [1] are presented in an Appendix.
1 Introduction
This document has been prepared for two very different audiences: for my fellow
students of GA, and for experts who are preparing materials for us, and need
to know which GA concepts we understand and apply readily, and which ones
we do not. In some sense, the solutions presented herein are “leftovers” from
work that was done to find more-efficient solutions, so the document is still in
somewhat of a rough condition, with a certain amount of repetition.
Readers are encouraged to study the following documents, GeoGebra work-
sheets, and videos before beginning:
“Rotations of Vectors via Geometric Algebra: Explanation, and Usage in
Solving Classic Geometric “Construction” Problems”
https://drive.google.com/file/d/0B2C4TqxB32RRdE5KejhQTzMtN3M/view?usp=sharing
“Answering Two Common Objections to Geometric Algebra”
As GeoGebra worksheet
As YouTube video.
“Geometric Algebra: Find unknown vector from two dot products”
As GeoGebra worksheet
As YouTube video
For an more-complete treatment of rotations in plane geometry, be sure to
read Hestenes D. 1999, pp. 78-92. His section on circles (pp. 87-89) is especially
relevant to the present document. Macdonald A. 2010 is invaluable in many
respects, and Gonz´alez Calvet R. 2001, Treatise of Plane Geometry through
Geometric Algebra is a must-read.
The author may be contacted at QueLaMateNoTeMate.webs.com.
6
7. 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)
7
8. 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).
8
9. 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
[?].
9
10. 3 The Problem of Apollonius, and Its CLP Lim-
iting Case
The famous “Problem of Apollonius”, in plane geometry, is to construct all of
circles that are tangent, simultaneously, to three given circles. In one variant
of that problem, one of the circles has infinite radius (i.e., it’s a line). The
Wikipedia article that’s current as of this writing has an extensive description
of the problem’s history, and of methods that have been used to solve it. As
described in that article, one of the methods reduces the “two circles and a line”
variant to the so-called “Circle-Line-Point” (CLP) limiting case:
Given a circle C, a line L, and a point P, construct the circles that
are tangent to C and L, and pass through P.
Figure 3.1: The CLP Limiting Case of the Problem of Apollonius: Given a
circle C, a line L, and a point P, construct the circles that are tangent to C and
L, and pass through P.
3.1 Observations, and Potentially Useful Elements of the
Problem
From the figure presented in the statement of the problem, we can see that
there are two types of solutions. That is, two types of circles that satisfy the
stated conditions:
• Circles that enclose C;
• Circles that do not enclose C.
As we will see later, there are actually two circles that do enclose the
given circle, and two that do not. We’ll begin by discussing circles that do not
10
11. enclose it. Most of our observations about that type will also apply, with little
modification, to circles that do enclose it.
Based upon our experience in solving other “construction problems” in-
volving tangency, a reasonable choice of elements for capturing the geometric
content of the problem is as shown below:
• Use the center point of the given circle as the origin;
Figure 3.2: Initial step in choosing elements for capturing the geometric content
of the problem. (See text for explanation.)
• Capture the perpendicular distance from c1’s center to the given line in
the vector h;
• Express the direction of the given line as ±ˆhi.
• Label the solution circle’s radius and its points of tangency with C and L
as shown in Fig. 3.3:
In deriving our solution, we’ll
use the same symbol —for
example, t —to denote both a
point and the vector to that
point from the origin. We’ll rely
upon context to tell the reader
whether the symbol is being
used to refer to the point, or to
the vector.
Now, we’ll express key features of the problem in terms of the elements
that we’ve chosen. First, we know that we can write the vector s as s =
h+λˆhi, where λ is some scalar. We can even identify that scalar, specifically, as
ˆt · ˆhi . We also know that the points of tangency t and s are equidistant (by
r2) from the center point of the solution circle. Combining those observations,
we can equate three expressions for the vector s:
s = (r1 + r2)ˆt + r2
ˆh = h + λˆhi = h + ˆt · ˆhi ˆhi. (3.1)
Examining that equation, we observe that we can obtain an expression for
11
12. Figure 3.3: Further steps in choosing elements for capturing the geometric
content of the problem. (See text for explanation.)
r2 in terms of known quantities by “dotting” both sides with ˆh:
(r1 + r2)ˆt + r2
ˆh · ˆh = h + λˆhi · ˆh
(r1 + r2)ˆt · ˆh + r2
ˆh · ˆh = h · ˆh + λ ˆhi · ˆh
(r1 + r2)ˆt · ˆh + r2 = h + 0;
∴ r2 =
h − r1
ˆt · ˆh
1 + ˆt · ˆh
. (3.2)
The denominator of the expression on the right-hand side might catch our
attention now because one of our two expressions for the vector s, namely
s = (r1 + r2)ˆt + r2
ˆh
can be rewritten as
s = r1
ˆt + r2
ˆt + ˆh .
That fact becomes useful (at least potentially) when we recognize that ˆt + ˆh
2
=
2 1 + ˆt · ˆh . Therefore, if we wish, we can rewrite Eq. (3.2) as
r2 = 2
h − r1
ˆt · ˆh
ˆt + ˆh
2
.
Those results indicate that we should be alert to opportunities to simplify
expressions via appropriate substitutions invoving ˆt + ˆh and 1 + ˆh · ˆt
As a final observation, we note that when a circle is tangent to other objects,
there will be many angles that are equal to each other. For example, the angles
whose measures are given as θ in Figs. 3.4 and 3.5:
12
13. Figure 3.4: A pair of equal angles that might provide a basis for solving the
problem via rotations.)
We also see in Fig. 3.5 that points T and S are reflections of each other
with respect to a line that passes through the center of the solution circle, and
has the direction of (t − s) i. References [1]-[6] showed how GA enables us to
make use of the reflections and rotations that we’ve noted.
Many of the ideas that we’ll employ here will also be used when we treat
solution circles that do enclose C.
3.1.1 Formulating a Strategy
Now, let’s combine our observations about the problem in a way that might
lead us to a solution. Our previous experiences in solving problems via vector
rotations suggest that we should equate two expressions for the rotation eθi
:
t − p
t − p
s − p
s − p
=
t − s
t − s
−ˆhi
=
s − t
s − t
ˆhi . (3.3)
We’ve seen elsewhere that we will probably want to transform that equation
into one in which some product of vectors involving our unknowns t and s is
equal either to a pure scalar, or a pure bivector. By doing so, we may find some
way of identifying either t or s.
We’ll keep in mind that although Eq. (3.3) has two unknowns (the vectors
t and s), our expression for r2 (Eq. (3.2)) enables us to write the vector s in
terms of the vector ˆt.
Therefore, our strategy is to
13
14. Figure 3.5: Another pair of equal angles that might provide a basis for solving
the problem via rotations. Note, too, that the points T and s are reflections of
each other with respect to the vector (t − s) i.
• Equate two expressions, in terms of the unknown vectors t and s, for the
rotation eθi
;
• Transform that equation into one in which on side is either a pure scalar
or a pure bivector;
• Watch for opportunities to simplify equations by substituting for r2; and
• Solve for our unknowns.
4 Solving the Problem
Our first solution, of which two variants are presented, uses rotations alone.
The second uses a combination of reflections and rotations. Reference [1] can
be studied for comparison.
4.1 Solution via Rotations Alone
4.1.1 Solving for the Solution Circles that Do Not Enclose the Given
Circle
We’ll start by examining Fig. 4.1, which is based upon Fig. 3.5. Using ideas
presented in Section 3.1, we can write
s − p
s − p
t − p
t − p
= eφi
= ˆh
i ˆt + ˆh
ˆt + ˆh
,
14
15. Figure 4.1: Elements used in finding the solution circles that do not enclose the
given one. The vector s has been written as (r1 + r2)ˆt + r2
ˆh, and we’ve made
use of the fact that i ˆt + ˆh has the same direction as i (s − t). (Compare Fig.
3.5.)
from which
s − p
s − p
t − p
t − p
= ˆh
i ˆt + ˆh
ˆt + ˆh
,
ˆh
s − p
s − p
t − p
t − p
=
i ˆt + ˆh
ˆt + ˆh
,
ˆh
s − p
s − p
t − p
t − p
=
i ˆt + ˆh
ˆt + ˆh
,
ˆh
s − p
s − p
t − p
t − p
ˆt + ˆh = i ˆt + ˆh , and
ˆh [s − p] [t − p] ˆt + ˆh = i ˆt + ˆh s − p t − p ,
the right-hand side of which is a bivector. Therefore,
ˆh [s − p] [t − p] ˆt + ˆh 0 = 0.
We can save ourselves some work by expanding the left-hand side in a way that
let’s us take advantage of what we’ve learned about the form in which GA can
express reflections:
ˆh [s − p] [t − p]ˆt 0 + ˆh [s − p] [t − p] ˆh 0 = 0. (4.1)
Note that the product ˆh [s − p] [t − p] ˆh is just the reflection of [s − p] [t − p]
with respect to ˆh, and is therefore equal to [t − p] [s − p]. Let’s find the scalar
part of that product, then return to the first term in 4.1. We’ll use the substitu-
tions t = r1
ˆt; r2 =
h − r1
ˆh · ˆt
1 + ˆh · ˆt
; r1 +r2 =
r1 + h
1 + ˆh · ˆt
; and s = (r1 + r2)ˆt+r2
ˆh
15
16. to write
[t − p] [s − p] = r1
ˆt − p
r1 + h
1 + ˆh · ˆt
ˆt +
h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh − p
= r1
ˆt − p
(r1 + h )ˆt + h − r1
ˆh · ˆt ˆh − 1 + ˆh · ˆt p
1 + ˆh · ˆt
.
That last step—putting everything in the last factor over the common
denominator 1 + ˆh · ˆt —is quite a natural one to take, but turns out to be
unnecessary. By taking it, I complicated solutions in other problems needlessly,
by forming awkward products like p · ˆt ˆh · ˆt . Later in this solution we’ll see
why that step is unnecessary, but for now, let’s go back to
[t − p] [s − p] = r1
ˆt − p
r1 + h
1 + ˆh · ˆt
ˆt +
h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh − p ,
then continue to expand the right-hand side:
[t − p] [s − p] = r1
ˆt − p
r1 + h
1 + ˆh · ˆt
ˆt +
h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh − p
=
r1 (r1 + h )
1 + ˆh · ˆt
ˆtˆt +
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆtˆh
−r1
ˆtp −
r1 + h
1 + ˆh · ˆt
pˆt −
h − r1
ˆh · ˆt
1 + ˆh · ˆt
pˆh + p2
,
the scalar part of which is
[t − p] [s − p] 0 =
r1 (r1 + h )
1 + ˆh · ˆt
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh · ˆt
−r1p · ˆt −
r1 + h
1 + ˆh · ˆt
p · ˆt −
h − r1
ˆh · ˆt
1 + ˆh · ˆt
p · ˆh + p2
.
16
17. Now, let’s find ˆh [s − p] [t − p]ˆt 0 from 4.1:
ˆh [s − p] [t − p]ˆt = ˆh
r1 + h
1 + ˆh · ˆt
ˆt +
h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh − p r1
ˆt − p ˆt
=
r1 (r1 + h )
1 + ˆh · ˆt
ˆhˆtˆtˆt −
r1 + h
1 + ˆh · ˆt
ˆhˆtpˆt
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆhˆhˆtˆt −
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆhˆhpˆt
−r1
ˆhpˆtˆt + ˆhppˆt
=
r1 (r1 + h )
1 + ˆh · ˆt
ˆhˆt −
r1 + h
1 + ˆh · ˆt
ˆhˆtpˆt
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
−
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
pˆt
−r1
ˆhp + p2 ˆhˆt.
The scalar part of ˆhˆtpˆt is ˆhˆt 2p · ˆt − ˆtp 0 = 2 p · ˆt ˆh · p − ˆh · p.
Therefore,
ˆh [s − p] [t − p]ˆt 0 =
r1 (r1 + h )
1 + ˆh · ˆt
ˆh · ˆt
−
r1 + h
1 + ˆh · ˆt
2 p · ˆt ˆh · p − ˆh · p
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
−
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
p · ˆt
−r1
ˆh · p + p2 ˆh · ˆt.
As we noted in 6.1,
ˆh [s − p] [t − p]ˆt 0 + ˆh [s − p] [t − p] ˆh 0 = 0.
Adding our the expressions that we’ve just developed for ˆh [s − p] [t − p]ˆt 0
17
18. and ˆh [s − p] [t − p] ˆh 0, we obtain
r1 (r1 + h )
1 + ˆh · ˆt
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
ˆh · ˆt
−r1p · ˆt −
r1 + h
1 + ˆh · ˆt
p · ˆt −
h − r1
ˆh · ˆt
1 + ˆh · ˆt
p · ˆh + p2
+
r1 (r1 + h )
1 + ˆh · ˆt
ˆh · ˆt −
r1 + h
1 + ˆh · ˆt
2 p · ˆt ˆh · p − ˆh · p
+
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
−
r1 h − r1
ˆh · ˆt
1 + ˆh · ˆt
p · ˆt
−r1
ˆh · p + p2 ˆh · ˆt = 0. (4.2)
Note the pairs of expressions in red and blue: those pairs sum, respectively, to
r1 (r1 + h ) and r1 h − r1
ˆh · ˆt . That fact is one of the reasons why we
didn’t need to put all of the terms over the common denominator 1+ ˆh·ˆt. (See
the discussion that follows 4.1.)
After simplification, 4.2 reduces to
2 (r1 + h ) p · ˆt − p2
− r1
2 ˆh · ˆt = 2 h r1 + r1
2
+ p2
.
We saw equations like this last one many times in Smith J A 2016. There,
we learned to solve those equations by grouping the dot products that involve
t into a dot product of t with a linear combination of known vectors:
2 (r1 + h ) p − p2
− r1
2 ˆh
A linear combination of ˆh and p
·ˆt = 2 h r1 + r1
2
+ p2
. (4.3)
The geometric interpretation of Eq. (4.3) is that 2 h r1 + r1
2
+ p2
is the
projection of the vector 2 (r1 + h ) p − p2
− r1
2 ˆh upon ˆt. Because we want
to find t, and know 2 (r1 + h ) p − p2
− r1
2 ˆh, we’ll transform Eq. (4.3)
into a version that tells us the projection of the vector t upon 2 (r1 + h ) p −
p2
− r1
2 ˆh.
First, just for convenience, we’ll multiply both sides of Eq. (4.3) by r1 h :
2 r1 h + h2
p − p2
− r1
2
h · t = 2h2
r1
2
+ r1 h r1
2
+ p2
.
Next, we’ll use the symbol “w” for the vector 2 r1 h + h2
p − p2
− r1
2
h ,
and write
w · t = 2h2
r1
2
+ r1 h r1
2
+ p2
.
18
19. Finally, because P w (t), the projection of the vector t upon w is (t · ˆw) ˆw,
we have
P w (t) =
2h2
r1
2
+ r1 h r1
2
+ p2
w
ˆw. (4.4)
As we learned in Smith J A 2016, Eq. (4.4) tells us that Eq (4.3) has two
solutions. That is, there are two circles that are tangent to L and pass through
the point P, and are also tangent to C without enclosing it:
Figure 4.2: The two solution circles that do not enclose the given circle, C.
Having identified P w (t), the points of tangency with C and L can be
determined using methods shown in Smith J A 2016, as can the equations for
the corresponding solution circles.
To round off our treatment of solution circles that don’t enclose C, we
should note that we derived our solution starting from equations that express
the relationship between C, L, P, and the smaller of the two solution circles.
You may have noticed that the larger solution circle does not bear quite the
same relationship to L, P, and C as the smaller one. To understand in what
way those relationships differ, please see the discussion in 6.
4.1.2 Solving for the Solution Circles that Do Enclose the Given
Circle
To introduce some new ideas and show how to employ useful GA identities,
we’ll follow a somewhat different route in this section than we did when finding
the solution circles that don’t enclose the given one. Instead of expressing s
—the point of tangency with the given line—as a linear combination of ˆt and
19
20. ˆh, we’ll express it as a linear combination of h and ˆhi: s = h + λˆh, where λ
is a scalar. Because λˆhi is the projection of (r1 − r3)ˆt upon ˆhi (Fig. 4.3), we
can identify λ, specifically, as (r1 − r3) ˆt · ˆhi .
Figure 4.3: The two solution circles that do enclose the given circle, C. The vec-
tor ˆt − ˆh i has the same direction as (t − s) i. Because λˆhi is the projection
of (r1 − r3)ˆt upon ˆhi, we can identify λ, specifically, as (r1 − r3) ˆt · ˆhi .
At this point, we can foresee needing to derive an expression for r3 that’s
analogous to that which we presented in Eq. 3.2 for r2. We’ll do so by equating
two expressions for s, then ”dotting” with ˆh:
(r1 − r3)ˆt + r3
ˆh = h + λˆh
(r1 − r3)ˆt + r3
ˆh · ˆh = h + λˆhi · ˆh
(r1 − r3)ˆt · ˆh + r3 = h
∴ r3 =
h − r1
1 − ˆh · ˆt
, and r1 − r3 =
r1 − h
1 − ˆh · ˆt
. (4.5)
Now we’re ready to begin to solve for t. We could start by equating
two expressions for eωi, but just to be different we’ll note that the product
t − p
t − p
s − p
s − p
, which is equal to eωi, is a rotation operator that brings
ˆt − ˆh i into alignment with ˆh:
ˆt − ˆh i
ˆt − ˆh
t − p
t − p
s − p
s − p
= ˆh,
20
21. which we then transform via
ˆt − ˆh i
ˆt − ˆh
t − p
t − p
s − p
s − p
= ˆh,
i
ˆt − ˆh
ˆt − ˆh
t − p
t − p
s − p
s − p
= −ˆh,
ˆt − ˆh
ˆt − ˆh
t − p
t − p
s − p
s − p
= iˆh,
ˆt − ˆh
ˆt − ˆh
t − p
t − p
s − p
s − p
ˆh = i,
ˆt − ˆh [t − p] [s − p] ˆh 0 = 0. (4.6)
In addition, we’ll use the Fourier expansion ˆt = ˆt · ˆh ˆh + ˆt · ˆhi ˆhi
to transform the factor ˆt − ˆh:
ˆt − ˆh = ˆt · ˆh ˆh + ˆt · ˆhi ˆhi
= ˆh · ˆt − 1 ˆh + ˆhi · ˆt ˆhi.
Now, we’ll substitute that expression in (4.6), and make an initial expan-
sion:
ˆt − ˆh [t − p] [s − p] ˆh 0 = 0,
ˆh · ˆt − 1 ˆh + ˆhi · ˆt ˆhi [t − p] [s − p] ˆh 0 = 0,
from which
ˆh · ˆt − 1 ˆh [t − p] [s − p] ˆh 0
+ ˆhi · ˆt ˆhi [t − p] [s − p] ˆh 0 = 0
and
ˆh · ˆt − 1 ˆh [t − p] [s − p] ˆh 0
− ˆhi · ˆt ˆh [t − p] [s − p] ˆh i 0 = 0.
Both of the terms within the angle brackets are reflections of the product
[t − p] [s − p] with respect to ˆh. Therefore,
ˆh · ˆt − 1 [s − p] [t − p] 0
− ˆhi · ˆt {[s − p] [t − p]} i 0 = 0,
from which
ˆh · ˆt − 1 [s − p] · [t − p]
− ˆhi · ˆt {[s − p] ∧ [t − p]} i = 0. (4.7)
21
22. We’ll treat each of those terms separately. From (4.5), and from s =
h + λˆhi = h + (r1 − r3) ˆt · ˆhi ˆhi, ˆh · ˆt − 1 [s − p] · [t − p] is
ˆh · ˆt − 1 h +
r1 − h
1 − ˆh · ˆt
ˆt · ˆhi ˆhi − p · r1
ˆt · ˆh ˆh + r1
ˆt · ˆhi ˆhi − p .
We needn’t put the terms in
curly brackets over the common
denominator 1 − ˆh · ˆt, because
that denominator is canceled by
the factor ˆh · ˆt − 1, leaving -1.
After expansion, that result simplifies to
ˆh · ˆt − 1 [s − p] · [t − p] = − h ˆh · p ˆh · ˆt − h ˆhi · p ˆhi · ˆt
+r1 h ˆh · ˆt
2
+ r1 h ˆhi · ˆt
2
+2r1
ˆhi · p ˆhi · ˆt − r1 h ˆh · ˆt
+ h ˆh · p − r1
2 ˆhi · ˆt
2
− r1
ˆh · p ˆh · ˆt
2
+r1
ˆh · p ˆh · ˆt − r1
ˆhi · p ˆhi · ˆt ˆh · ˆt
+p2 ˆh · ˆt − p2
.
We could simplify even further by recognizing that
ˆh · p ˆh · ˆt + ˆhi · p ˆhi · ˆt = p · ˆt,
so that the terms in red reduce to − h p · ˆt. However, we will not do so
because our goal is to obtain a solution in the form of a linear combination of
ˆh and ˆhi. The terms in blue are a different story: they sum to r1 h because
ˆh · ˆt
2
+ ˆhi · ˆt
2
= ˆt
2
= 1. Therefore,
ˆh · ˆt − 1 [s − p] · [t − p] = − h ˆh · p ˆh · ˆt − h ˆhi · p ˆhi · ˆt
+r1 h + 2r1
ˆhi · p ˆhi · ˆt − r1 h ˆh · ˆt
+ h ˆh · p − r1
2 ˆhi · ˆt
2
− r1
ˆh · p ˆh · ˆt
2
+r1
ˆh · p ˆh · ˆt − r1
ˆhi · p ˆhi · ˆt ˆh · ˆt
+p2 ˆh · ˆt − p2
. (4.8)
Now, we’ll treat the term ˆhi · ˆt {[s − p] ∧ [t − p]} i from Eq. (4.7). I’ll
go into some detail so that you may see the interesting uses of GA identities
that arise.
Using the same substitutions for r3 and λ as we did when deriving (4.8),
ˆhi · ˆt {[s − p] ∧ [t − p]} i is
ˆhi · ˆt h +
r1 − h
1 − ˆh · ˆt
ˆt · ˆhi ˆhi − p ∧ r1
ˆt − p .
22
23. Expanding that product using the identity a ∧ b ≡ [(ai) · b] i ≡ − [a · (bi)] i,
we find that
ˆhi · ˆt {[s − p] ∧ [t − p]} i = ˆhi · ˆt r1
ˆh ˆhi · ˆt i − ˆh ˆhi · p i
−
r1 (r1 − h ) ˆhi · ˆt ˆh · ˆt i
1 − ˆh · ˆt
−
(r1 − h ) ˆhi i · p ˆhi · ˆt i
1 − ˆh · ˆt
− r1
ˆh · ˆt (pi) · ˆh i
−r1
ˆhi · ˆt (pi) · ˆhi i i. (4.9)
As was the case when we expanded and simplified ˆh · ˆt − 1 [s − p]·[t − p],
the presence of 1 − ˆh · ˆt in the denominators of some of the terms is not
a problem because the only terms with that denominator contain the factor
ˆhi · ˆt in their numerators. Thus, when we multiply through by the factor
ˆhi · ˆt that’s outside the curly brackets, the numerators of those terms will
contain the factor ˆhi · ˆt
2
, which is equal to 1 − ˆh · ˆt
2
, and therefore to
1 − ˆh · ˆt 1 + ˆh · ˆt . The factor 1 − ˆh · ˆt in that product will cancel with
the 1 − ˆh · ˆt in the denominator, leaving 1 + ˆh · ˆt in the numerator.
Making use of that observation about the cancellation of the denominators
in Eq. (4.9), we can further expand and simplify terms in that equation to
obtain
ˆhi · ˆt {[s − p] ∧ [t − p]} = − r1 h ˆhi · ˆt
2
− h ˆhi · p ˆhi · ˆt
− r1 (r1 − h ) ˆh · ˆt − r1 (r1 − h ) ˆh · ˆt
2
+ (r1 − h ) ˆh · p + (r1 − h ) ˆh · p ˆh · ˆt
− r1 (pi) · ˆh ˆhi · ˆt ˆh · ˆt
−r1
ˆh · p ˆhi · ˆt
2
. (4.10)
In (Eq. (4.7)), we noted that
ˆh · ˆt − 1 [s − p] · [t − p] − ˆhi · ˆt {[s − p] ∧ [t − p]} i = 0.
Substituting in that equation the expressions for ˆh · ˆt − 1 [s − p]·[t − p] and
23
24. ˆhi · ˆt {[s − p] ∧ [t − p]} i in Eqs. (4.8) and (4.10),
h ˆh · p ˆh · ˆt − h ˆhi · p ˆhi · ˆt
+r1 h + 2r1
ˆhi · p ˆhi · ˆt − r1 h ˆh · ˆt
+ h ˆh · p − r1
2 ˆhi · ˆt
2
− r1
ˆh · p ˆh · ˆt
2
+r1
ˆh · p ˆh · ˆt − r1
ˆhi · p ˆhi · ˆt ˆh · ˆt
+p2 ˆh · ˆt − p2
+r1 h ˆhi · ˆt
2
− h ˆhi · p ˆhi · ˆt
−r1 (r1 − h ) ˆh · ˆt − r1 (r1 − h ) ˆh · ˆt
2
+ (r1 − h ) ˆh · p + (r1 − h ) ˆh · p ˆh · ˆt
−r1 (pi) · ˆh ˆhi · ˆt ˆh · ˆt − r1
ˆh · p ˆhi · ˆt
2
= 0.
In simplifying that equation, we again make use of the identity ˆh · ˆt
2
+
ˆhi · ˆt
2
= ˆt
2
= 1. The result is
2r1 h − h ˆh · p ˆh · ˆt − h ˆhi · p ˆhi · ˆt − r1
2
+ (2r1 − h ) ˆh · p ˆh · ˆt + (2r1 − h ) ˆhi · p ˆhi · ˆt
+p2 ˆh · ˆt − p2
− r1
2 ˆh · ˆt = 0.
As we did for the solution circles that don’t enclose the given, we transform the
equation that we’ve just obtained into one involving the dot product of t with
a linear combination of known vectors:
2 ( h − r1) ˆh · p + r1
2
− p2 ˆh
+ 2 ( h − r1) ˆhi · p ˆhi · t = 2r1 h − p2
− r1
2
. (4.11)
4.2 Solution via a Combination of Reflections and Rota-
tions
For more information on the ideas used in this solution, please see [2]. The key
ideas that we’ll use, with reference to Fig. 4.2, are:
• The vector s can be expressed as the sum of h and a scalar multiple (λ )
of ˆhi;
• Our goal will be to determine the values of λ for the four solution circles;
• The points s and t are reflections of each other with respect to the vector
i (s − t), which is the mediatrix of the base of the isosceles triangle tC2s;
24
25. • The isosceles triangle C1C2M is similar to tC2s. Therefore, the
points s and t are reflections of each other with respect to the vector
i h + r1
ˆh + λˆhi . The advantage of using that vector (as opposed to
s − t) for expressing reflections is that it is written in terms of λ, and has
the origin (C1) as its starting point.
• Similarly, the vectors ˆt and −ˆh are reflections of each other with respect
to i h + r1
ˆh + λˆhi .
Figure 4.4: Elements used in using a combination of reflections and rotations
to identify the two solution circles that do not enclose the given circle, C.
We’ll incorporate those observations in our usual technique of equating two
expressions for the same rotation, which in this case will be eθi:
t − p
t − p
s − p
s − p
=
i h + r1
ˆh + λˆhi
h + r1
ˆh + λˆhi
ˆh ;
∴ [t − p] [s − p] ˆh h + r1
ˆh + λˆhi = a bivector, and
ts − tp − ps + p2 ˆh h + r1
ˆh + λˆhi 0 = 0.
We’ll expand that result, initially, as
(ts) ˆh h + r1
ˆh + λˆhi 0
− (tp) ˆh h + r1
ˆh + λˆhi 0
− (ps) ˆh h + r1
ˆh + λˆhi 0
+ p2 ˆh h + r1
ˆh + λˆhi 0 = 0. (4.12)
25
26. To expand each of those four terms, we´ll write s as s = h+λˆhi = h ˆh+λˆhi,
and h + r1
ˆh + λˆhi as (r1 + h ) ˆh + λˆhi. In addition, we’ll write ˆt as the
reflection of −ˆh with respect to i h + r1
ˆh + λˆhi :
ˆt =
i h + r1
ˆh + λˆhi
h ˆh + r1
ˆh + λˆhi
−ˆh
i h + r1
ˆh + λˆhi
h ˆh + r1
ˆh + λˆhi
=
(r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
,
from which
t =
r1 (r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
. (4.13)
The factors in red constitute a
rotation operator. See 2.3.
Now, for the expansions of each of the four terms in (4.12).
(ts) ˆh h + r1
ˆh + λˆhi
=
r1 (r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
h ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
=
r1 (r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi h ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
= r1 (r1 + h ) ˆh + λˆhi h ˆh + λˆhi
∴ (ts) ˆh ˆh h + r1
ˆh + λˆhi 0 = r1
2
h + r1h2
+ r1λ2
. (4.14)
The factors in red constitute a
rotation operator. See 2.3.
(tp) ˆh h + r1
ˆh + λˆhi
=
r1 (r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
[p] ˆh (r1 + h ) ˆh + λˆhi
=
r1 (r1 + h ) ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi [p] ˆh (r1 + h ) ˆh + λˆhi
(r1 + h )
2
+ λ2
= r1 (r1 + h ) ˆh + λˆhi [p]
∴ (tp) ˆh h + r1
ˆh + λˆhi 0 = r1 (r1 + h ) ˆh · p − λr1
ˆhi · p . (4.15)
(ps) ˆh h + r1
ˆh + λˆhi
= [p] h ˆh + λˆhi ˆh (r1 + h ) ˆh + λˆhi
∴ (ps) ˆh h + r1
ˆh + λˆhi 0 = r1 h + h2
p · ˆh − λ (r1 + 2 h ) p · ˆh + λ2
p · ˆh.
(4.16)
26
27. p2 ˆh h + r1
ˆh + λˆhi =
(r1 + h ) p2
+ λp2
i
∴ p2 ˆh h + r1
ˆh + λˆhi 0 = (r1 + h ) p2
. (4.17)
If we wished, we could now
make the substitution
λ = (r1 + r2) ˆt · ˆhi (see
Fig. 3.3), then proceed to
transform this quadratic into
Eq. (4.1.1).
Using the scalar parts that we identified in Eqs. (4.14)-(4.17), we find that
(4.12) reduces to the following quadratic in λ:
r1 + p · ˆh λ2
− 2 (r1 + h ) p · ˆhi λ
+ (r1 + h ) p2
+ r1 h − (r1 + h )
2
p · ˆh = 0.
Its solutions are
λ =
(r1 + h ) p · ˆhi ± (p2 − r1
2) (r1 + h ) [(h − p)] · ˆh
r1 + p · ˆh
. (4.18)
The interpretation of that result is that there are two solution circles that don’t
enclose C, and each of those circles is tangent to L at its own distance λ from
the point at which h intersects L.
To find the solution circles that do enclose C, we start from Fig. 4.5.
Figure 4.5: Elements used in using a combination of reflections and rotations
to identify the solution circles that enclose the given circle, C.
This time, we must express ˆt as a reflection of ˆh with respect to i h − r1
ˆh + λˆhi .
From there, the procedure is the same as we used to find the solution circles
that do not enclose C. But note: ˆt for solution circles that do enclose C is the re-
flection of +ˆh with respect to i h − r1
ˆh + λˆhi , whereas ˆt for solution circles
27
28. that don’t enclose C is the reflection of −ˆh with respect to i h + r1
ˆh + λˆhi .
The result, for solution circles that do enclose C, is
λ =
(r1 − h ) p · ˆhi ± (p2 − r1
2) ( h − r1) [(h − p)] · ˆh
r1 − p · ˆh
. (4.19)
5 Literature Cited
References
[1] “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.
[2] “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.
[3] 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 .
[4] J. Smith, “A Very Brief Introduction to Reflections in 2D Geometric
Algebra, and their Use in Solving ‘Construction’ Problems”. Available
at http://vixra.org/abs/1606.0253.
[5] J. Smith, “Three Solutions of the LLP Limiting Case of the Problem
of Apollonius via Geometric Algebra, Using Reflections and Rotations”.
Available at http://vixra.org/abs/1607.0166.
[6] J. Smith, “Simplified Solutions of the CLP and CCP Limiting Cases of the
Problem of Apollonius via Vector Rotations using Geometric Algebra”.
Available at http://vixra.org/abs/1608.0217.
GeoGebra Worksheets and Related Videos (by title, in alphabetical order):
”Answering Two Common Objections to Geometric Algebra”
GeoGebra worksheet: http://tube.geogebra.org/m/1565271
YouTube video: https://www.youtube.com/watch?v=oB0DZiF86Ns.
”Geometric Algebra: Find unknown vector from two dot products”
GeoGebra worksheet:
http://tube.geogebra.org/material/simple/id/1481375
YouTube video:
https://www.youtube.com/watch?v=2cqDVtHcCoE
Books and articles (according to author, in alphabetical order):
28
29. Gonz´alez Calvet R. 2001. Treatise of Plane Geometry through Geometric
Algebra. 89.218.153.154:280/CDO/BOOKS/Matem/Calvet.Treatise.pdf.
Retrieved 30 December 2015.
Hestenes D. 1999. New Foundations for Classical Mechanics (Second
Edition). Kluwer Academic Publishers, Dordrecht/Boston/London.
Macdonald A. 2010. Linear and Geometric Algebra. CreateSpace Independent
Publishing Platform, ASIN: B00HTJNRLY.
Smith J. A. 2014. “Resoluciones de ’problemas de construcci´on’ geom´etricos
por medio de la geometr´ıa cl´asica, y el ´Algebra Geom´etrica”,
http://quelamatenotemate.webs.com/3%20soluciones
%20construccion%20geom%20algebra%20Cliffod.xps
2016. “Rotations of Vectors via Geometric Algebra: Explanation, and
Usage in Solving Classic Geometric “Construction” Problems”
https://drive.google.com/file/d/0B2C4TqxB32RRdE5KejhQTzMtN3M/view?usp=sharing
6 Appendix: The Rotations-Only Method Used
in the Original Version of this Document
6.1 Identifying the Solution Circles that Don’t Enclose C
Many of the ideas that we’ll employ here will also be used when we treat solution
circles that do enclose C.
6.1.1 Formulating a Strategy
Now, let’s combine our observations about the problem in a way that might
lead us to a solution. Our previous experiences in solving problems via vector
rotations suggest that we should equate two expressions for the rotation eθi
:
t − p
t − p
s − p
s − p
=
t − s
t − s
−ˆhi
=
s − t
s − t
ˆhi . (6.1)
We’ve seen elsewhere that we will probably want to transform that equation
into one in which some product of vectors involving our unknowns t and s is
equal either to a pure scalar, or a pure bivector. By doing so, we may find some
way of identifying either t or s.
We’ll keep in mind that although Eq. (6.1) has two unknowns (the vectors
t and s), our expression for r2 (Eq. (3.2)) enables us to write the vector s in
29
30. terms of the vector ˆt.
Therefore, our strategy is to
• Equate two expressions, in terms of the unknown vectors t and s, for the
rotation eθi
;
• Transform that equation into one in which on side is either a pure scalar
or a pure bivector;
• Watch for opportunities to simplify equations by substituting for r2; and
• Solve for our unknowns.
6.1.2 Transforming and Solving the Equations that Resulted from
Our Observations and Strategizing
For convenience, we’ll present our earlier figure again:
By examining that figure, we identified and equated two expressions for
the rotation eθi
, thereby obtaining Eq. (6.1):
t − p
t − p
s − p
s − p
=
s − t
s − t
ˆhi .
We noted that we might wish at some point to make the substitution
s = [r1 + r2]ˆt + r2
ˆh
= r1 +
h − r1
ˆt · ˆh
1 + ˆt · ˆh
ˆt +
h − r1
ˆt · ˆh
1 + ˆt · ˆh
ˆh. (6.2)
We also noted that we’ll want to transform Eq. (6.1) into one in which one
side is either a pure scalar or a pure bivector. We should probably do that
30
31. transformation before making the substitution for s. One way to effect the
transformation is by left-multiplying both sides of Eq. (6.1) by s − t, then by
ˆh, and then rearranging the result to obtain
ˆh [s − t] [t − p] [s − p] = s − t t − p s − p i (6.3)
This is the equation that we sought to obtain, so that we could now write
ˆh [s − t] [t − p] [s − p] 0 = 0. (6.4)
Next, we need to expand the products on the left-hand side, but we’ll want
to examine the benefits of making a substitution for s first. We still won’t, as
yet, write s in terms of ˆt. In hopes of keeping our equations simple enough
for us to identify useful simplifications easily at this early stage, we’ll make the
substitution
s = (r1 + r2)ˆt + r2
ˆh,
rather than making the additional substitution (Eq. (3.2)) for r2. Now, we can
see that s − t = r2
ˆt + ˆh . Using this result, and t = r1
ˆt, Eq. (6.4) becomes
ˆh r2
ˆt + ˆh r1
ˆt − p (r1 + r2)ˆt + r2
ˆh − p 0 = 0.
Now here is where I caused myself a great deal of unnecessary work in
previous versions of the solution by plunging in and expanding the product
that’s inside the 0 without examining it carefully. Look carefully at the last
factor in that product. Do you see that we can rearrange it to give the following?
ˆh r2
ˆt + ˆh r1
ˆt − p r2
ˆt + ˆh + r1
ˆt − p
After rearrangement
0 = 0.
That result is interesting, but is it truly useful to us? To answer that question,
let’s consider different ways in which we might expand the product, then find
its scalar part.
If we effect the multiplications in order, from left to right, we’re likely to
end up with a confusing mess. However, what if we multiply the last three
factors together? Those three factors, together, compose a product of the form
ab [a + b]:
r2
ˆt + ˆh
a
r1
ˆt − p
b
r2
ˆt + ˆh
a
+ r1
ˆt − p
b
.
The expansion of ab [a + b] is
ab [a + b] = aba + b2
a
= 2 (a · b) a − a2
b + b2
a (among other possibilites).
That expansion evaluates to a vector, of course. Having obtained the cor-
responding expansion of the product r2
ˆt + ˆh r1
ˆt − p r2
ˆt + ˆh + r1
ˆt − p , we’d
31
32. then “dot” the result with ˆh to obtain ˆh r2
ˆt + ˆh r1
ˆt − p r2
ˆt + ˆh + r1
ˆt − p 0.
We know, from the solutions to Problem 6 in Smith J A 2016 , that such a ma-
neuver can work out quite favorably. So, let’s try it.
Expanding r2
ˆt + ˆh r1
ˆt − p r2
ˆt + ˆh + r1
ˆt − p according to the identity
ab [a + b] = 2 (a · b) a − a2
b + b2
a, we obtain, initially,
2 r2
2 ˆt + ˆh · r1
ˆt − p ˆt + ˆh − r2
2 ˆt + ˆh
2
r1
ˆt − p + r2 r1
ˆt − p
2
ˆt + ˆh
When we’ve completed our expansion and dotted it with ˆh, we’ll set the result
to zero, so let’s divide out the common factor r2 now:
2 r2
ˆt + ˆh · r1
ˆt − p ˆt + ˆh − r2
ˆt + ˆh
2
r1
ˆt − p + r1
ˆt − p
2
ˆt + ˆh
Recalling that ˆt + ˆh
2
= 2 1 + ˆh · ˆt , the preceding becomes
2 r2
ˆt + ˆh · r1
ˆt − p ˆt + ˆh − 2r2 1 + ˆh · ˆt r1
ˆt − p + r1
ˆt − p
2
ˆt + ˆh .
This is the form that we’ll dot with ˆh. Having done so, the factor ˆt+ ˆh becomes
1 + ˆh · ˆt. Then, as planned, we set the resulting expression equal to zero:
2 r2
ˆt + ˆh · r1
ˆt − p 1 + ˆh · ˆt − 2r2 1 + ˆh · ˆt r1
ˆh · ˆt − ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0.
Next, we’ll rearrange that equation to take advantage of the relation r2 =
h − r1
ˆh · ˆt
1 + ˆh · ˆt
(see Eq. (3.2)). We’ll show the steps in some detail:
2 r2
ˆt + ˆh · r1
ˆt − p 1 + ˆh · ˆt − 2r2 1 + ˆh · ˆt r1
ˆh · ˆt − ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0.
2r2 1 + ˆh · ˆt ˆt + ˆh · r1
ˆt − p − r1
ˆh · ˆt + ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0
2
h − r1
ˆh · ˆt
1 + ˆh · ˆt
1 + ˆh · ˆt ˆt + ˆh · r1
ˆt − p − r1
ˆh · ˆt + ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0
2 h − r1
ˆh · ˆt ˆt + ˆh · r1
ˆt − p − r1
ˆh · ˆt + ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0
2 h − r1
ˆh · ˆt r1 − p · ˆt + r1
ˆh · ˆt − ˆh · p − r1
ˆh · ˆt + ˆh · p + r1
ˆt − p
2
1 + ˆh · ˆt = 0
2 h − r1
ˆh · ˆt r1 − p · ˆt + r1
ˆt − p
2
1 + ˆh · ˆt = 0.
Now that the dust has settled from the r2 substitution, we’ll expand r1
ˆt − p
2
,
then simplify further:
2 h − r1
ˆh · ˆt r1 − p · ˆt + r1
2
− 2r1p · ˆt + p2
1 + ˆh · ˆt = 0
−2r1
2
+ 2r1p · ˆt + r1
2
− 2r1
ˆt · p + p2 ˆh · ˆt − 2 h p · ˆt + 2 h r1 + r1
2
− 2r1p · ˆt + p2
= 0
p2
− r1
2 ˆh · ˆt − 2 (r1 + h ) p · ˆt + r1
2
+ p2
+ 2 h r1 = 0.
We saw equations like this last one many times in Smith J A 2016. There,
we learned to solve those equations by grouping the dot products that involve
t into a dot product of t with a linear combination of known vectors:
2 (r1 + h ) p − p2
− r1
2 ˆh
A linear combination of ˆh and p
·ˆt = 2 h r1 + r1
2
+ p2
. (6.5)
32
33. The geometric interpretation of Eq. (6.5) is that 2 h r1 + r1
2
+ p2
is the
projection of the vector 2 (r1 + h ) p − p2
− r1
2 ˆh upon ˆt. Because we want
to find t, and know 2 (r1 + h ) p − p2
− r1
2 ˆh, we’ll transform Eq. (6.5)
into a version that tells us the projection of the vector t upon 2 (r1 + h ) p −
p2
− r1
2 ˆh.
First, just for convenience, we’ll multiply both sides of Eq. (6.5) by r1 h :
2 r1 h + h2
p − p2
− r1
2
h · t = 2h2
r1
2
+ r1 h r1
2
+ p2
.
Next, we’ll use the symbol “w” for the vector 2 r1 h + h2
p − p2
− r1
2
h ,
and write
w · t = 2h2
r1
2
+ r1 h r1
2
+ p2
.
Finally, because P w (t), the projection of the vector t upon w is (t · ˆw) ˆw,
we have
P w (t) =
2h2
r1
2
+ r1 h r1
2
+ p2
w
ˆw. (6.6)
As we learned in Smith J A 2016, Eq. (6.6) tells us that Eq (6.5) has two
solutions. That is, there are two circles that are tangent to L and pass through
the point P, and are also tangent to C without enclosing it:
Having identified P w (t), the points of tangency with C and L can be
determined using methods shown in Smith J A 2016, as can the equations for
the corresponding solution circles.
To round off our treatment of solution circles that don’t enclose C, we
should note that we derived our solution starting from equations that express
33
34. the relationship between C, L, P, and the smaller of the two solution circles.
You may have noticed that the larger solution circle does not bear quite the
same relationship to L, P, and C as the smaller one. To understand in what
way those relationships differ, let’s examine the following figure.
By equating two expressions for the rotation eψi
, we’d find that
s − p
s − p
t − p
t − p
= ˆhi
t − s
t − s
.
Compare that result to the corresponding equation for the smaller of the solution
circles:
t − p
t − p
s − p
s − p
=
s − t
s − t
ˆhi .
We followed up on that equation by transforming it into one in which ˆh was at
one end of the product on the left-hand side. The result was Eq. (6.3):
ˆh [s − t] [t − p] [s − p] = s − t t − p s − p i.
We saw the advantages of that arrangement when we proceeded to solve for t.
All we had to do in order to procure that arrangement was to left-multiply both
sides of the equation
t − p
t − p
s − p
s − p
=
s − t
s − t
ˆhi
by s − t, and then by ˆh.
To procure a similar arrangement starting from the equation that we wrote
for the larger circle, using the angle ψ ,
s − p
s − p
t − p
t − p
= ˆhi
t − s
t − s
,
34
35. we could left-multiply both sides of the equation by ˆh, then right-multiply by
s − t, giving
ˆh [s − p] [t − p] [t − s] = s − p t − p t − s i.
Using the same ideas and transformations as for the smaller circle, we’d then
transform the product [s − p] [t − p] [t − s] into
r2
ˆt + ˆh + r1
ˆt − p r1
ˆt − p r2
ˆt + ˆh .
By comparison, the product that we obtained for the smaller circle was
r2
ˆt + ˆh r1
ˆt − p r2
ˆt + ˆh + r1
ˆt − p .
In both cases, the products that result from the expansion have the forms
ˆt + ˆh r1
ˆt − p ˆt + ˆh and r1
ˆt − p
2
ˆt + ˆh , so the same simplifications
work in both, and give the same results when the result is finally dotted with
ˆh and set to zero.
The above having been said, let’s look at another way of obtaining an equa-
tion, for the larger circle, that has the same form as Eq. (6.3). For convenience,
we’ll present the figure for the larger circle again:
Instead of beginning by equating two expressions for the rotation eψi
, we’ll
equate two expressions for e−ψi
:
t − p
t − p
s − p
s − p
=
t − s
t − s
ˆhi .
After left-multiplying both sides by t−s, then by ˆh, and rearranging, the result
would be identical to Eq. (6.3), except for the algebraic sign of the right-had
35
36. side, which —because it’s a bivector—would drop out when we took the scalar
part of both sides.
However, the difference in the sign of that bivector captures the geometric
nature of the difference between the relationships of the large and small circles
to L, P, and C. That difference in sign is also reflected in the positions, with
respect to the vector w, of the solution circles’ points of tangency t:
6.2 Identifying the Solution Circles that Enclose C
These solution circles can be found by modifying, slightly, the ideas that we
used for finding solution circles that don’t enclose C. Here, too, we’ll want to
express the radius r3 of the solution circle in terms of the vector t. Examining
the next figure,
36
37. we see that s = (r1 − r3)ˆt+r3
ˆh, and also that s = h+λˆhi. By equating those
two expressions, dotting both sides with ˆh, and then solving for r3, we find that
r3 =
h − r1
ˆt · ˆh
1 − ˆt · ˆh
.
As was the case when we found solution circles that didn’t enclose C, we’ll
want to equate expressions for two rotations that involve the unknown points
of tangency t and s. For example, through the angles labeled φ, below:
t − p
t − p
s − p
s − p
=
t − s
t − s
−ˆhi
=
s − t
s − t
ˆhi ,
Left-multiplying that result by s − t, and then by ˆh,
ˆh [s − t] [t − p] [s − p] = s − t t − p s − p i
∴ ˆh [s − t] [t − p] [s − p] 0 = 0,
which is identical to Eq. (6.4). To solve for t, we use exactly the same technique
that we did when we identified the solution circles that don’t surround C, with
r3 and 1 − ˆh · ˆt taking the place of r2 and 1 − ˆh · ˆt, respectively. The result is
z · t = 2h2
r1
2
− r1 h r1
2
+ p2
,
where z = 2 h2
− r1 h p − p2
− r1
2
h . Thus,
P z (t) =
2h2
r1
2
− r1 h r1
2
+ p2
z
ˆz. (6.7)
Again, there are two solution circles of this type:
37
38. 6.3 The Complete Solution, and Comments
There are four solution circles: two that enclose C, and two that don’t:
38