An Application of the Imaginary Number "i" Using Coordinate Geometry and Algebra

8/2/2019 An Application of the Imaginary Number "i" Using Coordinate Geometry and Algebra

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PROBLEMS IN APPLIED ALGEBRA AND BASIC TRIGONOMETRY


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PREFACE

Many students ask why do they have to study algebra and geometry. The reply usuallyinfers that these subjects are the building blocks of higher math. In a technical environment,algebra, geometry and trigonometry are the basic tools that are used to solve the problems.Basic Electricity is definitely an application where this is so very true. Engineering Graphics,Technical Drawing and Electronics are a few more examples of career application of appliedmath. Military technical schools seem to confirm the above observations in that most technicalpeople with a good command of algebra, geometry and basic trigonometry are quite successful intheir respective fields of technical training and in collegiate experiences.

Being a product of a military technical school (US Naval Schools Command), I found that

a solid high school math foundation in applied algebra, analytical geometry, trigonometry andphysics provided me with the tools required to begin my studies in the feld of electronics,mathematics and programming. The technical schools and colleges taught me many practicalmath problem solving techniques which I try to pass on to my students in my capacity as a mathtutor and math instructor.

The objective of this paper is to illustrate how some of these problem solving techniquesand methodologies are applied to algebra, coordinate geometry and vector problems. Sometimesthe formal presentation of math gets in the way of understanding its message. Many students

can quote you a math principle from a textbook while being completely clueless of what it meansor how to use it in a practical application. Many math students gain confidence when theyovercome their fear of making mistakes. They should be encouraged to try differentapproaches or techniques to problem solving. I call it Mathematical Innovation or Math R&Dwith a pencil. A piano player becomes a musician through practice and experimentation. Thesame analogy is true when applied to learning mathematics at any level. Students should beencouraged to try different methidologies to solve problems.

This treatment barely touches the tip of the iceberg with regard to applications ofimaginary numbers. The intent of this treatise is to enhance the students problem solving

techniques by applying some basic principles of analytical geometry, algebra and uses of theimaginary number to the solution of some common graph problems. I hope you will enjoy thistreatment and have as much fun as I did writing it.

Herb Norman Sr.Thornton, ColoradoMarch 20, 2012

AN APPLICATION OF THE IMAGINARY NUMBER iUSING COORDINATE GEOMETRY AND ALGEBRA

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CHAPTER #1SOLUTIONS USING BASIC TRIGONOMETRY

Solutions of Right Triangles

Section 1.01: Pythagorean TheoremGiven a Right Triangle; The square of the hypotenuse is equal to the sum of the squares of theother two sides.

c2 = a2 + b2Illustration:

c = hypotenusea & b are the other two sides that are perpendicular to each other

Problem #1.01-A Given a =4 and b =3; c2

= a2

+ b2

Find c, the hypotenusec2 = 42 + 32

c2 = 16 + 9c2 = 25

thenc = 25

c = 5

Problem #1.01-B Given c = 5 and b =3; c2 = a2 + b2

Find a, one of the perpendicular sides52 = a2 + 32

25 = a2 + 925 - 9 = a2

then

16 = a2

16 = a

4 = a


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Problem #1.01-C Given c = 5 and a =4; c2 = a2 + b2

Find b, one of the perpendicular sides

52

= 42

+ b2

25 = 16 + b225 - 16 = b2

then

9 = b2

9 = b

3 = b

Section 1.02: Basic Trigonometry Functions

Given a Right Triangle; the side opposite the 90O

is called the hypotenuse. The other two anglesare complementary angles. The other two sides are referred to as opposite and adjacent,depending on which of the complementary angles is referenced, determines the name of theside (opposite or adjacent).

The basic trigonometric functions of the reference angle () are Sine, Cosine, Tangent,Cosecant, Secant and Cotangent. The functions and their abbreviations are defined as follows:

Sine(i

) = Sin(i

) = Hypotenuse

Opposite

Co sin e(i) = Cos (i) =HypotenuseAdjacent

Tangent(i) = Tan(i) =AdjacentOpposite

Co secant(i) = Csc =Sin(i)

1=

Opposite

Hypotenuse

Secant(i) = Sec(i) =Cos (i)

1=

AdjacentHypotenuse

Co tangent(i) = Cot(i) =Tan(i)

1=

OppositeAdjacent


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The first three functions (Sine, Cosine and Tangent) are the basic three. Cosecant, Secant andCotangent are their respective reciprocals.

Calculate the trigonometric functions of angle (A) in the given right triangle below.

Problem #1.02-A Problem #1.02-BFind the Tangent of angle (A). Find the Cotangent of angle (A).

Tan(A) = AdjacentOpposite

Tan(A) =34

or Tan (A) = 1.3333

Cot(A) = Opposite

Adjacent

Cot(A) =43

or Cot(A) = 0.7500

Problem #1.02-C Problem #1.02-DFind the Sine of angle (A). Find the Cosecant of angle (A).

Sin(A) =Hypotenuse

Opposite

Sin(A) =54

or Sin (A) = 0.8000

Csc (A) =Opposite

Hypotenuse

Csc (A) =45

or Csc (A) = 1.2500

Problem #1.02-E Problem #1.02-FFind the Cosine of angle (A). Find the Secant of angle (A).

Cos (A) = Hypotenuse

Adjacent

Cos (A) =53

or Cos (A) = 0.6000

Sec (A) =Adjacent

Hypotenuse

Sec (A) =35

or Sec (A) = 1.6667


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Calculate the trigonometric functions of angle (B) in the given right triangle below:

Problem #1.02-G Problem #1.02-HFind the Tangent of angle (B). Find the Cotangent of angle (B).

Tan(B) =

AdjacentOpposite

Tan(B) =43

or Tan (B) = 0.7500

Tan (B) =AdjacentOpposite

Tan (B) =34

or Tan (B) = 1.3333

Problem #1.02-I Problem #1.02-J

Find the Sine of angle (B). Find the Cosecant of angle (B).Sin(B) =

HypotenuseOpposite

Sin(B) =53

or Sin (B) = 0.6000

Csc (B) =Opposite

Hypotenuse

Csc (B) =35

or Csc (B) = 1.6667

Problem #1.02-K Problem #1.02-LFind the Cosine of angle (B). Find the Secant of angle (B).

Cos (B) =

HypotenuseAdjacent

Cos (B) =

5

4or Cos (B) = 0.8000

Sec (B) =Adjacent

Hypotenuse

Sec (B) =

4

5or Sec (B) = 1.2500


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Section 1.03: Inverse Trigonometry Functions (The Angle )Given a Right Triangle and the Trigonometric Function; the Arc Function generates thereferenced angle (Degrees or Radians) as illustrated in the following Problems.

Once the Trigonometric Function is calculated, the Referenced Angle is generated through theArc Functions. The process is referred to as Inverse Trigonometric Functions and they arelisted below:

Trigonometric Function Inverse Trigonometric FunctionTan (A) = Opposite/Adjacent (A) = ArcTan (Opposite/Adjacent)Sin (A) = Opposite/Hypotenuse (A) = ArcSin (Opposite/Hypotenuse)Cos (A) = Adjacent/Hypotenuse (A) = ArcCos (Adjacent/Hypotenuse)Cot (A) = Adjacent/Opposite (A) = ArcCot (Adjacent/Opposite)Csc (A) = Hypotenuse/Opposite (A) = ArcCsc (Hypotenuse/Opposite)Sec (A) = Hypotenuse/Adjacent (A) = ArcSec (Hypotenuse/Adjacent)

Given the triangle is a Right Triangle, the acute angles are complementary and sum to 90o.

Under these conditions, the Complementary Angles can be expressed as ( and 90o

- ). If theReferenced Angle is designated (A), then its complement, (B) is equal to (90o - A).

Knowing this relationship simplifies the calculation of the Complementary Angles when solvingthe Right Triangle in the following problems.


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Generate the Referenced Angle (A), given the calculated trigonometric functions in the previousproblems (Problems 1.02-A through 1.02-L)

.Problem #1.03-A Problem #1.03-B

Using the Tangent of Angle (A). Using the CoTangent of Angle (A).

Tan (A) =AdjacentOpposite

Tan (A) =34

or Tan (A) = 1.3333

A^ h = ArcTan 34

a k = 53.13O

B^ h = 90O - 53.13O^ h = 36.87O

Cot(A) =Opposite

Adjacent

Cot(A) =43

or Cot(A) = 0.7500

(A) = ArcCot 43

a k = 53.13O

B^ h = 90O - 53.13O = 36.87O

Problem #1.03-C Problem #1.03-DUsing the Sine of Angle (A). Using the Cosecant of Angle (A).

Sin (A) =Hypotenuse

Opposite

Sin (A) =54

or Sin (A) = 0.8000

(A) = ArcSin

5

4

a k= 53.13

O

B^ h = 90O - 53.13O = 36.87O

Csc (A) =Opposite

Hypotenuse

Csc (A) =45

or Csc (A) = 1.2500

(A) = ArcCsc4

5

a k= 53.13

O

B^ h = 90O - 53.13O = 36.87O


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Generate the Referenced Angle (A) (Continued)

Problem #1.03-E Problem #1.03-F

Using the Cosine of Angle (A). Using the Secant of Angle (A).

Cos (A) =Hypotenuse

Adjacent

Cos (A) =53

or Cos (A) = 0.6000

(A) = ArcCos53a k = 53.13O

B^ h = 90O - 53.13O = 36.87O

Sec (A) =Adjacent

Hypotenuse

Sec (A) =35

or Sec (A) = 1.6667

(A) = ArcSec35a k = 53.13O

B^ h = 90O - 53.13O = 36.87O

Generate the Referenced Angle (B)

Problem #1.03-G Problem #1.03-HUsing the Tangent of Angle (B). Using the CoTangent of Angle (B).

Tan (B) =Adjacent

Opposite

Tan (B) =43

or Tan (B) = 0.7500

B^ h = ArcTan43a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O

Cot(B) =Opposite

Adjacent

Cot(B) =34

or Cot(B) = 0.7500

(B) = ArcCot34a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O


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Generate the Referenced Angle (B) (Continued)

Problem #1.03-I Problem #1.03-J

Using the Sine of Angle (B). Using Cosecant of Angle (B).Sin (B) =

HypotenuseOpposite

Sin (B) =53

or Sin (B) = 0.6000

(B) = ArcSin53a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O

Csc (B) =Opposite

Hypotenuse

Csc (B) =35

or Csc (B) = 1.6667

(B) = ArcCsc35a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O

Problem #1.03-K Problem #1.03-LUsing the Cosine of Angle (B). Using the Secant of Angle (B).

Cos (B) =Hypotenuse

Adjacent

Cos (B) =54

or Cos (B) = 0.8000

(B) = ArcCos54a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O

Sec (B) =Adjacent

Hypotenuse

Sec (B) =45

or Sec (B) = 1.2500

(B) = ArcSec45a k = 36.87O

A^ h = 90O - 36.87O^ h = 53.13O

Problem #1.03-M Problem #1.03-NReference Angle (A = 53.13O). Reference Angle (B = 36.87O).

Given Hypotenuse = 5 Given Adjacent = 4

Cos A^ h = HypotenuseAdjacent

Hypotenuse^ h Cos A^ h6 @ = Adjacent

5Cos 53.13O^ h = Adjacent

3.00 = Adjacent

Sin A^ h = HypotenuseOpposite

Hypotenuse^ h Sin A^ h6 @ = Opposite

5Sin 53.13O^ h = Opposite

4.00 = Opposite

B^ h = 90O - (A)

B^ h = 90O - (53.13O)

B^ h = 36.87O

Tan B^ h =AdjacentOpposite

Adjacent^ h Tan B^ h6 @ = Opposite

4Tan 36.87O^ h = Opposite

3.00 = Opposite

Cos B^ h = HypotenuseAdjacent

Hypotenuse =Cos B^ h

Adjacent

Hypotenuse = Cos 36.87O^ h4

Hypotenuse = 5.00

A^ h = 90O - (B)

A^ h = 90O - (36.87O)

A^ h = 53.13O


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CHAPTER #2SOLUTIONS USING ALGEBRA AND LINEAR EQUATIONS

Section 2.01: Coordinate System & the Distance FormulaBy transposing the same problem to a two dimensional coordinate system, we can explorealgebraic methodologies with regard to solutions. The following Line Graph forms a RightTriangle of the same dimensions as the one in the previous chapter.

Basic Equations of Linear SystemsSlope Intercept Form: y = mx+ b

Slope (m): m =x2- x1y2- y1

i = ArcTan m^ h

Y-Intercept: b

X-Intercept: xo = m- b ; given y = 0

Point-Slope Form: y- yp = m x- xp^ h

Angle ()between 2 Slopes (m1 & m2) Tan b^ h = 1 - m1m2m2- m1

The Slope (m) or Tangent() is the angle between the X-Axis and the linear function (AB) atthe point of the X-Intercept, B (0 ,4) where Xo = 4. The Y-Intercept occurs at the point A(0,3 ) where b = 3. A Right Triangle (ACB) with Height (CA = 3) and Base (CB = 4).


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Distance Formula: D = x2- x1^ h2+ y2- y1^ h2

The distance formula is used to calculate the straight line distance between two points. Giventhe two points are the X and Y intercepts, [ (XO, 0) & (0, b) ] respectively. If the two points are

the X and Y intercepts, then the straight line distance lies on the hypotenuse, (A, B)

Problem #2.01-A

Given: The X-Intercept is located at B (0 ,4) Line y = -4

3x + 3

The Y-Intercept occurs at A (0,3 ) ( )m Tangent i=

D = x2 - x1^ h2

+ y2 - y1^ h2

AB = 0 - 3^ h2 + 4 - 0^ h2

AB = - 3^ h2 + 4 2

AB = 25

AB = 5

Slope = m^ h =- xOb

m =-4

3

i = ArcTan m^ h

i = ArcTan -4

3a ki =-36.87

O

(A) = i

A^ h = - 53.13O

(B) = 90O

- (A)

(B) = 90O

- (- 53.13O)

B^ h = 143.13O or - 36.87O


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Section 2.02: Reconfiguration and Intersecting LinesReconfigure the 3-4-5 Right Triangle so the hypotenuse is the base, the opposite and adjacentsides are perpendicular to each other and intersect at a point, C (x, y). Label the figure so that

the vertex of Angle (B) is at the Origin, B (0, 0) and has a positive slope toward C (x, y) andforms a straight line (BC), which is above the x-axis. Construct the vertex of Angle (A) at (5, 0)so that its side (AC) lies on a straight line with C (x, y) and intersects the line (BC) at 90O. Seethe figure below..

Problem #2.02-A

B^ h = ArcTan B^ h

mB = Tan B^ h

mB =4

3

yBC = mBx + bBC

yBC =4

3 x given bBC = 0

mB =4

3

A^ h = ArcTan A^ h

mA = Tan A^ h

mA = -3

4and x0 = - mA

bAC= 5

yAC = mAx + bAC

yAC = -3

4x +

3

20given bAC = - mAx0

mA = -3

4

bAC =3

20


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yBC = yAC at C xC, yC^ h xC = x

43x = -

34x +

320 yC = yBC = 4

3x

34

x +43

x =3

20yC = 4

35

16a k

1216x + 9x

=3

20yC = 5

12

1225x

=3

20C xC, yC^ h = 5

16,

512a k

x =5

16

The Distance Formula can be used to confirm the lengths (BC = 4), (AC = 3) and (AB = 5). TheVertex C (Xc, Yc) can be confirmed to be a Right Angle by comparing the slopes.The Base Line (BA) from Origin B (0, 0) to A (5, 0) is 5 units by way of subtraction of thecoordinates since they lie on the same axis. XA - XB = Delta x; x = 5

Length of Line (BC): Length of Line (AC):

B (0,0) to C 516

, 512

a kBC = xB - xC^ h

2+ yB - yC^ h

2

BC = 0 - 516^ h2 + 0 - 5

12^ h2

BC = 25256 + 25

144

BC = 25400

BC = 4

A (5, 0) to C 516

, 512

a kAC= xA - xC^ h

2+ yA - yC^ h

2

AC= 5 - 516^ h2 + 0 - 5

12^ h2

AC= 2581 + 25

144

AC= 25225

AC= 3

Two slopes are perpendicular to each other if one slope is the negative reciprocal of the other.

If slope (mB) slope (mA) then mA =-mB

1

Refer to the slopes in the above problem: mB =4

3and mA =-

3

4

Since the slopes (mB) and (mA) are negative reciprocals, (mB) slope (mA). These slopes

intersect at the Vertex (C) @ 90O.


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Section 2.03: Area & Enclosed PolygonsRefer to the figure used in Problem #2.02-A

C xC, yC^ h =516

,512a k

Problem #2.03-A

Given the 3-4-5 Triangle BAC: Vertices: B(0,0), A(5,0), C(16/5, 12/5)The Base (BA =5) and the Height (Yc =12/5), find the AREA.Area = (Base)(Height)/2: By calculator, Area = 6 sq-units

Using the Determinant and Vertices to calculate the Area :

Area3=2

1

xB yB 1

xA yA 1

xC yC 1

=2

1

0 0 1

5 0 1

165

125 1

=2

112^ h - 0^ h6 @ = 6

2^ h Area3^ h = Base^ h Height^ h =

xB yB 1

xA yA 1

xC yC 1


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Given any POLYGON ENCLOSED BY (n 3) VERTICES;Vertices defined by (x1, y1), (x2, y2), (x3, y3), ... , (xn, yn), (x1, y1). If the polygon is aTriangle (n = 3), the procedure shown above BAC, also works. However, if n > 3,

Problem #2.03-A can be solved by using the following methodology.

2^ h Area^ h =x1 y1

x2 y2+

x2 y2

x3 y3+

x3 y3

x4 y4+ . .. +

xn yn

x1 y1

Note:The last determinant references the nth vertex (xn,yn) and the 1st (x1,y1).Twice the rea is equal to the sum of n (2 x 2) determinants. Each determinantconsists of overlapping (x, y) vertices in sequence, ending with the 1st vertex.

Problem #2.03-B

Given Vertices: B(0,0), A(5,0), C(16/5, 12/5), find the AREA.

2^ h Area^ h =0 0

5 0+

5 0

165

125

+

165

125

0 0

2^ h Area^ h = 0 - 0^ h + 12 - 0^ h + 0 - 0^ h

2^ h Area^ h = 12

Area = 6


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CHAPTER #3SOLUTIONS USING THE COMPLEX PLANE

Section 3.01: Coordinate System & Imaginary NumbersBy placing the Right Triangle in the Complex Plane, the Y-Axis becomes the Imaginary Axis andthe X-Axis becomes the Real Number Axis. The complex number system provides a few moretools that can be used to analyze and solve the Right Triangle. By using the complex model, thesides and angles of the triangle can be calculated by using the product and ratio properties ofcomplex numbers. Transformation and rotation of points is obtained by the use of complexproperties.

Given a vertical Vector of length (3) expressed as ( 3i ) on the Imaginary Axis and ahorizontal Vector of length (4) expressed as (4) on the Real Axis; The Vector which is the(Magnitude or Hypotenuse) of length (5) will be the Result in the direction ( ), also referredto as the Argument with respect to the Real Axis in the Complex Plane.

Vector Length Typical Designations Angle Complex ExpressionHypotenuse

Modulus

Absolute

Magnitude

rii, A i , Z , R

Argument

i = Arg (z)

a + bi or a+ jb

x + iy or x+ jy


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There are Four Formats used to express complex numbers in the complex plane.

RECTANGULAR: x+ iy 4 + i3

POLAR : A i 5 36. 87O

TRIGONOMETRIC: A Cos i^ h+ i Sin i^ h6 @ 5 Cos 36.87O^ h+ i Sin 36.87O^ h6 @

EXPONENTIAL: Aeii 5ei36.87

O

There are advantages for using one format over another. In most cases the application dictatesthe format. There are some conditions which require transferring from one format during theprocess of solving.

x+

jy=

Ze

ji

Given:

x= Cos i^ h & y = Sin i^ h

Z= x2 + y2

Z Cos i^ h+ jSin i^ h6 @

The PRODUCTof two complex numbers generates the Sum of the two arguments.

ProductPropertyZ1e

ji^ h Z2ej}^ h = Z1Z2ej i+}^ h

Z1eji^ h Z2ej}^ h = Z1Z2 Cos i+ }^ h+ jSin i+ }^ h6 @

Z1eji^ h Z2ej}^ h = Z1Z2 i + }^ h

The RATIO of two complex numbers generate the Difference between the arguments.Ratio Property

Z2ej}Z

1

e

ji

= Z2Z1

a kej i-}^ h

Z2ej}

Z1eji

=Z2Z1a k Cos i - }^ h+ jSin i- }^ h6 @

Z2ej}

Z1eji

=Z2Z1a k i - }^ h


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De Moivres Theorem ZnThis theorem enables us to find the power of a complex number in Polar or Trigonometricformat very easily.

Z

n=

x+

jy^ h

n

Zn = Zn

Cos i^ h+ jSin i^ h6 @n

Zn= Z

nCos ni^ h+ jSin ni^ h6 @

Section 3.02: Applications - Triangle Solution Methodologies

Problem 3.02AGiven the vertices of ABC, Solve Completely for Sides and Angles.


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Step #1Identify and label all vertices with coordinates derived from the original figure.

Coordinate D (4, 0) is derived from Coordinates B (4, 3) and A (0, 0)Coordinate E (5, 3) is derived from Coordinates B (4, 3) and C (5, 8.66)Coordinate F (0, 8.66) is derived from Coordinates A (0, 0) and C (5, 8.66)

Step #2Define Coordinates with complex numbers.

A = 0 + j0 B = 4 + j3 C= 5 + j8.66

D = 4 + j0 E= 5 + j3 F= 0 + j8.66


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Step #3Given ADB defined by Point B; B = 4 + j3

Modulus |AB| Exponential Format Angle ABD

AB = 42

+ 32

AB = 5

BAD = ArcTan4

3a kBAD = 36.87

O

B (4,3)

AB = 4 + j3

AB = 5ej36.87O

AB = 5 36.87O

ABD = 90O

- BAD

ABD = 90O

- 36.87O

ABD = 53.13O

Step #4Given EBC defined by Points B (4, 3), C (5, 8.66) and E(5, 3).Side BC connecting B = 4 + j3 to C = 5 + j8.66.Side BC simplifies to: C - B = (x, jy) = (5-4, j8.66-j3): BC = (1, j5.66)

Modulus |BC| Exponential Format Angle BCE

3BC = 3x^ h2 + 3y^ h2

3BC = 1 2 + 5.662

3BC = 33.04 = 5.75

CBE = ArcTan1

5.66a kCBE = 79.98O

B (4,3) to C(5,8.66)

3x = 1 & 3y = 5.66BC =3x + j3y

BC = 1 + j5.66

BC = 5.75ej79.98O

BC = 5.75 79.98O

BCE = 90O

- CBE

BCE = 90O

- 79.98O

BCE = 10.02O


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Step #4 (Continued)

Step #5Given ABC, the original triangle, defined by Points A, B and C.We also have two of the three sides and some information about each angle.First lets take the same approach as the previous steps.

Modulus |AC| Exponential Format Angle CAB

AC = 52

+ 8.662

AC = 10

CAD = ArcTan5

8.66a kCAD = 60

O

C(5,8.66)

AC= 5 + j8.66

AC= 10ej60O

AC= 10 60O

CAB = CAD - BAD

CAB = 60O

- 36.87O

CAB = 23.13O


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Step #5 (Continued)Angle ABC Angle BCA

ABC = 360O

- 90O

- CBE- ABD

ABC = 270

O

- 79.98

O

- 53.13

O

ABC = 136.89O

BCA = 180O

- ABC - CAB

BCA = 180

O

- 136.89

O

- 23.13

O

BCA = 19.98O

COMPLETE SOLUTION


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Section 4.01: Shifting a Geometric Figure in a Complex PlaneGiven the vertices of ABC, SHIFT the triangle to position ABC within the Complex Plane by

adding a SHIFT CONSTANT a + bi^ h to each Vertex.

The figure below shows the Rt ABC with Vertices A(0,0), B(4,0) and C(4,3i): SHIFTED toVertex C( -4, -3i ), B(0, -3i) and A(0,0) respectively. The new vertices of ABC, are calculatedby adding the Negative of Vertex C(4,3i) to each of the original vertices.The SHIFT CONSTANT = -4 - 3i, shifting the object 4 units left and 3 units down.

The SHIFT Constant = -4 - 3i

The SHIFT Constant in a COPLEX PLANE can be calculated by:1) Selecting one of the original vertices V0 x0 + y0 i^ h

2) Select the destination or Shifted vertex position VS xS+ yS i^ h

3) The SHIFT Constant == VS- V0

= xS+ yS i^ h- x0 + y0 i^ h

In this figure, select V0 as the Vertex C(4,3i) to be SHIFTED to VS Vertex C( 0, 0i ).Following the procedure outlined above, calculate the SHIFT Constant.

1) Selecting one of the original vertices V0 4 + 3i^ h

2) Select the destination or Shifted vertex position VS 0 + 0i^ h

3) The SHIFT Constant == VS - V0

= 0 + 0i^ h - 4 + 3i^ h

= - 4 - 3i^ h

4) Add the Shift Constant to each of the vertices

A to VERTEX C B to VERTEX B C to VERTEX A

AC'B' = CAB + (- 4 - 3i)

AC'B' = 0 + 0i - 4 - 3i

AC'B' = - 4 - 3i

Vertex C' = (- 4, - 3i)

C'B'A = CBA + (- 4 - 3i)

C'B'A = 4 + 0i - 4 - 3i

C'B'A = 0 - 3i

Vertex B' = (0, - 3i)

C'AB' = ACB + (- 4 - 3i)

C'AB' = 4 + 3i - 4 - 3i

C'AB' = 0 + 0i

Vertex A = (0,0i)


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See the figure below to show the new positions as a result of shifting.

This methodology can be used to shift the object anywhere in the Complex Plane or CartesianPlane. To maintain consistency, this model uses the imaginary numbers and the Complex Plane

even though x,y coordinates work well when shifting points within the Cartesian Plane.


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The next figure shows an example of shifting the object to Quadrant II but, 1 unit above theReal Axis and 2 units left of the Imaginary Axis.

SHIFT CONSTANT = -6 + 1i

VERTEX A VERTEX B VERTEX CC'A'B' = CAB + (- 6 + i)

C'A'B' = 0 + 0i - 6 + i

C'A'B' = - 6 + i

Vertex A' = (- 6, i)

A'B'C' = ABC + (- 6 + i)

A'B'C' = 4 + 0i - 6 + i

A'B'C' = - 2 + i

Vertex B' = (- 2, i)

A'C'B' = ACB + (- 6 + i)

AC'B' = 4 + 3i - 6 + i

AC'B' = - 2 + 4i

Vertex C' = (- 2, + 4i)


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Section 4.02: Shifting a Straight Line y = mx + bGiven a Point = (4,0) and Slope m = 3/4 are defined by the Straight Line Equation; y = 3x/4 - 3.

Problem 4.02A:The graph below shows how to shift a straight line function vertically.Add a shift value to the y-intercept, b: y = mx + (b shift): + shift up & - shift down.

VERTICAL SHIFT

Given Equation: y = 3x/4 - 3 y = 3x/4 - 3 is Shifted Vertical +2: y = 3x/4 - (3 + 2)

y = 3x/4 - 3 is Shifted Vertical -2: y = 3x/4 - (3 - 2)


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Problem 4.02B:The graph below shows how to shift a straight line function horizontally.Add a shift value to the x variable, x: y = m (x shift) + b:

+ shift LEFT & - shift RIGHT

HORIZONTAL SHIFT

Given Equation: y = 3x/4 - 3

y = 3x/4 - 3 is Shifted Horizontal (Right) +2: y = 3(x - 2)/4 - 3

y = 3x/4 - 3 is Shifted Horizontal (Left) -2: y = 3(x + 2)/4 - 3


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Section 5.01: Rotating a Geometric FigureGiven the vertices of ABC, ROTATE the triangle to position ABC within the Complex Planeby Multiplying each of the Vertices by a ROTATION Constant a+ bi^ h.

The figure below shows the Rt ABC with Vertices A(0,0i), B(4,0i) and C(4,3i): ROTATED toVertex A(0,0), B(-4, 0i) and C( -4, -3i ) respectively. The new vertices of ABC, are calculatedby multiplying each of the original vertices by the ROTATION Constant.

The ROTATION Constant = -1 + 0i

The ROTATION Constant in a COPLEX PLANE can be calculated by:1) Selecting one of the original vertices V0 x0 + y0 i^ h

2) Select the respective rotated vertexVR xR+ yR i

^ h3) The Counter-Clockwise Rotation Constant =

V0VR=

x0+ y0 i^ hxR+ yR i^ h

To rotate Clockwise use the reciprocal: VRV0=

xR+ yR i^ hx0 + y0 i^ h

In this figure, lets select Vertex C(4,3i) being ROTATED to Vertex C( -4, -3i ).Following the procedure outlined above, calculate the Rotation Constant.

VR = - 4 - 3i^ h and V C = 4 + 3i^ h

Rotation Cons tan t=VC

VCR=

4 + 3i^ h- 4 - 3i^ h

VC

VCR=

4 + 3i^ h- 4 - 3i^ h

:

4 - 3i^ h4 - 3i^ h

VC

VCR=

16 + 9

- 1 6 - 9=

25

- 25= - 1

and thus

VA

= 0 + 0i^ h

and VAR

= VA

- 1^ h

VAR = 0 + 0i^ h - 1^ h

VAR = 0 + 0i^ h

VB = 4 + 0i^ h and VBR = VB - 1^ h

VBR = 4 + 0i^ h - 1^ h

VBR = - 4 + 0i^ h


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The Rotation of Vertices with known destination:

Rotation FROM Vertices: A = ( 0, 0i ), B = ( 4, 0i ), C = ( 4, 3i )

Rotation TO Vertices:A = ( 0, 0i ), B = (-4,0i), C = ( -4, -3i )

Calculation of Rotation Constant:Counter-Clockwise = Ratio of Vertex (Destination) to Vertex (Original)Clockwise = Ratio of Vertex (Original) to Vertex (Destination)


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OBJECT ROTATION From 0O to 45O & 90O w/r ORIGINRight Triangle ABC Base = Line AB Height = BC Hypotenuse = AC

Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4).Position #2 Defined by vertices A(3/2, 3/2), B(6/2, 6/2) & C(2/2, 10/2).Position #3 Defined by vertices A(0,3), B(0,6) & C(-4,6).

Problem #5.02-AGiven initial Position #1, calculate the vertices of each position (Position #2 @ = 45O) and(Position #3 @ 90O) as the object is rotated counterclockwise with respect to the origin fromits initial position (Position #1) in the complex plane. Verify that the objects maintain theirinitial dimensions.


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All three triangles are congruent to each other due to equal segments: That is the Basesare EQUAL in all positions; The Heights are Equal in all positions; and the Hypotenuses areEqual in all positions. Congruency validates using the Unit Vector in the calculation of the

Rotation Constant for each position angle.

0O 30O 45O 60O 90OCosine 1 3/2 1/2 1/2 0Sine 0 1/2 1/2 3/2 1Tangent 0 1/3 1 3

Unit Vector Expression for Complex Number (a + bi) = cos() + i sin () = 0O: a = 1 i b = 0i

= 45O: a = 1/2 i b = i/2 = 90O: a = 0 i b = 1i

Rotation Constant ( = 45O): k = (1/2 + i/2) ( = 90O): k = ( 0 + i)Rotated Vertex (A) V,A = k VA Given VA = (3, 0) V,A = k VA Given VA = (3, 0)

V45,A = [cos() + i sin ()] ( 3 + 0i) V45,A = [cos() + i sin ()] ( 3 + 0i)V45,A = (1/2 + 1/2i) ( 3 + 0i) V45,A = (0 + i) ( 3 + 0i)V45,A = (3/2 + 3i/2) V45,A = (0 +3i)

V45,A = (3/2 , 3/2) V45,A = (0 , 3)

Rotated Vertex (B) V,B = k VB Given VB = (6, 0) V,B = k VB Given VB = (6, 0)V45,B = [cos() + i sin ()] (6 + 0i) V45,B = [cos() + i sin ()] (6 + 0i)V45,B = (1/2 + 1/2i) ( 6 + 0i) V45,B = (0 + i) ( 6 + 0i)V45,B = (6/2 + 6i/2) V45,B = (0 + 6i)

V45,B = (6/2 , 6/2) V45,B = (0 , 6)

Rotated Vertex (C) V,C = k VC Given VC = (6, 4) V,C = k VC Given VC = (6, 4)V45,C = [cos() + i sin ()] ( 6 + 4i) V45,C = [cos() + i sin ()] ( 6 + 4i)V45,C = (1/2 + 1/2i) ( 6 + 4i) V45,C = (0 + i) ( 6 + 4i)

V45,C = (2/2 + 10i/2) V45,C = (-4 + 6i) V45,C = (2/2 , 10/2) V45,C = (-4 , 6)

Compare calculated vertices to the Vertices given for Positions #1, #2 & #3Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4).Position #2 Defined by vertices A(3/2, 3/2), B(6/2, 6/2) & C(2/2, 10/2).Position #3 Defined by vertices A(0,3), B(0,6) & C(-4,6).


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Problem #5.02-BGiven initial Position #1, calculate the vertices of each position at = 45O and 90Ocounterclockwise from initial using the XY Coordinate Plane. Let X & Y = New_Rotated

Coordinates x, y = coordinates of A, B & C Vertices at Position #1 Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4)

X = x Cos() - y Sin() Y = x Sin() + y Cos()

Rotation = 45O Rotation = 90OVertex A (3, 0) X = x Cos() - y Sin() X = x Cos() - y Sin()

X = 3 Cos(45O) - 0 Sin(45O) X = 3 Cos(90O) - 0 Sin(90O)X = (3) (1/2) - (0) (1/2) X = (3) (0) - (0) (1)X = (3/2) X = (0)

Y = x Sin() + y Cos() Y = x Sin() + y Cos()Y = 3 Sin(45O) + 0 Cos(45O) Y = 3 Sin(90O) + 0 Cos(90O)Y = 3 (1/2) + 0 (1/2) Y = 3 (1) + 0 (0)Y = (3/2) Y = (3)

A (3/2 , 3/2) A (0 , 3)

Rotation = 45O Rotation = 90OVertex B (6, 0) X = x Cos() - y Sin() X = x Cos() - y Sin()

X = 6 Cos(45

O

) - 0 Sin(45

O

) X = 6 Cos(90

O

) - 0 Sin(90

O

)X = (6) (1/2) - (0) (1/2) X = (6) (0) - (0) (1)X = (6/2) X = (0)

Y = x Sin() + y Cos() Y = x Sin() + y Cos()Y = 6 Sin(45O) + 0 Cos(45O) Y = 6 Sin(90O) + 0 Cos(90O)Y = 6 (1/2) + 0 (1/2) Y = 6 (1) + 0 (0)Y = (6/2) Y = (6)

A (6/2 , 6/2) A (0 , 6)

Rotation = 45O Rotation = 90OVertex C (6, 4) X = x Cos() - y Sin() X = x Cos() - y Sin()

X = 6 Cos(45O) - 4 Sin(45O) X = 6 Cos(90O) - 4 Sin(90O)X = (6) (1/2) - (4) (1/2) X = (6) (0) - (4) (1)X = (2/2) X = (-4)


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Rotation = 45O Rotation = 90OVertex C (6, 4) Continued:

Y = x Sin() + y Cos() Y = x Sin() + y Cos()

Y = 6 Sin(45O

) + 4 Cos(45O

) Y = 6 Sin(90O

) + 4 Cos(90O

)Y = 6 (1/2) + 4 (1/2) Y = 6 (1) + 4 (0)Y = (10/2) Y = (6)

A (2/2 , 10/2) A (-4 , 6)

Review Graphics:

OBJECT ROTATION From 0O to 45O & 90O w/r ORIGIN

Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4).Position #2 Defined by vertices A(3/2, 3/2), B(6/2, 6/2) & C(2/2, 10/2).Position #3 Defined by vertices A(0,3), B(0,6) & C(-4,6)


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Section #5.03: Rotation Using Matrix MethodologiesProblem #5.03AGiven initial Position #1, calculate the vertices of each position at = 45O and 90O

counterclockwise from initial using the XY Coordinate Plane. Let X & Y = New_RotatedCoordinatesx, y = coordinates of A, B & C Vertices at Position #1

Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4) Position #1 Defined in the Complex Plane A(3+0i), B(6+0i) & C(6+4i)

SIMPLEX MATRIX Solution in the Complex Plane

V = Output Vertex Matrix (1 x 2), given n = number of (x,y) verticesk = Unit Rotation Constant given = Angle of Rotation

V = Input Vertex Matrix (n x 1), given n = number of (x,y) vertices

V'6 @ = k : V6 @ Given,k = Cos i^ h+ i Sin i^ h

k = x+ yi V'6 @ = x + yi^ h

x1 + iy1

x2 + iy2

x3 + iy3

...

xn + iyn

R

T

SSSSSSS

V

X

WWWWWWW

i = 45O

i = 90O

V'A

V'B

V'C

R

T

SSSS

V

X

WWWW

= Cos 45O^ h + i Sin 45O^ h" ,

3 + 0i

6 + 0i

6 + 4i

R

T

SSSS

V

X

WWWW

V'A

V'B

V'C

R

T

SSSS

V

X

WWWW

= 1 2 +i

2" ,

3 + 0i

6 + 0i

6 + 4i

R

T

SSSS

V

X

WWWW

V'A

V'B

V'C

R

T

SSSS

V

X

WWWW

=

32 +

3i2

62 +

6i2

22 +

10i2

R

T

SSSS

V

X

WWWW

=

32 ,

3i2

62 ,

6i2

22 ,

10i2

R

T

SSSS

V

X

WWWW

V'A

V'B

V'C

R

T

SSSS

V

X

WWWW

= Cos 90O^ h + i Sin 90O^ h" ,

3 + 0i6 + 0i

6 + 4i

R

T

SSSS

V

X

WWWW

V'A

V'B

V'C

R

T

SSSS

V

X

WWWW

= 0 + i" ,

3 + 0i

6 + 0i

6 + 4i

R

T

SSSS

V

X

WWWW

V'A

V'B

V'C

R

T

S

SSS

V

X

W

WWW =

0 + 3i

0 + 6i- 4 + 6i

R

T

S

SSS

V

X

W

WWW =

0 , 3i

0 , 6i- 4 , 6i

R

T

S

SSS

V

X

W

WWW

Convert Complex Matrix [V], (1 x 3) to (2 x 3) by defining (x) as the Real Value of the

Complex Number and defining (y) as the Imaginary Value of the Complex Number.


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Section #5.04: Rotation About a Referenced VertexProblem #5.04-A (Rotation About a Referenced Vertex)Given initial Position #1, calculate the vertices of each position at = 45O and 90O

counterclockwise from initial using the XY Coordinate Plane. Let X & Y = New_RotatedCoordinatesx, y = coordinates of A, B & C Vertices at Position #1

Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4) Reference Vertex: VRef = A(3,0)

R = Cos() + i Sin() V = R [ Vx - VRef ] + VRef

V = [ Cos() + i Sin() ] [ Vx - VRef ] + VRef VRef = A(3,0)V = [ x + yi ] [ Vx - VRef ] + VRef


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R = Cos() + i Sin()V = [ Cos() + i Sin() ] [ Vx - VRef ] + VRefV = [ x + yi ] [ Vx - VRef ] + VRef

VRef = A(3,0) Rotation = 45O Rotation = 90OR = ( 0.707, 0.707i ) R = ( 0,1i )

Vc = C (8, 3)Vc @ 45O = R [(8 + 3i) - (4 + 0i) ] + (4 + 0i) Vc @ 90O = R [(8 + 3i) - (4 + 0i) ] + (4 + 0i)Vc @ 45O = R [(8 + 3i) - (4) ] + (4) Vc @ 90O = R [(8 + 3i) - (4) ] + (4)Vc @ 45O = 4.71 + 4.95 Vc @ 90O = 1 + 4i

Vc @ 45O = (4.71 , 4.95) Vc @ 90O = (1 , 4)

VB = C (8, 0) VB @ 45O = R [(8 + 0i) - (4 + 0i) ] + (4 + 0i) VB @ 90O = R [(8 + 0i) - (4 + 0i) ] + (4 + 0i) VB @ 45O = R [(8 + 0i) - (4) ] + (4) VB @ 90O = R [(8 + 0i) - (4) ] + (4) VB @ 45O = 6.83 + 2.83 VB @ 90O = 4 + 4i VB @ 45O = (6.83 , 2.83) VB @ 90O = (4 , 4)


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Section #5.05: Rotate About a Referenced Vertex Using MaticesSolution #5.05-A_Matrix (Rotation About a Referenced Vertex-Detailed Load)Given initial Position #1, calculate the vertices of each position at = 45O and 90O

counterclockwise from initial using the XY Coordinate Plane. Let X & Y = New_RotatedCoordinates represented by the matrix (V). Each ROW of the matrix, V defines the verticesgiven the respective angle of rotation, .

x, y = coordinates of A, B & C Vertices at Position #1 Position #1 Defined by vertices A(3,0), B(6,0) & C(6,4) Reference Vertex: VRef = A(3,0)

Position #2 - 45O Rotation from Position #1Position #3 - 90O Rotation from Position #1

Based on the Vertex Positioning procedure:Rotation Constant: R = Cos() + i Sin()

Rotation Constant: R = x + yiV = [ Cos() + i Sin() ] [ Vx - VRef ] + VRef

VRef = Reference Vertex (Pivot Point)

V = n x m Output Matrix - A matrix ROW defines a set of vertices for a Position #n = Nbr of Angles or Positions (i): ROWSm = Nbr of Vertices in the object: COLUMNS

lV6 @ =

V1A V1B V1C ...

V2A V2B V2C ...

V3A V3B V3C ...

...

VnA VnB VnC ...

R

T

SSSSSSS

V

X

WWWWWWW

R = n x n Input Matrix - A Square Diagonal Matrix of Rotational Constants

R6 @ =

Cos i1^ h

+ i Sin i1^ h

0 0 ....

0 Cos i2^ h + i Sin i2^ h 0 ....

0 0 Cos i3^ h + i Sin i3^ h ...

. . . .

0 0 0 Cos in^ h + i Sin in^ h

R

T

SSSSSSSS

V

X

WWWWWWWW

=

x1 + y1 i 0 0 ...

0 x2 + y2 i 0 ...

0 0 x3 + y3 i ...

...

0 0 0 ... xn + yn i

R

T

SSSSSSS

V

X

WWWWWWW


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Vx = n x m Input Matrix - 1st of Row the matrix is comprised of the vertices of thefundamental figure (Position #1) and is duplicated for each of the remaining rows.Vx matrix is equivalent to the Transposition of the product of the (m x m) diagonal

of the Position #1 Vertex and the (m x n) Unit Matrix.n #m6 @ = m #m : m # n6 @T

Vx6 @ =

V1A V1B V1C ...

V1A V1B V1C ...

V1A V1B V1C ...

...

V1A V1B V1C ...

R

T

SSSSSSS

V

X

WWWWWWW

=

V1A 0 0 . . .

0 V1B 0 ...

0 0 V1C ...

...

0 0 0 . . .V1m

R

T

SSSSSSS

V

X

WWWWWWW

1 1 1 ...1

1 1 1 ...1

1 1 1 ...1

...

1 1 1 ...1

R

T

SSSSSSS

V

X

WWWWWWW

R

T

SSSSSSS

V

X

WWWWWWW

T

VRef = n x m Input Matrix - Pivot or Referenced Vertex populated to each matrix element.

The equivalent VRef matrix can be produced by the product of the Reference Vertex,(k) and the (n x m) Unit Matrix.

n #m6 @ = k n #m6 @

VRef6 @ =

VRef VRef VRef ...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSSS

V

X

WWWWWWW

= VRef

1 1 1 ...

1 1 1 ...

1 1 1 ...

...

1 1 1 ...

R

T

SSSSSSS

V

X

WWWWWWW

Therefore:V = [ Cos() + i Sin() ] [ Vx - VRef ] + VRef

V'6 @ =

Cos i1^ h + i Sin i1^ h 0 0 ....

0 Cos i2^ h + i Sin i2^ h 0 ....

0 0 Cos i3^ h + i Sin i3^ h ...

. . . .

0 0 0 Cos in^ h + i Sin in^ h

R

T

SSSSSSSS

V

X

WWWWWWWW

:

V1A V1B V1C ...

V1A V1B V1C ...

V1A V1B V1C ...

...

V1A V1B V1C ...

R

T

SSSSSSS

V

X

WWWWWWW

-

VRef VRef VRef ...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSSS

V

X

WWWWWWW

R

T

SSSSSSS

V

X

WWWWWWW

+

VRef VRef VRef ...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSSS

V'6 @ =

x1 +

y1

i0 0 ...

0 x2 + y2 i 0 ...

0 0 x3 + y3 i ...

...

0 0 0 ... xn + yn i

R

T

SSSSSSS

V

X

WWWWWWW

:

V1A V

1B V

1C

...

V1A V1B V1C ...

V1A V1B V1C ...

...

V1A V1B V1C ...

R

T

SSSSSSS

V

X

WWWWWWW

-

VRe

f VRe

f VRe

f...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSSS

V

X

WWWWWWW

R

T

SSSSSSS

V

X

WWWWWWW

+

VRe

f VRe

f VRe

f...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSSS

V

X

WWWWWWW


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Solution #5.05-B_By Matrix (Rotation About a Referenced Vertex - Summary Load)Position #1: Vertices: A(3,0), B(6,0) & C(6,4) 1 = 0O 1 = 0

Position #2: 2 = 45O 1 = /4

Position #3:

3 = 90O

1 =/2

Reference Vertex VRef = A(3,0)

Populate the matrices

V'6 @ =

x1 + y1 i 0 0 ...

0 x2 + y2 i 0 ...

0 0 x3 + y3 i ...

...

0 0 0 ... xn + yn i

R

T

SSSSSS

S

V

X

WWWWWW

W

:

V1A V1B V1C ...

V1A V1B V1C ...

V1A V1B V1C ...

...

V1A V1B V1C ...

R

T

SSSSSS

S

V

X

WWWWWW

W

-

VRef VRef VRef ...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSS

S

V

X

WWWWWW

W

R

T

SSSSSS

S

V

X

WWWWWW

W

+

VRef VRef VRef ...

VRef VRef VRef ...

VRef VRef VRef ...

...

VRef VRef VRef ...

R

T

SSSSSS

S

V

X

WWWWWW

W

V'6 @ =

1 0 0

0 12

+ i2

0

0 0 i

R

T

SSSS

V

X

WWWW

:

3 6 6 + 4i

3 6 6 + 4i

3 6 6 + 4i

R

T

SSSS

V

X

WWWW

-

3 3 3

3 3 3

3 3 3

R

T

SSSS

V

X

WWWW

R

T

SSSS

V

X

WWWW

+

3 3 3

3 3 3

3 3 3

R

T

SSSS

V

X

WWWW

V'6 @ =

3 6 6 + 4i

3 3 +2

3+ 3i3 -

2

1- 7i

3 3 + 3i - 1 + 3i

R

T

SS

SS

V

X

WW

WW

=

3 6 6 + 4i

3 5.12 + 2.12i 2.93 + 4.95i

3 3 + 3i - 1 + 3i

R

T

SS

SS

V

X

WW

WW

A B C

V'6 @ =

Position #1

Position #2

Position #3

R

T

SSSS

V

X

WWWW

=

3, 0^ h 6, 0^ h 6, 4^ h

3, 0^ h 5.12, 2.12^ h 2.93, 4.95^ h

3, 0^ h 3, 3^ h - 1 , 3^ h

R

T

SSSS

V

X

WWWW


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Solution #5.05-C_By Matrix (Rotation About a Referenced Vertex)Position #1: Vertices: A(4,0), B(8,0) & C(8,3) 1 = 0O 1 = 0

Position #2: 2 = 45O 1 = /4

Position #3:

3 = 90O

1 =/2

Reference Vertex VRef = A(4,0)

Populate the matrices

V'6 @ =

1 0 0

0 12

+ i2

0

0 0 i

R

T

SSSS

V

X

WWWW

:

4 8 8 + 3i

4 8 8 + 3i

4 8 8 + 3i

R

T

SSSS

V

X

WWWW

-

4 4 4

4 4 4

4 4 4

R

T

SSSS

V

X

WWWW

R

T

SSSS

V

X

WWWW

+

4 4 4

4 4 4

4 4 4

R

T

SSSS

V

X

WWWW

V'6 @ =

4 8 8 + 3i

4 4 + 2 + 2i^ h 2 4 +2

1+ 7i

4 4 + 4i 1 + 4i

R

T

SSSS

V

X

WWWW

=

4 8 8 + 3i

4 6.83 + 2.83i 4.71 + 4.95i

4 4 + 4i 1 + 4i

R

T

SSSS

V

X

WWWW

A B C

V'6 @ =

Position #1

Position #2

Position #3

R

T

SSSS

V

X

WWWW=

4,0^ h 8,0^ h 8,3^ h

4,0^ h 6.83, 2.83^ h 4.71, 4.95^ h4,0^ h 4,4^ h 1,4^ h

R

T

S

SSS

V

X

W

WWW


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Section #5.06: Rotate About Multiple VerticesProblem #5.06-A (Rotation of Triangle ABC About 2 Referenced Vertices - R1=R2)In the previous sections in this chapter, the calculations describe rotations about the origin or

another vertex. This section explores the calculations which describe sequential & simultaneousrotations about two vertices. The triangle is rotated about the origin and then rotated aboutone of its vertices. That is; Object #1 is rotated from = 0O to 90O counterclockwise fromPosition #1 to Position #3. The Problem is: Calculate the end vertices @ Position #3.

The first assumption is with rotation constants equal (R1=R2). R1= Rotation Constant about the Origin

R2= Rotation Constant about a Vertex (This Example: Vertex A)


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Problem 5.06A - Restated:Calculate the end vertices @ Position #3.Given: Position #1 @ Vertex A(3,0)

Rotate Object #1 about Origin (0,0) to Position #2Given: Position #2:Rotate Object #2 about Vertex A to Position #3

Given: R2 = R1 = R

The results are:Position #2 @ Vertex A(0,3):

Position #1 (@ 0O) Defined by vertices A(3,0), B(6,0) & C(6,4)Position #2 (@ 45O) Defined by vertices A(0,3), B(0,6) & C(-4,6)Position #3 (@ 90O) Defined by vertices A(0,3), B(-3,-3) & C(-3,-1)

Basic Rotation Model: V = R [ Vx - VRef ] + VRefResulting Vertex: Vout = VVin = initial vertices: Vx = R1 [ Vin - VRef#1 ] + VRef#1Substitutions: Vout = R2 [R1 ( Vin - VRef#1 ) + VRef#1 ] + VRef#2Distributions: Vout = R2 [R1Vin - R1VRef#1 + VRef#1 ] + VRef#2

Results: Vout = R2 R1 Vin - R2 R1 VRef#1 + R2 VRef#1 + VRef#2

Vout = R2 R1 ( Vin - VRef#1 ) + R2 VRef#1 + VRef#2

Given:R2 = R1 = R: Vout = R2Vin - R2VRef#1 + R VRef#1 + VRef#2 Vout = R2( Vin - VRef#1 ) + R VRef#1 + VRef#2

,

,

,

,,

V

A

B

C

i

i

i

V i

V i R R R i

V R R V V R V V

V i V V iV V

3 0

6 0

6 4

3 0

6 0

6 4

3 0 3 0 20 0 0 0

IN

Ref#1

Ref#2 1 2

OUT 1 2 IN Ref#1 2 Ref#1 Ref#2

OUT2

IN Ref#1 Ref#1 Ref#2

"

"

"

i r

= =

+

+

+

= + == + = = =

= - + +

= - + +

^ ^^ ^

^

^

h hh h

h

h

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

( , )

( , )

( , )

V i

i

i

i

i

i

i

i

i

i

i

i

i

i

V

i

i

i

i

i

i

i

i

i

i

i

i

V

i

i

i

A

B

C

3 0

6 0

6 4

3 0

3 0

3 0

3 0

3 0

3 0

0 0

0 0

0 0

1

0 0

3 0

3 4

0 3

0 3

0 3

0 0

3 0

3 4

0 3

0 3

0 30 3

3 3

3

0 3

3 3

3 1

OUT2

OUT

OUT "

=

+

+

+

-

+

+

+

+

+

+

+

+

+

+

+

= -

+

+

+

+

+

+

+

=

+

- +

- -

+

+

+

+

=

+

- +

- -

-

- -

J

L

KKK

J

L

KK

K

^

N

P

OOO

N

P

OO

O

h

R

T

SSSS

R

T

SSSS

R

T

SSSS

R

T

SSSS

R

T

SS

SS

R

T

SS

SS

R

T

SS

SS

R

T

SS

SSR

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

X

WW

WW

V

X

WW

WW

V

X

WW

WW

V

X

WW

WWV

X

WWWW

V

X

WWWW


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Problem 5.06B - Reference Vertex B(6,0):90O Rotation of OBJECT #1 to OBJECT #2


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Problem 5.06B - Reference ORIGIN Vertex O(0,0):90O Rotation of OBJECT #2 to OBJECT #3


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Problem 5.06B - Calculation of OUTPUT Vertices (OBJECT #3)

,

,,

V

A

B

C

i

i

i

3 0

6 06 4

3 0

6 06 4

IN = =

+

++

^

^^

h

hh

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

( , )( , )

R R R i

V i

V B i

90 1 90

0 0 0 0 0 06 0 6 0 6 0

O XO O

RO

O

RXO

= = = = =

= + == = + =

^^

hh

V R R V R R V R V V

V R R V V R V V

V R V V RV V

OUT O X IN O X RX O RX RO

OUT O X IN RX O RX RO

OUT

2

IN RX RX RO

= - + +

= - + +

= - + +

^

^

h

h

6 6

6 6

6 6

@ @

@ @

@ @

V i

i

i

i

i

i

i

i

i

i

i

i

i

i

3 0

6 0

6 4

6 0

6 0

6 0

6 0

6 0

6 0

0 0

0 0

0 0

OUT

2=

+

+

+

-

+

+

+

+

+

+

+

+

+

+

+

J

L

KKK

N

P

OOO

R

T

SSSS

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

i

i

i

i

i

i

1

3 0

0 0

0 4

6

6

6

OUT = - +

- +

+

+

+^ h

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

i

i

i

i

i

i

i

i

i

3 0

0 0

0 4

6

6

6

3 6

0 6

0 2

OUT =

+

+

-

+ =

+

+

+

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

,

,

,

.

,

,

V

A x y

B x y

C x y

3 6

0 6

0 2

OUT

A A

B B

C C

= =

^

^

^

^

^

^

h

h

h

h

h

h

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Refer to the previous graphic illustration to identify the input and output coordinates.The circular arcs indicate the path of the points (vertices) given the rotation completion ateach pivot point VRX and VRO. The rotation is incremented in angular steps (open dots) on thegraph. Compare VOUTMatrix to the vertices of OBJECT #3 in the graphic illustration.


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Problem 5.06C - Reference Vertex C(6,4):90O Rotation of OBJECT #1 to OBJECT #2

90O Rotation of OBJECT #2 to OBJECT #3


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Problem 5.06C - Calculation of OUTPUT Vertices (OBJECT #3)

,

,,

V

A

B

C

i

i

i

3 0

6 06 4

3 0

6 06 4

IN = =

+

++

^

^^

h

hh

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

( , )( , )

R R R i

V i

V C i

90 1 90

0 0 0 0 0 06 4 6 4

O XO O

RO

O

RX

= = = = =

= + == = +

^^

hh

V R R V R R V R V V

V R R V V R V V

V R V V RV V

OUT O X IN O X RX O RX RO

OUT O X IN RX O RX RO

OUT

2

IN RX RX RO

= - + +

= - + +

= - + +

^

^

h

h

6 6

6 6

6 6

@ @

@ @

@ @

V i

i

i

i

i

i

i

i

i

i

i

i

i

i

3 0

6 0

6 4

6 4

6 4

6 4

6 4

6 4

6 4

0 0

0 0

0 0

OUT

2=

+

+

+

-

+

+

+

+

+

+

+

+

+

+

+

J

L

KKK

N

P

OOO

R

T

SSSS

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

i

i

i

i

i

i

1

3 4

0 4

0 0

4 6

4 6

4 6

OUT = - +

- -

-

+

+

- +

- +

- +

^ h

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

i

i

i

i

i

i

i

i

i

3 4

0 4

0 0

4 6

4 6

4 6

1 10

4 10

4 6

OUT =

+

+

+

+

- +

- +

- +

=

- +

- +

- +

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

,

,

,

,

,

,

V

A x y

B x y

C x y

1 10

4 10

4 6

OUT

A A

B B

C C

= =

-

-

-

^

^

^

^

^

^

h

h

h

h

h

h

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Refer to the previous graphic illustration to identify the input and output coordinates.The circular arcs indicate the path of the points (vertices) given the rotation completion ateach pivot point VRX and VRO. The rotation is incremented in angular steps (open dots) on thegraph. Compare VOUTMatrix to the vertices of OBJECT #3 in the graphic illustration.


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Section #5.07: Rotate About Multiple Vertices - R1R2

Problem #5.07-A :Rotation of Triangle ABC About 2 Referenced Vertices - R1R2

In the previous sections in this chapter, the calculations describe rotations about the origin or

another vertex. This section explores the calculations which describe sequential & simultaneousrotations about two vertices. The triangle is rotated about the origin at on angle R1 and thenrotated about one of its vertices at on angle R2.

EXAMPLE

The first assumption is with rotation constants not equal (R1R2) R1=RORG Rotation Constant about the Origin 45O

R2=ROBJ Rotation Constant about the object Vertex 60O (This Example: Vertex A)

The Basic Rotation Model: Vout = R2 R1 Vin - R2 R1 VRef#1 + R2 VRef#1 + VRef#2 Vout = R2 R1 ( Vin - VRef#1 ) + R2 VRef#1 + VRef#2

Becomes:Vout = RORG ROBJ Vin - RORG ROBJ VRef-OBJ + RORG VRef-OBJ + VRef-ORG

Vout = RORG ROBJ ( Vin - VRef-OBJ ) + RORG VRef-OBJ + VRef-ORGGiven:

Vout = Coordinates Vertices of the OUTPUT MATRIXR1 = RX = ROBJ = Unit Rotation Constant of the OBJECT VERTEXR2 = RO = RORG = Unit Rotation Constant of the ORGIN VERTEX

VRef#1 = VRX = VRef-OBJ = Reference Vertex of the OBJECT VRef#2 = VRO = VRef-ORG = Reference Vertex of the ORGIN

Becomes:Vout = RORG ROBJ VIN - RORG ROBJ VRef-OBJ + RORG VRef-OBJ + VRef-ORG

Vout = RORG ROBJ ( VIN - VRef-OBJ ) + RORG VRef-OBJ + VRef-ORG

Symbolic:Vout = RO RX VIN - RO RX VRX + RO VRX + VRO

Vout

= RO

RX

( VIN

- VRX

) + RO

VRX

+ VRO


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Model 5.07: Rotation of Triangle ABC About 2 Referenced Vertices - RO RXThis model explores the algebra which describe sequential & simultaneous rotations about twovertices. The triangle is rotated about the origin VRO at on angle RO and then rotated about one

of its vertices VRX at on angle RX.

V R R V R R V R V V OUT O X IN O X RX O RX RO= - + +6 6@ @

V R R V V R V V OUT O X IN RX O RX RO= - + +^ h6 6@ @

Given:

,

,

,

V

A x y

B x y

C x y

x i y

x i y

x i y

IN

A A

B B

C C

A A

B B

C C

= =

+

+

+

^

^

^

h

h

h

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

( , )

,

R V i

R V x y x i y

Z x y

1 0 0 0 0 0 0

1

O O O ROO

X X X RX RX RX RX RX

2 2

i i

i i

= = = + =

= = = +

= +

^

^ ^

h

h h

Using:

V R R V V R V V OUT O X IN RX O RX RO= - + +^ h6 6@ @

V

x i y

x i y

x i y

x i y

x i y

x i y

x i y

x i y

x i y

i

i

i

1 1 1

0 0

0 0

0 0

OUT O X

A A

B B

C C

RX RX

RX RX

RX RX

O

RX RX

RX RX

RX RX

i i i=

+

+

+

-

+

+

+

+

+

+

+

+

+

+

+

J

L

KKK

^ ^ ^

N

P

OOO

h h h

R

T

SSSS

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

X

WWWW

V

x x i y y

x x i y y x x i y y

Z Tan

y

x1 1OUT O X

A RX A RX

B RX B RX

C RX C RX

O RX

1 RX

RXi i i= +

- + -

- + -

- + -

+

-

^ ^

^ ^^ ^ ^ c

h h

h hh h h m

R

T

S

SSS6

Documents

An Application of the Imaginary Number "i" Using Coordinate Geometry and Algebra