Universidad De Las Américas

UDLA

Hector Oña

31-12-2023

IEAZ4841-2850_2851-ROBOTICA INDUSTRIAL

Portafolio de Ejercicios Progreso 2

𝜃𝑥 = 45 𝜃30 = 30
1 0 0 𝐶𝑜𝑠(𝜃) 0 𝑆𝑖𝑛(𝜃)
Rx= [ 0 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃)] Ry=[ 0 1 0 ]
0 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0 𝐶𝑜𝑠(𝜃)
1 0 0 𝐶𝑜𝑠(30) 0 𝑆𝑖𝑛(30)
Rx= [0 𝐶𝑜𝑠(45) −𝑆𝑖𝑛(45)] Ry=[ 0 1 0 ]
0 𝑆𝑖𝑛(45) 𝐶𝑜𝑠(45) −𝑆𝑖𝑛(30) 0 𝐶𝑜𝑠(30)
1 0 0 √3 1
√2 √2 0
2 2
Rx= 0 2
−
2 Ry= 0 1 0
√2 √2 1 √3
[0 2 2 ] [− 2 0 2 ]

Matrix Rotación= Rx*Ry

𝑅 = 𝑅𝑥 ∗ 𝑅𝑦

1 0 0 √3 1
√2 √2 0
0 − 2 2
𝑅= 2 2 ∗ 0 1 0
√2 √2 1 √3
[0 − 0
2 2 ] [ 2 2]

√3 1
0
2 2
1 1 √3
𝑅= −
2√2 √2 2√2
1 1 √3
−
[ 2 √2 √2 2√2 ]
𝜃𝑤 = 30 𝜃𝑣 = 45
𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0 𝐶𝑜𝑠(𝜃) 0 𝑆𝑖𝑛(𝜃)
Rw= [ 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) 0] Rv=[ 0 1 0 ]
0 0 1 −𝑆𝑖𝑛(𝜃) 0 𝐶𝑜𝑠(𝜃)
𝐶𝑜𝑠(30) −𝑆𝑖𝑛(30) 0 𝐶𝑜𝑠(45) 0 𝑆𝑖𝑛(45)
Rw= [ 𝑆𝑖𝑛(30) 𝐶𝑜𝑠(30) 0] Rv=[ 0 1 0 ]
0 0 1 −𝑆𝑖𝑛(45) 0 𝐶𝑜𝑠(45)
√3 1 √2 √2
− 0 0
2 2 2 2
Rw= 1 √3 Rv= 0 1 0
0 √2 √2
2 2
[0 0 1] [− 2
0
2 ]

Matriz Rotacion R=Rv*Rw

√3 1 √2 √2
− 0 0
2 2 2 2
𝑅 = 1 √3 ∗ 0 1 0
0 √2 √2
2 2 − 0
[0 0 1 ] [ 2 2]
√6 1 √6
−
4 2 4
2 √3 √2
𝑅= √
4 2 4
√ 2 √2
−
[ 2 0
2]

𝜃𝑥 = 60 𝜃𝑦 = 45 𝜃𝑧 = 30
1 0 0 𝐶𝑜𝑠(𝜃) 0 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0
Rx= [0 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃)] Ry=[ 0 1 0 ] Rz= [ 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) 0]
0 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0 𝐶𝑜𝑠(𝜃) 0 0 1
 1 0 0 𝐶𝑜𝑠(45) 0 𝑆𝑖𝑛(45) 𝐶𝑜𝑠(30) −𝑆𝑖𝑛(30) 0
Rx= [0 𝐶𝑜𝑠(60) −𝑆𝑖𝑛(60)] Ry=[ 0 1 0 ] Rz= [ 𝑆𝑖𝑛(30) 𝐶𝑜𝑠(30) 0]
0 𝑆𝑖𝑛(60) 𝐶𝑜𝑠(60) −𝑆𝑖𝑛(45) 0 𝐶𝑜𝑠(45) 0 0 1
1 0 0 √2 √2 √3 1
1 √3 0 − 0
2 2 2 2
Rx= 0 2
−
2 Ry= 0 1 0 Rz= 1 √3
√3 1 √2 √2
0
2 2
[0 2 2 ] [− 2
0
2] [0 0 1]

𝑅 = 𝑅𝑥 ∗ 𝑅𝑦 ∗ 𝑅𝑧

01 0 √2 √2 √3 1
1 √3 0 − 0
0 − 2 2 2 2
𝑅= 2 2 ∗ 0 1 0 ∗ 1 √3
√3 1 √2 √2 0
0 − 0 2 2
[ 2 2 ] [ 2 2] [0 0 1]

√6 √2 √2
−
4 4 2
3√2 1 ( )
√2 − 2 ∗ √3 √6
𝑅= + − −
8 4 8 4
(√2 − 2) ∗ √3 √2 3 √2
[− 8 8 4
+
4 ]

Necesitamos encontrar S de la matriz, esto se hace para las componentes (x,y,z)

𝑎 32 − 𝑎 23
𝑆𝑋 =
2𝑠𝑖𝑛(𝜃)
𝑎13 − 𝑎 31
𝑆𝑌 =
2𝑠𝑖𝑛(𝜃)
𝑎 21 − 𝑎12
𝑆𝑍 =
2𝑠𝑖𝑛(𝜃)
Angulo de rotación 𝜃

𝑎11 + 𝑎 22 + 𝑎 33 − 1
𝜃 = 𝑐𝑜𝑠 −1
2

Angulo de rotación 𝜃
𝑎11 + 𝑎 22 + 𝑎 33 − 1
𝜃 = 𝑐𝑜𝑠 −1
2
0.866 + 0.866 + 1 − 1
𝜃 = 𝑐𝑜𝑠 −1
2
𝜃 = 30°

𝑎32 −𝑎23 0−0

𝑆𝑥 = = =0
2𝑠𝑖𝑛𝜃 2sin (30)
𝑎13 −𝑎31 0−0
𝑆𝑦 = 2𝑠𝑖𝑛𝜃
= 2sin (30)= 0
𝑎21 −𝑎12 0.5+0.5
𝑆𝑧 = 2𝑠𝑖𝑛𝜃
= 2sin (30)
=1
Angulo de rotación 𝜃
𝑎11 + 𝑎 22 + 𝑎 33 − 1
𝜃 = 𝑐𝑜𝑠 −1
2
1 + 0.866 + 0.866 − 1
𝜃 = 𝑐𝑜𝑠 −1
2
𝜃 = 30°

0.5+0.5
𝑆𝑥= 2sin (30) = 1
0−0
𝑆𝑦 = =0
2sin (30)
0+0
𝑆𝑧 = 2sin (30)
=0

En base a la matriz propuesta el ángulo será 30 grados y rotara en X

𝑎11 + 𝑎 22 + 𝑎 33 − 1
𝜃 = 𝑐𝑜𝑠 −1
2
−1
0.866 + 0.866 + 1 − 1
𝜃 = 𝑐𝑜𝑠
2
𝜃 = 30°
 0 −0
𝑆𝑥 = =0
2𝑠(30)
0−0
𝑆𝑦 = =0
2𝑠(30)
0.5 + 0.5 1
𝑆𝑧 = = 1 =1
2𝑠(30) 2( )
2

𝜃𝑧 = 30
𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0
Rz= [ 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) 0 ]
0 0 1
𝐶𝑜𝑠(30) −𝑆𝑖𝑛(30) 0
Rz= [ 𝑆𝑖𝑛(30) 𝐶𝑜𝑠(30) 0]
0 0 1
√3 1
− 0
2 2
Rz= 1 √3
0
2 2
[0 0 1]

Matriz Homogénea

√3 1
−
2 2 0 0
𝑀= 1 √3 0 0
2 2
0 0 1 0
[ 0 0 0 1]

A. Matriz trasnformacion aTb

Pa=(2, 4, 0)
1 0 𝟎 𝟐
𝑅𝑃𝑎 = [0 1 𝟎 𝟒]
0 0 𝟏 𝟎
0 𝟎 𝟎 𝟏
 √𝟑 𝟏
−
𝟐 𝟐 𝟎 𝟎 𝟏 𝟎 𝟎 𝟐
𝑨 𝟏 √𝟑 𝟎 𝟎 ∗ [𝟎 𝟏 𝟎 𝟒]
𝑩𝑻 = 𝟎 𝟎 𝟏 𝟎
𝟐 𝟐
𝟎 𝟎 𝟏 𝟎 𝟎 𝟎 𝟎 𝟏
[ 𝟎 𝟎 𝟎 𝟏]

√𝟑 𝟏
− 𝟎 √𝟑 − 𝟐
𝟐 𝟐
𝑨 𝟏 √𝟑 𝟎 𝟐√ 𝟑 + 𝟏
𝑩𝑻 =
𝟐 𝟐
𝟎 𝟎 𝟏 𝟎
[ 𝟎 𝟎 𝟎 𝟏 ]

B. Matriz Inversa bTa

√𝟑 𝟏
− 𝟎
𝟐 𝟐
Rz= 𝟏 √𝟑
𝟎
𝟐 𝟐
[𝟎 𝟎 𝟏]
Pa=(2, 4, 0)

𝑹𝒂 = 𝑹𝒛 ∗ 𝑷𝒂

√𝟑 𝟏
− 𝟎
𝟐 𝟐 𝟐
𝑹𝒂 = 𝟏 √𝟑 ∗ [𝟒 ]
𝟎 𝟎
𝟐 𝟐
[𝟎 𝟎 𝟏]

√𝟑 − 𝟐
𝑹𝒂 = [𝟐√𝟑 + 𝟏]
𝟎

√𝟑 𝟏
− 𝟎 −𝟐
𝟐 𝟐
𝑩 𝟏 √𝟑 𝟎 −𝟒
𝑨𝑻 =
𝟐 𝟐
𝟎 𝟎 𝟏 𝟎
[ 𝟎 𝟎 𝟎 𝟏]
 𝟎 𝟎 𝟏 −𝟏
𝟏 √𝟑 𝟏 𝟏+𝟐√𝟑
𝟎 * { 𝟐 }= //
𝟐 𝟐 𝟐
√𝟑 𝟏 −𝟏 −√𝟑+𝟐
{− 𝟐 𝟐
𝟎} { 𝟐 }

Quaternion
𝒂𝟏𝟏 + 𝒂𝟐𝟐 + 𝒂𝟑𝟑 − 𝟏
𝜽 = 𝐜𝐨𝐬−𝟏
𝟐
𝟎 + √𝟐 + 𝟎 − 𝟏
𝟑
𝜽 = 𝐜𝐨𝐬−𝟏
𝟐
𝜽 = 𝟏. 𝟔𝟑° = 𝟗𝟑. 𝟑°

𝟏
𝒂𝟑𝟐 −𝒂𝟐𝟑 −𝟎
𝑺𝒙 = 𝟐𝒔𝜽
= 𝟐𝐬𝐢𝐧 (𝟗𝟑.𝟑) = 0.25
𝟐

√𝟑
𝒂𝟏𝟑 −𝒂𝟑𝟏 𝟏−(− )
𝑺𝒚 = 𝟐𝒔𝜽
= 𝟐𝐬𝐢𝐧 (𝟗𝟑.𝟑)=0.93
𝟐

𝟏
𝒂𝟐𝟏 −𝒂𝟏𝟐 −()
𝑺𝒛 = 𝟐𝒔𝜽
= 𝟐
𝟐𝐬𝐢𝐧 (𝟗𝟑.𝟑)
=0.25
MODULO=

|𝒖| = √𝟎. 𝟐𝟓𝟐 + 𝟎. 𝟗𝟑𝟐 + 𝟎. 𝟐𝟓𝟐

|𝒖| = 𝟎. 𝟗𝟗
𝟎. 𝟐𝟓
{𝟎. 𝟗𝟑}
𝟎. 𝟐𝟓

𝟏.𝟔𝟑
𝟎. 𝟐𝟓 𝟏.𝟔𝟑
𝒒 = 𝑪𝒐𝒔 ( ) + {𝟎. 𝟗𝟑}*𝐬𝐢𝐧 ( )
𝟐 𝟐
𝟎. 𝟐𝟓
𝒒 = 𝟎. 𝟔𝟖 + 𝟎. 𝟏𝟖𝟑𝒊 + 𝟎. 𝟔𝟖𝟑 𝒋 + 𝟎. 𝟏𝟖𝟑𝒌
𝒒 ∗= 𝟎. 𝟔𝟖 − 𝟎. 𝟏𝟖𝟑𝒊 − 𝟎. 𝟔𝟖𝟑𝒋 − 𝟎. 𝟏𝟖𝟑𝒌

𝒒𝟐 = −𝟏. 𝟑𝟔𝟔 − 𝟎. 𝟑𝟔𝟔𝒊 + 𝟏. 𝟕𝟑𝟐 𝒋 − 𝟏𝒌

𝒒 ∗= 𝟎. 𝟔𝟖 − 𝟎. 𝟏𝟖𝟑𝒊 − 𝟎. 𝟔𝟖𝟑𝒋 − 𝟎. 𝟏𝟖𝟑𝒌
Determinacion Angulo

√𝟑 + 𝟏 + √𝟑 − 𝟏
𝜽 = 𝐜𝐨𝐬−𝟏 𝟐 𝟐
𝟐
√𝟑
𝜽 = 𝒄𝒐𝒔−𝟏 ( )
𝟐
𝜽 = 𝟑𝟎
 𝒂𝟑𝟐 −𝒂𝟐𝟑 𝟎−𝟎
𝑺𝒙 = = =0
𝟐𝒔𝜽 𝟐𝐬𝐢𝐧 (𝟑𝟎)
𝟏 𝟏
𝒂𝟏𝟑 −𝒂𝟑𝟏 −(− )
𝑺𝒚 = 𝟐𝒔𝜽
= 𝟐𝐬𝐢𝐧 (𝟑𝟎)=1
𝟐 𝟐

𝒂𝟐𝟏 −𝒂𝟏𝟐 𝟎−𝟎

𝑺𝒛 = 𝟐𝒔𝜽
= 𝟐𝐬𝐢𝐧 (𝟑𝟎)
=0

Parametros: 𝜽 = 𝟑𝟎 ; 𝑲 = (𝟎 𝟏 𝟎)𝑻

Movimiento
Mx= 3
My= 10
Mz= -1

1 0 0 3
0 cos(30) −sin (30) 𝑀𝑦
𝑇=[ ]
0 sin(30) cos(30) 𝑀𝑧
0 0 0 1

1 0 0 3
√3 1
0 − 10
2 2
𝑇=
1 √3
0 −1
2 2
[0 0 0 1]
17. 𝑃 𝐵 = (0, −1, 2) 𝑇
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑃 𝐴
0
−1
𝑃𝐴 = 𝑇 ∗ [ ]
2
1
1
0 0 3
√3 1 0
0 − 10
𝐴 2 2 −1
𝑃 = ∗[ ]
1 √3 2
0 −1 1
2 2
[0 0 0 1 ]

3
18 − √3
𝑃𝐴 = 2
2√3 − 3
2
[ 1 ]
19.
Punto A = (6,2,4)
Eje de rotación Z
𝜃𝑧 = 90
𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0
Rz= [ 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) 0 ]
0 0 1
𝐶𝑜𝑠(90) −𝑆𝑖𝑛(90) 0
Rz= [ 𝑆𝑖𝑛(90) 𝐶𝑜𝑠(90) 0]
0 0 1
0 1 0
Rz= [−1 0 0]
0 0 1

Matriz Homogenea A
0 1 0 𝐴𝑥
−1 0 0 𝐴𝑦]
𝑀𝑎 = [
0 0 1 𝐴𝑧
0 0 0 1

0 1 0 6
𝑀𝑎 = [−1 0 0 4]
0 0 1 4
0 0 0 1

20.
Punto B = (-2, 9, 7)
No existe rotación

1 0 0
𝑅 = [0 1 0]
0 0 1

Matriz Homogenea B
 1 0 0 𝐵𝑥
𝐵𝑦 ]
𝑀𝑏 = [0 1 0
0 0 1 𝐵𝑧
0 0 0 1

1 0 0 −2
𝑀𝑏 = [0 1 0 9]
0 0 1 7
0 0 0 1

21.

Punto C = (2, 3, -4)

Hay dos rotaciones, una en x y otra en Y

𝜃𝑥 = 180 𝜃𝑦 = −90
1 0 0 𝐶𝑜𝑠(𝜃) 0 𝑆𝑖𝑛(𝜃)
Rx= [ 0 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃)] Ry=[ 0 1 0 ]
0 𝑆𝑖𝑛(𝜃) 𝐶𝑜𝑠(𝜃) −𝑆𝑖𝑛(𝜃) 0 𝐶𝑜𝑠(𝜃)
1 0 0 𝐶𝑜𝑠(−90) 0 𝑆𝑖𝑛(−90)
Rx= [0 𝐶𝑜𝑠(180) −𝑆𝑖𝑛(180)] Ry=[ 0 1 0 ]
0 𝑆𝑖𝑛(180) 𝐶𝑜𝑠(180) −𝑆𝑖𝑛(−90) 0 𝐶𝑜𝑠(−90)
1 0 0 0 0 −1
Rx= [0 −1 0 ] Ry= [0 1 0 ]
0 0 −1 1 0 0

R= Rx*Ry

1 0 0 0 0 −1
𝑅 = [0 −1 0 ] ∗ [0 1 0 ]
0 0 −1 1 0 0
 0 0 −1
𝑅 = [ 0 −1 0 ]
−1 0 0

Matriz Homogénea C

0 0 −1 𝐶𝑥
𝐶𝑦]
𝑀𝑐 = [ 0 −1 0
−1 0 0 𝐶𝑧
0 0 0 1

0 0 −1 2
𝑀𝑐 = [ 0 −1 0 3 ]
−1 0 0 −4
0 0 0 1

2. Resolver por matrices, quaternios y quaternios duales.

cos (π) −sen (π) 10 −1 0 10

M1 = [sen (π) cos (π) 4 ] = [ 0 −1 4 ]
0 0 1 0 0 1
3π
M2 -> B T B
Ct = [−10 5] y CRz = rad.
4
3π 3π −√2 −√2
cos ( ) −sen ( ) −10 −10
4 4 2 2
M1 = 3π 3π = √2 −√2
sen ( ) cos ( ) 5 5
4 4 2 2
[ 0 0 1 ] [ 0 0 1 ]
La posición del punto C:
P C = M1 ∗ M2 ∗ p
−√2 −√2
−10
−1 0 10 2 2 −2
C
P = [ 0 −1 4 ] ∗ √2 −√2 ∗[ 3 ]
0 0 1 5 0
2 2
[ 0 0 1 ]
√2
+ 40
2
P C = 5√2
−1
2
[ 1 ]

Quaterniones
S = (0,0,1)
θ θ
q = C2 + S ∗ S 2
0
θ θ π π
q1 = C 21 + S ∗ S 21 = C2 + S ∗ S 2 = 0 + [0 ] ∗ 1 = 0 + 0i + 0j + k
1
3π 3π 0
θ θ
q2 = C 22 + S ∗ S 22 = C 42 + S ∗ S 42 = 0.382683 + [0] ∗ 0.92388
1
= 0.382683 + 0i + 0j + 0.92388k

Traslación:
t1 = 0 + 10i + 4j + 0k
t 2 = 0 − 10i + 5j + 0k

P C = q1 ( q2 ∗ p ∗ q2 ∗ + t 2 ) ∗ q1 ∗ + t1
P C = 0 + 20.71i + 2.54j + 1k

Quaterniones duales
ϵrxyz = h0 ∗ ϵruvw ∗ ̅̅̅̅
h0 ∗

t i ∗ qi
hi = qi + ε
2
t1 ∗ q1 (0 + 10i + 4j + 0) ∗ (0 + 0i + 0j + k)
h1 = q1 + ε = (0 + 0i + 0j + k) + ε
2 2
= (0 + 0i + 0j + k) + ε(0 + 2.5i − 5j + 0k)
 t 2 ∗ q2
h2 = q2 + ε
2
= (0.382683 + 0i + 0j + 0.92388k)
(0 − 10i + 5j + 0k) ∗ (0.382683 + 0i + 0j + 0.92388k)
+ε
2
= (0.382683 + 0i + 0j + 0.92388k) + ε(0 + 0.396857i − 5.57611j
+ 0k)

Dual conjugado:
̅̅̅̅
h1 ∗ = (0 − 0i − 0j − k) − ε(0 + 2.5i − 5j + 0k)
̅̅̅̅
h2 ∗ = (0.382683 − 0i − 0j − 0.92388k) − ε(0 + 0.396857i − 5.57611j + 0k)

El punto C resulta:
ϵC = h1 ∗ h2 ∗ ϵP ∗ ̅̅̅̅
h1 ∗ ∗ ̅̅̅̅
h2 ∗
Donde ϵP = (1 + 0i + 0j + 0k) + ε(0 − 2i + 3j + 1k))
ϵC = ((0 + 0i + 0j + k) + ε(0 + 4i − 10j + 0k)) ∗ ((0 + 0i + 0j + k) + ε(0 + 0.79257i
− 11.152215j + 0k)) ∗ ϵP ∗ ((0 − 0i − 0j − k) − ε(0 + 4i − 10j + 0k))
∗ ( (0.382683 − 0i − 0j − 0.92388k) − ε(0 + 0.79257i − 11.152215j
+ 0k))
ϵC = ( (1 + 0i + 0j + 0k) + ε(0 + 20.71i + 2.54j + 1k)

Matrices

1 0 0 1 0 0
( )
R x, 90 = [ 0 C(90° ) −S(90)] = [0 0 −1]
0 S(90) C(90) 0 1 0
Translación en p (8, -4, 12)
 1 0 0 8
XYZ
TUVW = [0 0 −1 −4]
0 1 0 12
0 0 0 1
1 0 0 8 −3
XYZ
rUVW = [0 0 −1 −4] ∗ [ 4 ]
0 1 0 12 −11
0 0 0 1 1
5
XYZ 7
rUVW =[ ]
16
1
Quaterniones
Quaternion de traslación p (8, -4, 12)
qt = 0 + 8i − 4j + 12k

Quaternion de rotación
S = i + 0j + 0k
90 90
qr = cos ( ) + S ∗ sin ( )
2 2
√ 2 √ 2
qr = + i + 0j + 0k
2 2

√2 √2
qr ∗ = − i − 0j − 0k
2 2

p = 0 − 3i + 4j − 11k
rxyz = qr ∗ p ∗ qr ∗
√2 √2 3√2
− 0 0
2 2 2
√2 √2 0 −3√2
0 0
−3
qr ∗ p = 2 2 ∗[ ]= 2
√2 √2 4 15√2
0 0 − −11
2 2 2
√ 2 √ 2 −7 √2
[ 0 0
2 2 ] [ 2 ]
3√2 3√2 15√2 7√2
−
2 2 2 2 √2
3√2 3√2 7√2 15 √2 2 0
−
qr ∗ p ∗ qr ∗ = 2 2 2 2 ∗ √2 = [−3]
15√2 −7√2 3√2 3√2 − 11
2
2 2 2 2 0 4
−7√2 15√2 3√2 [ 0 ]
√2
[ 2 − −
2 2 2 ]
∗
(qr ∗ p ∗ qr ) + qt = (0 − 3i + 11j + 4k) + (0 + 8i − 4j + 12k)
rxyz = 0 + 5i + 7j + 16k
Quaterniones duales
∈rxyz = h0 ∗ ∈ruvw ∗ ̅̅̅̅
h0 ∗
t∗q
h0 = q + ε
2
√2 √2
√2 √2 (0 + 8i − 4j + 12k) ∗ ( + i + 0j + 0k)
h0 = ( + i + 0j + 0k) + ε 2 2
2 2 2
√ 2 √ 2
h0 = ( + i + 0j + 0k) + ε(−2√2 + 2√2i + 2√2j + 2√2k)
2 2
̅̅̅̅ √2 √2
h0 ∗ = ( + i + 0j + 0k) − ε(−2√2 − 2√2i − 2√2j − 2√2k)
2 2

∈ruvw = (1 + 0i + 0j + 0k) + ε(0 − 3i + 4j − 11k)

√2 √2 √2 √2
h0 ∗ ∈ruvw = ( + i + 0j + 0k) ∗ (1 + 0i + 0j + 0k) + ε(( + i + 0j + 0k)
2 2 2 2
∗ (0 − 3i + 4j − 11k) + (−2√2 + 2√2i + 2√2j + 2√2k)
∗ (1 + 0i + 0j + 0k))
h0 ∗ ∈ruvw = (0.71 + 0.71i + 0.0j + 0.0k) + ε(−0.71 + 0.71i + 13.44j + 0.71k)
√2 √2
h0 ∗ ∈ruvw ∗ ̅̅̅̅
h0 ∗ = (0.71 + 0.71i + 0.0j + 0.0k) ∗ ( + i + 0j + 0k) + ε((0.71
2 2
+ 0.71i + 0.0j + 0.0k) ∗ (2√2 + 2√2i + 2√2j + 2√2k) + (−0.71 + 0.71i
√2 √2
+ 13.44j + 0.71k) ∗ ( + i + 0j + 0k))
2 2
rxyz = (1 + 0i + 0j + 0k) + ε(0 − 5i + 7j − 16k)

Matrices
 C( −90°) −S(−90°) 0 0 1 0
R(z, −90°) = [ S(−90°) C(−90°) 0] = [−1 0 0]
0 0 1 0 0 1

0 1 0 4
rxyz = [−1 0 0] ∗ [ 8 ]
0 0 1 12
8
rxyz = [−4]
12

Quaterniones

θ θ
q = C2 + S ∗ S 2
0
−90 −90
q = C 2 + [0] ∗ S 2
1
√2 √2
q= + 0i + 0j − k
2 2

p = q ∗ p ∗ q∗
√2 √2 √2 √2
rxyz = ( + 0i + 0j − k) ∗ (0 + 4i + 8j + 12k) ∗ ( − 0i − 0j + k)
2 2 2 2
√2 √2
0 0 −
2 2
√ 2 √ 2 0 6√2
0 0
4
= 2 2 ∗ [ ] = 6√2
√2 √2 8 2√2
0 − 0 12 [
2 2 6√2]
√2 √2
[− 2 0 0
2 ]
√2
6√2 −6√2 −2√2 −6√2 2 0
6√2 6√2 −6√2 2√2 0 8
= ∗ =[ ]
2√2 6√2 6√2 −6√2 0 −4
[6√2 −2√2 6√2 6√2 ] √2 12
[2]

Quaterniones duales
ϵrxyz = h0 ∗ ϵruvw ∗ ̅̅̅̅
h0 ∗

t∗q
h0 = q + ε
2
√2 √2
√2 √2 (0 + 4i + 8j + 12k) ∗ ( + 0i + 0j − k)
h0 = ( + 0i + 0j − k) + ε 2 2
2 2 2
 √2 √2
t ∗ q = (0 + 4i + 8j + 12k) ∗ ( + 0i + 0j − k)
2 2
√2
0 −4 −8 −12 2 6√2
4 0 −12 8 ] 0
t ∗q = [ ∗ = −2√2
8 12 0 −4 0 6√2
12 −8 4 0 √2 [ 6√2 ]
[− 2 ]
3√2
t∗q
= −√2
2 3√2
[ 3√2 ]
√2 √2
h0 = ( + 0i + 0j − k) + ε(3√2 − √2i + 3√2j + 3√2k)
2 2

̅̅̅̅ √2 √2
h0 ∗ = ( − 0i − 0j + k) − ε(3√2 − √2i + 3√2j + 3√2k)
2 2

ϵruvw = (1 + 0i + 0j + 0k) + ε(0 + 4i + 8j + 12k))

ϵrxyz = h0 ∗ ϵruvw ∗ ̅̅̅̅

h0 ∗
√2 √2
ϵrxyz = ( + 0i + 0j − k) + ε(3√2 − √2i + 3√2j + 3√2k) ∗ ((1 + 0i + 0j + 0k) + ε(0
2 2
√2 √2
+ 4i + 8j + 12k)) ∗ (( − 0i − 0j + k) − ε(3√2 − √2i + 3√2j
2 2
+ 3√2k))

∈rxyz = (1 + 0i + 0j + 0k) + ε(0 + 8i − 4j + 12k)

Matrices
0.2366124 0.9501203 0.2031894 −10
A A
BR = [0.8820099 −0.2977589 0.3652371 ] ; Bt = [ 5 ]
0.4075207 0.0927954 −0.9084690 2

0.2366124 0.9501203 0.2031894 −10

A 0.8820099 −0.2977589 0.3652371 5 ]
BT =[
0.4075207 0.0927954 −0.9084690 2
0 0 0 1

1 0 0 1 0 0 0
B
CR = [0 Cπ −Sπ] = [0 −1 0 ] ; B
Ct = [0]
0 Sπ Cπ 0 0 −1 5

1 0 0 0
B 0 −1 0 0]
CT = [ 0 0 −1 5
0 0 0 1
PA = A B
BT CTP
C
 0.2366124 0.9501203 0.2031894 −10 1 0 0 0 1
0.8820099 −0.2977589 0.3652371 5 ] [0 −1 0 0] [0 ]
PA = [
0.4075207 0.0927954 −0.9084690 2 0 0 −1 5 1
0 0 0 1 0 0 0 1 1
P A = (−8.9506 7.3430 −1.2264)

Quaterniones
θ θ
q = C2 + S ∗ S 2

0.2366124 0.9501203 0.2031894

A
BR = [0.8820099 −0.2977589 0.3652371 ]
0.4075207 0.0927954 −0.9084690
0.2366124 − 0.2977589 − 0.9084690 − 1
θ = cos −1 = 170
2

0.0927954 − 0.3652371
Sx = = −0.78446
2s(170)
0.2031894 − 0.4075207
Sy = = −0.58834
2s (170)
0.8820099 − 0.9501203
Sz = = −0.19611
2s(170)
S = (−0.78446, −0.58834, −0.19611)

A 170 170
Bq =C 2
+ (−0.78446, −0.58834, −0.19611) ∗ S 2

A
Bq = −0.98427 + 0.13812i + 0.10359j + 0.03453k
A ∗
Bq = −0.98427 − 0.13812i − 0.10359j − 0.03453k

B π π
Cq = C2 + (1,0,0) ∗ S2
B
Cq = 0 + 1i + 0j + 0k
B ∗
Cq = 0 − 1i − 0j − 0k

A
Bt = 0 − 10i + 5j + 2k
B
Ct = 0 + 0i + 0j + 5k
 P C = 0 + 1i + 0j + 1k

PA = A B CB ∗ B A ∗ A
Bq( CqP Cq + Ct) Bq + Bt

P A = (−0.98427 + 0.13812i + 0.10359j + 0.03453k) [(0 + 1i + 0j + 0k)(0 + 1i + 0j + 1k)(0 − 1i

− 0j − 0k) + (0 + 0i + 0j + 5k)](−0.98427 − 0.13812i − 0.10359j − 0.03453k)
+ (0 − 10i + 5j + 2k)

P A = 0 − 8.95i + 7.34j − 1.23k

Quaterniones duales
θ θ
q = C2 + S ∗ S 2

0.2366124 0.9501203 0.2031894

A
BR = [0.8820099 −0.2977589 0.3652371 ]
0.4075207 0.0927954 −0.9084690
0.2366124 − 0.2977589 − 0.9084690 − 1
θ = cos −1 = 170
2
0.0927954 − 0.3652371
Sx = = −0.78446
2s(170)
0.2031894 − 0.4075207
Sy = = −0.58834
2s (170)
0.8820099 − 0.9501203
Sz = = −0.19611
2s(170)
S = (−0.78446, −0.58834, −0.19611)

A 170 170
Bq =C 2
+ (−0.78446, −0.58834, −0.19611) ∗ S 2

A
Bq = −0.98427 + 0.13812i + 0.10359j + 0.03453k
A
Bt = 0 − 10i + 5j + 2k
A A
A Bt ∗ Bq
Bh =A
Bq + ε
2
A
Bh = [0.0871 − 0.781i − 0.586j − 0.195k − ε2.247 − ε0.338i − ε1.540j + ε4.971k]
̅̅̅̅̅
A ∗
Bh = [0.0871 + 0.781i + 0.586j + 0.195k + ε2.247 − ε0.338i − ε1.540j + ε4.971395k]

B π π
Cq = C2 + (1,0,0) ∗ S2
 B
Cq = 0 + 1i + 0j + 0k
B
Ct = 0 + 0i + 0j + 5k
B
B C t∗ B
Cq
Ch =B
Cq + ε
2
B
Ch = [0 + 1.0i + 0.0j + 0.0k + ε0.0 + ε0.0i + ε2.5j + ε0.0k]
̅̅̅̅̅
B ∗
Ch = [0 − 1.0i − 0.0j − 0.0k − ε0.0 + ε0.0i + ε2.5j + ε0.0k]

PA = A B ̅̅̅̅̅∗ ̅̅̅̅̅
CB A ∗
Bh Ch P Ch Bh

P A = [1 + 0i − 0.0j + 0.0k + ε0 − ε8.950i + ε7.343j − ε1.226k]

La rotación y translación:
π
R z0 = θ1 = rad = 30°
6
Tx1 = a1 = 3ft
π
R z2 = θ2 = rad = 45°
4
Tx3 = a2 = 2ft
π
R z4 = θ3 = rad = 60°
3
 P ′ = (a3,0) = (1,0)

Matrices
M0 -> T=0 y R z0 = 30°
√3 1
C(30°) −S(30°) 0 0 − 0 0
2 2
M0 = [ S(30°) C(30°) 0 0] = 1 √3
0 0
0 0 1 0 2 2
0 0 0 1 0 0 1 0
[0 0 0 1]
M1 ->Tx1 = 3ft y R z2 = 45°

√2 √2
C(45°) −S(45°) 0 3 − 0 3
2 2
M1 = [ S(45°) C(45°) 0 0] = √2 √2
0 0
0 0 1 0 2 2
0 0 0 1 0 0 1 0
[0 0 0 1]
M2 ->Tx3 = 2ft y R z4 = 60°

1 √3
C(60°) −S(60°) 0 2 − 0 2
2 2
M1 = [ S(60°) C(60°) 0 0] = √3 1
0 0
0 0 1 0 2 2
0 0 0 1 0 0 1 0
[0 0 0 1]

R = M0 ∗ M1 ∗ M2 ∗ P′
√3 1 √2 √2 1 √3
− 0 0 − 0 3 − 0 2 1
2 2 2 2 2 2
R = 1 √3 0 √ 2 √ 2 √ 3 1 [0]
0 ∗ 0 0 ∗ 0 0 ∗ 0
2 2 2 2 2 2
0 0 1 0 0 0 1 0 0 0 1 0 1
[0 0 0 1] [ 0 0 0 1] [ 0 0 0 1]
1
−
√2 2.4086
1 4.1390 ]
R= =[
√2 0
0 0
[ 0 ]

Quaterniones
S0 = S2 = S4 = 0i + 0j + k
 θi θi
qi = C ( ) + Si ∗ S ( )
2 2
θ1 θ1 30 0 30
q0 = C ( ) + S0 ∗ S ( ) = C ( ) + [0] ∗ S ( ) = 0.96592 + 0i + 0j + 0.25881k
2 2 2 2
1
θ2 θ2 45 0 45
q2 = C ( ) + S2 ∗ S ( ) = C ( ) + [0] ∗ S ( ) = 0.92388 + 0i + 0j + 0.38268k
2 2 2 2
1
θ3 θ3 60 0 60
q4 = C ( ) + S4 ∗ S ( ) = C ( ) + [0] ∗ S ( ) = 0.86602 + 0i + 0j + 0.5k
2 2 2 2
1

t1 = 0 + 3i + 0j + 0k
t 3 = 0 + 2i + 0j + 0k
P′ = 0 + 1i + 0j + 0k
P = q0 q2 q4 ∗ P ′ ∗ q4 ∗ + t 3 ) ∗ q2 ∗ + t1 ) ∗ q0 ∗
0 ( (
P 0 = 0 + 2.41i + 4.14j + 0k

Quaterniones duales

ϵrxyz = h0 ∗ ϵruvw ∗ ̅̅̅̅

h0 ∗

t i ∗ qi
hi = qi + ε
2

t 0 ∗ q0
h0 = q0 + ε
2
h0 = (0.96592 + 0i + 0j + 0.25881k)
(0 + 0i + 0j + 0k) ∗ (0.96592 + 0i + 0j + 0.25881k)
+ε
2
h0 = (0.96592 + 0i + 0j + 0.25881k) + ε(0 + 0i − 0j + 0k)

t1 ∗ q2
h2 = q2 + ε
2
h2 = (0.92388 + 0i + 0j + 0.38268k)
(0 + 3i + 0j + 0k) ∗ (0.92388 + 0i + 0j + 0.38268k)
+ε
2
h2 = (0.92388 + 0i + 0j + 0.38268k) + ε(0 + 0.1.3858i − 0.57403j + 0k)

t 3 ∗ q4
h4 = q4 + ε
2
(0 + 2i + 0j + 0k) ∗ (0.86602 + 0i + 0j + 0.5k)
h4 = (0.86602 + 0i + 0j + 0.5k) + ε
2
h4 = (0.86602 + 0i + 0j + 0.5k) + ε(0 + 0.86603i − 0.5j + 0k)
 ̅̅̅̅
h0 ∗ = (0.96592 − 0i − 0j − 0.25881k) − ε(0 + 0i − 0j + 0k)
̅̅̅̅
h2 ∗ = (0.92388 − 0i − 0j − 0.38268k) − ε(0 + 0.1.3858i − 0.57403j + 0k)
̅̅̅̅
h4 ∗ = (0.86602 − 0i − 0j − 0.5k) − ε(0 + 0.86603i − 0.5j + 0k)

ϵC = h0 ∗ h2 ∗ h4 ∗ ϵP ∗ ̅̅̅̅
h0 ∗ ∗ ̅̅̅̅
h2 ∗ ∗ ̅̅̅̅
h4 ∗

Donde ϵP = (1 + 0i + 0j + 0k) + ε(0 − 2i + 3j + 1k))

ϵC = 1 + 0i + 0j + 0k + ε(0 + 2.41i + 4.14j + 0k)

Matrices

Traslación: A
Bt = (−10 2 10 )
π
Rotación 1: A
BR1 => θ = rad; S = (1 0 0 )
10
π
Rotación 2: A
BR 2 => θ = rad; S = (0 1 0)
10
 Traslación: B
Ct = ( 1 10 −10)

Posición de la cámara: P C = ( 0 0 2)
1 0 0
π π
A 0 c −s
BR1 = 10 10
π π
[0 s 10 c 10 ]
π π
c 0 s
10 10
A
BR 2 = 0 1 0
π π
−s
[ 10 0 c
10]
1 0 0 π π
π π c 0 s
0 c −s 10 10 0.9511 0 0.3090
A
BR = 10 10 ∗ 0 1 0 = [ 0.0955 0.9511 −0.2939]
π π π π −0.2939 0.3090 0.9045
[0 s 10 c 10 ] [ −s 10 0 c 10]
0.9511 0 0.3090 −10
A 0.0955 0.9511 −0.2939 2 ]
BT =[
−0.2939 0.3090 0.9045 10
0 0 0 1
1 0 0 1
B 0 1 0 10 ]
CT = [0 0 1 −10
0 0 0 1
PA = A B
BT CTP
C

0.9511 0 0.3090 −10 1 0 0 1 0

0.0955 0.9511 −0.2939 2 ] [0 1 0 10 ] [0 ]
PA = [
−0.2939 0.3090 0.9045 10 0 0 1 −10 2
0 0 0 1 0 0 0 1 1
−11.5211
13.9572 ]
PA =[
5.5602
1
P A = (−11.5211 13.9572 5.5602 )

Quaterniones

TA
Bt = (0 − 10i + 2j + 10k)
A π
BR1 => θ = rad; S = (1 0 0 )
10
 A π π
Bq1 =C + (1 0 0) S
2 ∗ 10 2 ∗ 10
A
Bq1 = (0.9877 + 0.1564i + 0j + 0k)
A ∗
Bq1 = (0.9877 − 0.1564i − 0j − 0k)
A π
BR 2 => θ = rad; S = (0 1 0)
10

A π π
Bq2 =C + (0 1 0) S
2 ∗ 10 2 ∗ 10
A
Bq2 = (0.9877 + 0i + 0.1564j + 0k)
A ∗
Bq2 = (0.9877 − 0i − 0.1564j − 0k)
B
Ct = (0 + 1i + 10j − 10k)

P C = (0 + 0i + 0j + 2k)
∗ ∗
PA = A A c B A A A
Bq1 ( Bq2 (P + Ct) Bq2 ) Bq1 + Bt

P A = (0 − 11.52i + 13.96j + 5.56k)

Quaterniones duales
A
Bt = (0 − 10i + 2j + 10k)
A π
BR1 => θ = rad; S = (1 0 0 )
10

A π π
Bq1 =C + (1 0 0) S
2 ∗ 10 2 ∗ 10
A
Bq1 = (0.9877 + 0.1564i + 0j + 0k)
A A
A Bt ∗ Bq1
=A BH1
Bq1 + ε
2
A
H
B 1 = [0.988 + 0.156i + 0.0j + 0.0k + ε0.782 − ε4.938i + ε1.769j + ε4.782k]
̅̅̅̅̅̅
A ∗
BH1 = [0.988 − 0.156i − 0.0j − 0.0k − ε0.782 − ε4.938i + ε1.769j + ε4.782k]

A π
BR 2 => θ = rad; S = (0 1 0)
10

A π π
Bq2 =C + (0 1 0) S
2 ∗ 10 2 ∗ 10
A
Bq2 = (0.9877 + 0i + 0.1564j + 0k)
 A A
B Bt ∗ Bq1
CH2 =Aq
B 1 + ε
2
B
CH2 = [0.988 + 0.0i + 0.156j + 0.0k + ε0.0 + ε0.0i + ε0.0j + ε0.0k]
̅̅̅̅̅̅
B ∗
CH2 = [0.988 − 0.0i − 0.156j − 0.0k − ε0.0 + ε0.0i + ε0.0j + ε0.0k]
B
Ct = (0 + 0i + 0j + 0k + ε0 + ε1i + ε10j − ε10k)

P C = (1 + 0i + 0j + 0k + ε0 + ε0i + ε0j + ε2k)

B ̅̅̅̅̅̅
B ∗ ̅̅̅̅̅̅
A ∗
PA = A B c
Bq1 ( CH2 (P + Ct) CH2 ) BH1

P A = (1 + 0i + 0j + 0k + ε0 − ε11.521i + ε13.957j + ε5.560k)

 Tc = (0.1, 1, 1.6)m
R zo = 180°
R xo = −90°
pp = (2, −1.6, −0.1)m
Matrices
M0 -> Tc = (0.1, 1, 1.6)m y R z0 = 180°
C(180°) −S(180°) 0 0.1 −1 0 0 0.1
0 −1 0 1]
M0 = [ S(180°) C(180°) 0 1 ] = [
0 0 1 1.6 0 0 1 1.6
0 0 0 1 0 0 0 1
M1 -> T = (0, 0, 0)m y R x0 = −90°
1 0 0 0 1 0 0 0
0 C(−90°) −S(−90) 0
M1 = [ ] = [0 0 1 0]
0 S(−90) C(−90°) 0 0 −1 0 0
0 0 0 1 0 0 0 1
oM = M0 ∗ M1
c

−1 0 0 0.1 1 0 0 0 −1 0 0 0.1
oM = [ 0 −1 0 1 ]∗ [ 0 0 1 0]=[ 0 0 −1 1]
c
0 0 1 1.6 0 −1 0 0 0 −1 0 1.6
0 0 0 1 0 0 0 1 0 0 0 1
R = ocM ∗ pp
−1 0 0 0.1 2
0 0 −1 1 −1.6
R= [ ]∗ [ ]
0 −1 0 1.6 −0.1
0 0 0 1 1
−1.9
1.1 ]
R=[
3.2
1
Quaterniones
q0 = 0 + 0.1i + 1j + 1.6k
180 180
q1 = C ( ) + ⃗S ∗ S ( ) → ⃗S = 0i + 0j + 1k
2 2
q1 = 0 + 0i + 0j + 1k
−90 −90
q2 = C ( ) + ⃗S ∗ S ( ) → ⃗S = 1i + 0j + 0k
2 2
√2 √2
q2 = − i + 0j + 0k
2 2
pp = 0 + 2i − 1.6j − 0.1k
R = (q1 ∗ (q2 ∗ pp ∗ q′2 ) ∗ q1′ ) + q0
 R = ((0 + 0i + 0j + 1k)

√2 √2 √2 √2
∗ (( − i + 0j + 0k) ∗ (0 + 2i − 1.6j − 0.1k) ∗ ( + i − 0j − 0k))
2 2 2 2

∗ (0 − 0i − 0j − 1k)) + (0 + 0.1i + 1j + 1.6k)

√2 √2
0 0
2 2
√2 √2 0 1.4142
− 0 0
2 2 2 ] = [ 1.4142 ]
q2 ∗ pp = ∗[
√2 √2 −1.6 − 1.2021
0 0 − −0.1 1.0607
2 2
√2 √2
[ 0 0
2 2 ]

√2
1.4142 −1.4142 1.202 −1.0607 2 0
1.4142 1.4142 −1.0607 − 1.202 ] √2 = [ 2 ]
q2 ∗ pp ∗ q′2 = [ ∗
− 1.202 1.0607 1.4142 −1.4142 − 0.1
2
1.0607 1.202 1.4142 1.4142 0 1.6
[0]
0 0 0 −1 0 −1.6
(q2 ∗ pp ∗ q′2 ) 0 0 −1 0 ] [ 2 ] 0.1 ]
q1 ∗ =[ ∗ = [
0 1 0 0 − 0.1 2
1 0 0 0 1.6 0
−1.6 −0.1 −2 0 0 0
(q1 ∗ (q2 ∗ pp ∗ q′2 ) ∗ q1′ ) = [ 0.1 −1.6 0 2 ]∗[ 0 ] = −2
[ ]
2 0 −1.6 −0.1 0 0.1
0 −2 0.1 −1.6 −1 1.6
(q1 ∗ (q2 ∗ pp ∗ q′2 ) ∗ q′1 ) + q0 = (0 − 2i + 0.1j + 1.6k) + (0 + 0.1i + 1j + 1.6k)

R = 0 − 1.9i + 1.1j + 3.2k

Quaterniones duales
H0 -> To = 0 + 0.1i + 1j + 1.6k y R z0 = 180°
180 180
q0 = C ( ) + ⃗S ∗ S ( ) → ⃗S = 0i + 0j + 1k
2 2
q0 = 0 + 0i + 0j + 1k
To ∗ q0
H0 = q0 + ε
2
(0 + 0.1i + 1j + 1.6k) ∗ (0 + 0i + 0j + 1k)
H0 = (0 + 0i + 0j + 1k) + ε
2
 0 −0.1 −1 −1.6 0 −1.6
0.1 0 −1.6 1 ] ∗[ ] = [ 1 ]
0
To ∗ q0 = [
1 1.6 0 −0.1 0 −0.1
1.6 −1 0.1 0 1 0
−0.8
To ∗ q0 0.5 ]
=[
2 −0.05
0

H0 = (0 + 0i + 0j + 1k) + ε(−0.8 + 0.5i − 0.05j + 0k)

H1 -> T1 = 0 + 0i + 0j + 0k y R x0 = −90°
−90 −90
q1 = C ( ) + ⃗S ∗ S ( ) → ⃗S = 1i + 0j + 0k
2 2
√2 √2
q2 = − i + 0j + 0k
2 2
T1 ∗ q1
H1 = q1 + ε
2
√2 √2
√2 √ 2 (0 + 0i + 0j + 0k) ∗ ( − i + 0j + 0k)
H1 = ( − i + 0j + 0k) + ε 2 2
2 2 2
√2 √2
H1 = ( − i + 0j + 0k) + ε(0 + 0i + 0j + 0k)
2 2
∈= (1 + 0i + 0j + 0k) + ε(0 + 2i − 1.6j − 0.1k)

R = H0 ∗ H1 ∗ ϵ ∗ ̅̅̅̅̅
H1 ∗ ∗ ̅̅̅̅̅
H0 ∗
̅̅̅̅̅
H0 ∗ = (0 + 0i + 0j + 1k) − ε(−0.8 − 0.5i + 0.05j − 0k)

̅̅̅̅̅ √2 √2
H1 ∗ = ( − i + 0j + 0k) − ε(0 − 0i − 0j − 0k)
2 2

R = ((0 + 0i + 0j + 1k) + ε(−0.8 + 0.5i − 0.05j + 0k))

√2 √2
∗ (( − i + 0j + 0k) + ε(0 + 0i + 0j + 0k)) ∗ ((1 + 0i + 0j + 0k) + ε(0 + 2i
2 2

√2 √2
− 1.6j − 0.1k)) ∗ (( − i + 0j + 0k) − ε(0 − 0i − 0j − 0k))
2 2
∗ ((0 + 0i + 0j + 1k) − ε(−0.8 − 0.5i + 0.05j − 0k))

√2 √2
H0 ∗ H1 = (0 + 0i + 0j + 1k) ∗ ( − i + 0j + 0k) + ε((0 + 0i + 0j + 1k) ∗ (0 + 0i + 0j + 0k)
2 2
√2 √2
+ (−0.8 + 0.5i − 0.05j + 0k) ∗ ( − i + 0j + 0k))
2 2
H0 ∗ H1 = (0 + 0i − 0.71j + 0.71k) + ε(−0.21 + 0.92i − 0.04j − 0.4k)
 H0 ∗ H1 ∗ ϵ = (0 + 0i − 0.71j + 0.71k) ∗ (1 + 0i + 0j + 0k) + ε((0 + 0i − 0.71j + 0.71k) ∗ (0 + 2i
− 1.6j − 0.1k) + (−0.21 + 0.92i − 0.04j − 0.4k) ∗ (1 + 0i + 0j + 0k))
H0 ∗ H1 ∗ ϵ = (0 + 0i − 0.71j + 0.71k) + ε(−1.27 + 2.13i + 1.38j + 1.38k)

H0 ∗ H1 ∗ ϵ ∗ ̅̅̅̅̅
H1 ∗
√2 √2
= (0 + 0i − 0.71j + 0.71k) ∗ ( − i + 0j + 0k) + ε((0 + 0i − 0.71j + 0.71k)
2 2
√2 √2
∗ (0 + 0i + 0j + 0k) + (−1.27 + 2.13i + 1.38j + 1.38k) ∗ ( − i + 0j + 0k))
2 2

H0 ∗ H1 ∗ ϵ ∗ ̅̅̅̅̅
H1 ∗ = (0 + 0i + 0j + 1k) + ε(−2.4 + 0.61i + 1.95j + 0k)

H0 ∗ H1 ∗ ϵ ∗ ̅̅̅̅̅
H1 ∗ ∗ ̅̅̅̅̅
H0 ∗
= (0 + 0i + 0j + 1k) ∗ (0 + 0i + 0j + 1k) + ε((0 + 0i + 0j + 1k) ∗ (+0.8 + 0.5i
− 0.05j + 0k) + (−2.4 + 0.61i + 1.95j + 0k) ∗ (0 + 0i + 0j + 1k))

H0 ∗ H1 ∗ ϵ ∗ ̅̅̅̅̅
H1 ∗ ∗ ̅̅̅̅̅
H0 ∗ = (1 + 0i + 0j + 0k) + ε(0 − 1.9i + 1.1j + 3.2k)

R = (1 + 0i + 0j + 0k) + ε(0 − 1.9i + 1.1j + 3.2k)