2IV60 Computer Graphics
Basic Math for CG
Jack van Wijk
TU/e
Overview
• Coordinates, points, vectors
• Matrices
H&B A-1
nD Space
nD Space: n (typically)
n : number of dimensions
Examples:
• 1D space: time, along a line or curve
• 2D space: plane, sphere
• 3D space: the world we live in
• 4D space: 3D + time
H&B A-2
Coordinates
2D Cartesian coordinates:
x
y
(x,y)
(x,y)
y
x
Standard
Screen (output, input)
H&B A-1
Polar coordinates
Coordinate transform ation
y
from ( r , ) to ( x, y ) space :
x  r cos 
(x,y)
y  r sin
Angle : in radians
r

x
H&B A-1
3D coordinates 1
3D Cartesian coordinates:
z
x
(x,y,z)
y
Right-handed
z
(x,y,z)
x
y
Left-handed
H&B A-1
3D coordinates 2
3D Cartesian coordinates:
z (middle)
x (thumb)
(x,y,z)
y (index)
Right-handed
z (middle)
(x,y,z)
y (index)
x (thumb)
Left-handed
H&B A-1
3D coordinates 3
Cylinder coordinates:
(x,y,z)
z
x

y
x  r cos
y  r sin 
zz
H&B A-1
3D coordinates 4
Spherical coordinates:
(x,y,z)
z

r
x

y
x  r cos  sin
y  r sin sin
z  r cos 
 : azimuth or longitude
 : elevation or latitude
Also :    / 2  
 : colatitude
H&B A-1
Points
• Point: position in nD space
• Notation: P (H&B), also P, p, p en p
• (x,y,z) (H&B), also (x1, x2, x3), (Px, Py, Pz),
(r, , z), ( r, , ), …
H&B A-2
Vectors 1
• Vector:
– “arrow”
– multiple interpretations (displacement, velocity,
force, …)
– has a magnitude and direction
– has no position
• Notation: V (H&B), also V, v, v en v
• (Vx, Vy, Vz) (H&B), also (x, y, z), (x1, x2, x3)
H&B A-2
Vectors 2
y
V  P2  P1
 ( x2  x1 , y2  y1 )
 (Vx ,Vy )
P2
y2
P1
y1
V: directed line segment, or
difference between two points
x1
x2
x
H&B A-2
Vectors 3
Length of a vector:
V  Vx2  V y2 (2D : P ythagoras
)
V  V V V
2
x
2
y
2
z
(3D)
H&B A-2
Vectors 4
Direction of a vector: Direction angles.
Vy
Vx
Vz
cos  , cos  
, cos 
V
V
V
z

Unit vector V :
V
V
|V|

x
y

Magnitude info is removed, direction is kept.
H&B A-2
Vector addition
Add components, put vector head to tail
y
y
W
V+W
W
V
V
x
x
V  W  (Vx  Wx ,Vy  Wy ,Vz  Wz )
H&B A-2
Scalar multiplication of vector
Multiplication components with scalar s
y
y
2V
V
x
x
sV  (sVx ,sVy , sVz )
H&B A-2
Combining vector operations
Infinite line through P with direction V:
L(t) = P + Vt, t 
y
V
P
t
x
t : parameter along line
t [a, b]: line segment
H&B A-2
Vector multiplication 1
Scalar product or dot product:
Product of parallel components, gives 1 real value
W
V
V  W  V W cos
 VxWx  V yWy  VzWz

|W| cos 
|V|
H&B A-2
Vector multiplication 2
Scalar product:
V  W  V W cos
 VxWx  V yW y  VzWz
Commutative : V  W  W  V
Distributive : V  (W  Z)  V  W  V  Z
Associative : V  (W  Z)  (V  W)  Z ??
H&B A-2
Vector multiplication 3
Vector product or cross product: gives a vector (in 3D)
n perpendicular to V and W
 (V yWz  VzW y ,
VW
n
V  W  n V W sin
W
VzWx  VxWz ,

VxWy  V yWx )
V
H&B A-2
Vector multiplication 5
Scalar product:
Anticommutative : V  W   W  V
Not associative : V  (W  Z)  (V  W)  Z
Distributive :
V  (W  Z)  V  W  V  Z
H&B A-2
Vector multiplication 6
Scalar product:
• scalar
• Test if vectors are
perpendicular
• cos
• project,…
Vector product:
• vector
• Get a vector
perpendicular to two
given vectors
• sin
• surface area,…
H&B A-2
Exercise 1
• Given a point P.
• Requested: Reflect a point Q with respect to P.
Q
W
y
W
x
P
Q’
W=P–Q
Q’ = Q + 2W
= 2P – Q
or: = P + (P – Q )
We don’t need coordinates!
Exercise 1
• Given a point P.
• Requested: Reflect a point Q with respect to P.
Q
P
Q’
Alternative
P is halfway Q and Q’:
P = (Q + Q’)/2
2P = Q + Q’
Q’ = 2P – Q
Exercise 2
• Given a line L: L(t) = P + Vt .
• Requested: Reflect a point Q with respect to L.
Q
W
y
V
P
t
x
W
L(t)
Q’
We know:
Q’ = Q + 2 W
W = L(t) – Q
W. V = 0
Exercise 2
We know:
L(t) = P + Vt
Q’ = Q + 2 W
W = L(t) – Q
W. V = 0
Substitute to get t:
(L(t) – Q).V = 0
(P + Vt – Q).V = 0
V .V t + (P – Q).V = 0
t = ((Q – P).V) / (V .V)
Then:
Q’ = Q + 2 (P + Vt – Q)
= 2P – Q + V((Q – P).V / V .V)
Steps to be made
• Write down what you know
• Eliminate, substitute, etc. to get he result
• Check the result
– Does it make sense?
– Is there a simpler derivation?
Exercise 3
• Given a triangle with vertices P, Q en R,
where the angle PQR is perpendicular.
• Requested: Rotate the triangle around the
line PQ over an angle  . What is the new
P
position R’ of R?
Q

R’
R
Circle in space
y
C()
C()
r

P A

A
B
B
x
P
|A|=1, |B|=1, A.B = 0
C() = (r cos , r sin , 0)
= (0,0,0) + r cos  (1,0,0) + r sin  (0,1,0)
= P + r cos  A + r sin  B
Exercise 3
A = (R – Q) / |R – Q|
B = (R – Q)(P – Q) / | (R – Q)(P – Q) |
R’ = Q + |R – Q| cos  A + |R – Q| sin  B
P
B
Q

R’
A
R
Use:
• Scalar product, cross product
• coordinate independent definitions
• vector algebra
Don’t use:
• arccos, arcsin
• y = f(x)
Very short intro to Linear Algebra
System of linear equations:
u  2x  3y  4z
v  x  5 y  3z
w  5x  y  z
Such systems occur in many, many applications.
They are studied in Linear Algebra.
H&B A-5
Very short intro to Linear Algebra
System of linear equations:
u  2x  3y  4z
v  x  5 y  3z
w  5x  y  z
Typical questions:
- Given u, v, w, what are x, y, z?
- Can we find a unique solution?
H&B A-5
Very short intro to Linear Algebra
System of linear equations:
u  2x  3y  4z
v  x  5 y  3z
w  5x  y  z
Crucial in computer graphics:
- Transforming geometric objects
- Change of coordinates
H&B A-5
Example transformation
T ransformation P  P':
P : ( x, y )
P'  xU  yV, with
U  (3,2)
V  (2,3)
In coordinates :
x'  3x  2 y
y '  2 x  3 y
y
P: (x,y)
x
y’
P’
V
y
U
x
x’
H&B A-5
What is a matrix?
Matrix:
- Mathematical objects with operations
Matrix in computer graphics:
- Defines a coordinate frame
- Defines a transformation
- Handy tool for manipulating transformations
H&B A-5
Matrix
Matrix: rectangular array of elements
Element: quantity (value, expression, function, …)
Examples:
 3.60  0.01 2.1  e
 5.46 0.00 1.6 ,  2 x

 e
x
x
,
2
x 
a1
a2
 x
a3  ,  y 
 z 
H&B A-5
Matrix
r  c matrix: r rows, c columns
mij : element at row i and column j.
 m11 m12  m1c 
m

m

m
22
2c 
M   21
 

 


 mr1 mr 2  mrc 
H&B A-5
Matrix
r  c matrix: r rows, c columns
mij : element at row i and column j.
 m11 m12  m1c 
m

m

m
22
2c 
M   21
 

 


 mr1 mr 2  mrc  r  c elements
H&B A-5
Matrix
r  c matrix: r rows, c columns
mij : element at row i and column j.
 m11 m12  m1c 
m

m

m
22
2c 
M   21
 

 


 mr1 mr 2  mrc 
c columns
H&B A-5
Matrix
r  c matrix: r rows, c columns
mij : element at row i and column j.
 m11 m12  m1c 
m

m

m
21
22
2c  r rows

M
 

 


 mr1 mr 2  mrc 
H&B A-5
Matrix
Column vector:
matrix with c =1
 v1 
v 
2

V

 
 vr 
Row vector:
matrix with r =1
V  v1 v2  vc 
Scalar:
matrix with r=1and c=1
V  v 
Used for vectors
(and points)
H&B A-5
Matrix
Matrix as collection of column vectors:
u x
M  U V W  u y
u z
vx
vy
vz
wx 
wy 
wz 
W
Matrix contains an axis-frame:
V
U
H&B A-5
Operations on matrices
• Multiplication with scalar
– simple
• Addition
– simple
• Matrix-matrix multiplication
– More difficult, but the most important
H&B A-5
Scalar Matrix multiplication
Matrix M multiplied with scalar s:
 sm11 sm12 
 sm
sm

21
22
A  sM  
 


 smr1 smr 2 
3x 12 
x
example : 
 3

 6 2 y
2
sm1c 
sm2c 
, i.e., aij  smij
 

smrc 
4
y 
H&B A-5
Scalar Matrix addition
 a11 a12
a
21 a22

M AB
 


 ar1 ar 2
 a11  b11
a  b
  21 21



 ar1  br1
 a1c   b11
 a2c  b21

   
 
 arc   br1
a12  b12 
a22  b22

ar 2  br 2
b12  b1c 
b22  b2c 

 

br 2  brc 
a1c  b1c 
 a2c  b2c 
, i.e., mij  aij  bij



 arc  brc 
3x 12  4  10 3x  4 2 
example : 





6
2
y
2
0
8
2
y

 
 

H&B A-5
Scalar Matrix addition
 a11 a12
a
21 a22

M AB
 


 ar1 ar 2
 a11  b11
a  b
  21 21



 ar1  br1
 a1c   b11 b12 
 a2c  b21 b22 

   

 
 arc   br1 br 2 
a12  b12  a1c  b1c 
a22  b22  a2c  b2c 




ar 2  br 2  arc  brc 
b1c 
b2c 
 

brc 
, i.e., mij  aij  bij
• Just add elements pairwise
• A and B must have the same number of rows and columns
• Generalization of vector and scalar addition
H&B A-5
Matrix Matrix multiplication
 a11
a
C  AB   21
 

amn
n
cij 
aik bkj

k 1
a12
a22

am 2
a1n 
 a2 n 
 

 amn 

i = 2 ; k
 b11 b12  b1q 
b

b

b
2q 
 21 22
 

 


b p1 b p 2  b pq 
j = 1 ; k
• cij: dot product of row vector ai and column vector bj
• #columns A must be the same as #rows B: n = p
• C: m  q matrix
H&B A-5
Matrix Matrix multiplication
Example:
3
 2
 2  1  3  3 2  2  3  4  11 16
 1  4 1 2   1 1  4  3  1 2  4  4   13  16

 3 4 
 



 0
 0  1  5  3 0  2  5  4  15
5
20
i=3
j=2
i=3, j=2
H&B A-5
Matrix Matrix multiplication
Example:
4
A  1 2 3 , B  5
6
AB  
BA  
H&B A-5
Matrix Matrix multiplication
Example:
 4
A  1 2 3 , B  5
6
 4
AB  1 2 3 5  1  4  2  5  3  6  32
6
BA  
H&B A-5
Matrix Matrix multiplication
Example:
 4
A  1 2 3 , B  5
AB  BA
6
 4
AB  1 2 3 5  1 4  2  5  3  6  32
6
 4
4  1 4  2 4  3 4 8 12
BA  5 1 2 3  5  1 5  2 5  3  5 10 15
6
6  1 6  2 6  3 6 12 18
H&B A-5
Matrix Matrix multiplication
• AB  BA
Matrix matrix multiplication is not commutative!
Order matters!
• A(B+C) = AB+AC
Matrix matrix multiplication is distributive
H&B A-5
Example transformation sequence
(coordinate version)
X '  X : x'  px  qy,
X ' '  X ': x' '  ax'by' ,
y '  rx  sy;
y ' '  cx ' dy'.
T hen X ' '  X :
x' '  a ( px  qy )  b( rx  sy),
y ' '  c( px  qy )  d (rx  sy);
or
x' '  ( ap  br ) x  ( aq  bs) y ,
y ' '  (cp  dr ) x  (cq  ds) y.
Unclear!
Error-prone!
H&B A-5
Example transformation sequence
(matrix vector version)
 x'  p
X ' X :    
 y '  r
 x' ' a
X ' '  X ':    
 y ' '  c
q  x 
;



s   y
b   x'
.



d   y '
T hen X ' '  X :
 x' ' a b   p q   x  ap  br aq  bs   x 
 y ' '   c d   r s   y    cp  dr cq  ds   y 
  

  
 
H&B A-5
Example transformation sequence
(compact matrix vector version)
X '  X : X'  AX ;
X ' '  X ': X''  BX'
T hen X ' '  X :
X''  BAX
Here some more detail :
 x
 x'
 x' '
X    , X' '    , X' '    ,
 y
 y '
 y ' '
 p q
a b 
A
;B 
.


r s
c d 
Matrix vector notation allows for compactness
and genericity!
H&B A-5
Matrix Transpose
M
 a11 a12  a1c 
a

a

a
2c 
  21 22
 

 


a
a

a
r2
rc 
 r1
Transpose matrix:
Interchange rows and columns.
r  c matrix M  c  r matrix MT
 a11 a21  ar1 
a

a

a
12
22
r
2
 , i.e., mT  m
MT  
ji
ij
 

 


a
a

a
2c
rc 
 1c
1 4 
1 2 3 
2 5 ,
examples : 




4
5
6


3 6
T
[x
y
 x
z ]T   y 
 z 
H&B A-5
Matrix Inverse
Simple algebra puzzle. Let ax = b. What is x?
ax  b
Multiplyleft and right with a 1
a 1ax  a 1b Use a 1a equals 1
x  a 1b Done
Linear algebra puzzle. Let MU = V. What is U?
MU  V
Multiplyleft and right with M 1
M 1MU  M 1V Use M 1M equals I
U  M 1V Done!
H&B A-5
Matrix Inverse
Inverse of a n  n (square) matrix M is denoted M 1 , with
MM 1  M 1M  I,
where I is the identity matrix :
1
0
I


0
0  0
1
0
1 if i  j
, i.e., I ij  
.



0 if i  j

0  1
I is a diagonal matrix, with all 1' s on the diagonal.
M 1 " undoes" the effect of M,
multiplying with I has no effect.
H&B A-5
Matrix Inverse examples
41  1/ 4
1
2  1
3 2

5 4   5 / 2 3 / 2




2  1 3  2  2  5 / 2  3 1  2  3 / 2 1 0
3 2 
5 4  5 / 2 3 / 2  5  2  4  5 / 2  5 1  4  3 / 2  0 1


 
 

H&B A-5
Matrix Inverse examples
a b 
c d 


a b 
c d 


1
 ?
1
1  d  b


a 
ad  bc  c
a b   d  b  ad  bc  ab  ba 
1 0

 (ad  bc)
 c d   c



a

ab

ba

cb

ad
0
1


 



1
2 1
 4 2 ?


Does not exist, cannot be inverted.
H&B A-5
Matrix Inverse
- Does not always exist
- In general: if determinant = | M | = 0, matrix
cannot be inverted
- Inverse for n = 1 and n = 2: easy
- Inverse for higher n: use library function
- Important special case: orthonormal matrix.
H&B A-5
Overview
• Coordinates, points, vectors
– definitions, operations, examples
• Matrices
– definitions, operations, examples