SLIDE 1
Points, Scalars and Vectors
Points, vectors defined relative to a coordinate system Point: Location in coordinate system Example: Point (5,4) Cannot add or scale points
y x
(5,4) (0,0) 5 4
Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for - - PowerPoint PPT Presentation
Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for - - PowerPoint PPT Presentation
Computer Graphics (CS 543) Lecture 3 (Part 1): Linear Algebra for Graphics (Points, Scalars, Vectors) Prof Emmanuel Agu Computer Science Dept. Worcester Polytechnic Institute (WPI) Points, Scalars and Vectors Points, vectors defined
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SLIDE 3
Vectors
Magnitude Direction NO position Can be added, scaled, rotated CG vectors: 2, 3 or 4 dimensions
Length
Angle
SLIDE 4
Vector-Point Relationship
Subtract 2 points = vector
v = Q – P
point + vector = point
P + v = Q
P Q v
SLIDE 5
Vector Operations
Define vectors
) , (
3 2 , 1
a a a a ) , (
3 2 , 1
b b b b ) , (
3 3 2 2 , 1 1
b a b a b a b a
Then vector addition:
a a+b b
SLIDE 6
Vector Operations
Define scalar, s Scaling vector by a scalar
) , , (
3 2 1
s a s a s a s a )) ( ), ( ), ( (
3 3 2 2 1 1
b a b a b a b a
Note vector subtraction:
a b a-b a 2.5a
SLIDE 7
Vector Operations: Examples
Scaling vector by a scalar
For example, if a=(2,5,6) and b=(-2,7,1)
and s=6, then
) , , (
3 2 1
s a s a s a s a
- Vector addition:
) , (
3 3 2 2 , 1 1
b a b a b a b a ) 7 , 12 , ( ) , (
3 3 2 2 , 1 1
b a b a b a b a ) 36 , 30 , 12 ( ) , , (
3 2 1
s a s a s a s a
SLIDE 8
Affine Combination
Given a vector Affine combination: Sum of all components = 1 Convex affine = affine + no negative component
i.e
1 .........
2 1
n
a a a
negative non a a a
n
,......... ,
2 1
) ,..., , (
3 2 , 1 n
a a a a a
SLIDE 9
Magnitude of a Vector
Magnitude of a Normalizing a vector (unit vector) Note magnitude of normalized vector = 1. i.e
2 2 2 2 1
.......... | |
n
a a a a magnitude vector a a a ˆ
1 ..........
2 2 2 2 1
n
a a a
SLIDE 10
Magnitude of a Vector
Example: if a = (2, 5, 6) Magnitude of a Normalizing a
65 6 5 2 | |
2 2 2
a
65 6 , 65 5 , 65 2 ˆ a
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11
Convex Hull
Smallest convex object containing P1,P2,…..Pn Formed by “shrink wrapping” points
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Dot Product (Scalar product)
Dot product, For example, if a=(2,3,1) and b=(0,4,-1)
then
3 3 2 2 1 1
........ b a b a b a d b a ) 1 1 ( ) 4 3 ( ) 2 ( b a 11 1 12
SLIDE 13
Properties of Dot Products
Symmetry (or commutative): Linearity: Homogeneity: And
a b b a ) ( ) ( b a b a s s b c b a b c a ) (
b b b
2
SLIDE 14
Angle Between Two Vectors
c x b y
b
c
b b
sin , cos b b b
c c
sin , cos c c c
cos c b c b
Sign of b.c: b.c > 0
c b b
b.c = 0
b c c
b.c < 0
SLIDE 15
Angle Between Two Vectors
Problem: Find angle b/w vectors b = (3,4) and c = (5,2) Step 1: Find magnitudes of vectors b and c Step 2: Normalize vectors b and c
5 4 , 5 3 ˆ b
29 2 , 29 5 c ˆ
5 25 4 3 | |
2 2
b 29 2 5 | |
2 2
c
SLIDE 16
Angle Between Two Vectors
Step 3: Find angle as dot product Step 4: Find angle as inverse cosine
29 2 , 29 5 5 4 , 5 3 ˆ ˆ c b
326 . 31 ) 85422 . cos(
c b ˆ ˆ
85422 . 29 5 23 29 5 8 29 5 15 ˆ ˆ c b
SLIDE 17
Standard Unit Vectors
, , 1 i
Define
, 1 , j
1 , , k
So that any vector,
k j i v c b a c b a , ,
y z x i j k
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Cross Product (Vector product)
z y x
a a a , , a
z y x
b b b , , b
If Then
k j i b a ) ( ) ( ) (
x y y x x z z x y z z y
b a b a b a b a b a b a
Remember using determinant
z y x z y x
b b b a a a k j i
Note: a x b is perpendicular to a and b
SLIDE 19
Cross Product
Note: a x b is perpendicular to both a and b a x b a b
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Cross Product (Vector product)
8 , 1 , 4 b
Then
k j i b a ) 3 ( ) 8 24 ( ) 2 (
Using determinant
Calculate a x b if a = (3,0,2) and b = (4,1,8)
2 , , 3 a
8 1 4 2 3 k j i
k j i 3 16 2
SLIDE 21
Normal for Triangle using Cross Product Method
p0 p1 p2 n plane n·(p - p0 ) = 0 n = (p2 - p0 ) ×(p1 - p0 ) normalize n n/ |n| p Note that right-hand rule determines outward face
SLIDE 22
Newell Method for Normal Vectors
Problems with cross product method:
calculation difficult by hand, tedious If 2 vectors almost parallel, cross product is small Numerical inaccuracy may result
Proposed by Martin Newell at Utah (teapot guy)
Uses formulae, suitable for computer Compute during mesh generation Robust!
p0 p2 p1
SLIDE 23
Newell Method Example
Example: Find normal of polygon with vertices
P0 = (6,1,4), P1=(7,0,9) and P2 = (1,1,2)
Using simple cross product:
((7,0,9)-(6,1,4)) X ((1,1,2)-(6,1,4)) = (2,-23,-5)
PO P2 P1 (6,1,4) (7,0,9) (1,1,2) P1 - P0 P2 - P0
SLIDE 24
Newell Method for Normal Vectors
Formulae: Normal N = (mx, my, mz)
) ( 1 ) ( i next i N i i next i x
z z y y m
) ( 1 ) ( i next i N i i next i y
x x z z m
) ( 1 ) ( i next i N i i next i z
y y x x m
SLIDE 25
Newell Method for Normal Vectors
Calculate x component of normal
) ( 1 ) ( i next i N i i next i x
z z y y m
4 1 6 2 1 1 9 7 4 1 6 z y x
P0 P1 P2 P0
2 11 13 ) 6 )( ( ) 11 )( 1 ( ) 13 )( 1 (
x x x
m m m
SLIDE 26
Newell Method for Normal Vectors
Calculate y component of normal
4 1 6 2 1 1 9 7 4 1 6 z y x
P0 P1 P2 P0
23 14 56 65 ) 7 )( 2 ( ) 8 )( 7 ( ) 13 )( 5 (
y y y
m m m
) ( 1 ) ( i next i N i i next i y
x x z z m
SLIDE 27
Newell Method for Normal Vectors
Calculate z component of normal
4 1 6 2 1 1 9 7 4 1 6 z y x
P0 P1 P2 P0
5 10 6 1 ) 2 )( 5 ( ) 1 )( 6 ( ) 1 )( 1 (
z z z
m m m
) ( 1 ) ( i next i N i i next i z
y y x x m
Note: Using Newell method yields same result as Cross product method (2,-23,-5)
SLIDE 28
Finding Vector Reflected From a Surface
a = original vector
n = normal vector
r = reflected vector
m = projection of a along n
e = projection of a orthogonal to n
m e
- m
r n a e
m a r m e r m a e 2
Θ1 Θ2 Note: Θ1 = Θ2
SLIDE 29
Forms of Equation of a Line
Two-dimensional forms of a line
Explicit: y = mx +h Implicit: ax + by +c =0 Parametric:
x(a) = ax0 + (1-a)x1 y(a) = ay0 + (1-a)y1
Parametric form of line
More robust and general than other forms Extends to curves and surfaces Po P1 α Pα 1 - α
SLIDE 30
Convexity
An object is convex iff for any two points in the
- bject all points on the line segment between these
points are also in the object
P Q Q P convex not convex
SLIDE 31