Transcript Basic mathematics for geometric modeling

Basic mathematics for geometric
modeling
Coordinate Reference Frames
• Cartesian Coordinate (2D)
y
(x, y)
x
• Polar coordinate
r

Relationship : polar & cartesian
Y
P
P
y
r

x
x
Use trigonometric, polar  cartesian
x = r cos  , y = r sin 
Cartesian  polar
r =  x2 + y2,  = tan-1 (y/x)
r

y
x
3D cartesian coordinates
y
y
x
z
x
z
Right-handed 3D coordinate system
POINT
•
•
•
•
The simplest of geometric object.
No length, width or thickness.
Location in space
Defined by a set of numbers (coordinates)
e.g P = (x, y) or P = (x, y, z)
• Vertex of 2D/ 3D figure
VECTOR
• distance and direction
• Does not have a fixed location in space
• Sometime called “displacement”.
VECTOR
• Can define a vector as the difference
between two point positions.
y
y2
V
y1
Q
P
x1
V=Q–P
= (x2 – x1, y2 – y1)
= (Vx, Vy)
Also can be expressed as
V = Vxi + Vyj
x2
x
Component form
VECTOR : magnitude &
direction
• Calculate magnitude using the Pythagoras
theorem  distance
– |V| =  Vx2 + Vy2
• Direction
–  = tan-1 (Vy/Vx)
VECTOR : magnitude &
direction
V
Q
• Example 1
• If P(3, 6) and Q(6, 10). Write vector V in
component form.
• Answer
• V = [6 - 3, 10 – 6]
= [3, 4]
VECTOR : magnitude &
direction
• Example 1 (cont)
• Compute the magnitude and direction of
vector V
• Answer
• Magnitud |V| =  32 + 42
•
= 25 = 5
• Direction  = tan-1 (4/3) = 53.13
Unit Vector
• As any vector whose magnitude is equal to
one
• V= V
|V|
• The unit vector of V in example 1 is
= [Vx/|V| , Vy/|V|]
= [3/5, 4/5]
VECTOR : 3D
y
• Vector Component
– (Vx, Vy, Vz)
• Magnitude
Vy
V
x
Vz
– |V| =  Vx2 + Vy2 + Vz2
• Direction
Vx
z
–  = cos-1(Vx/|V|),  = cos-1(Vy/|V|),  =cos-1(Vz/|V|)
• Unit vector
• V = V = [Vx/|V|, Vy/|V|, Vz/|V|]
|V|
Scalar Multiplication
• kV = [kVx, kVy, kVz]
• If k = +ve  V and kV are in the same direction
• If k = -ve  V and kV are in the opposite
direction
• Magnitude |kV| = k|V|
Scalar Multiplication
• Base on Example 1
• If k = 2, find kV and the magnitudes
• Answer
• kV = 2[3, 4] = [6, 8]
• Magnitude |kV|=  62 + 82 = 100 = 10
•
= k|V| = 2(5) = 10
Vector Addition
y
y
V
U+V
U
x
U
V
x
• Sum of two vectors is obtained by adding
corresponding components
• U = [Ux, Uy, Uz], V = [Vx, Vy, Vz]
• U + V = [Ux + Vx, Uy + Vy, Uz + Vz]
Vector Addition
Q
P
P
Q
• Example
• If vector P=[1, 5, 0], vector Q=[4, 2, 0]. Compute
P+Q
• answer
• P + Q = [1+4, 5+2, 0+0] = [5, 7, 0]
Vector Addition & scalar
multiplication properties
•
•
•
•
•
U+V=V+U
T + (U + V) = (T + U) + V
k(lV) = klV
(k + l)V = kV + lV
k(U + V) = kU + kV
Scalar Product
• Also referred as dot product or inner product
• Produce a number.
• Multiply corresponding components of the two
vectors and add the result.
• If vector U = [Ux, Uy, Uz], vector V = [Vx, Vy,
Vz]
• U . V = UxVx + UyVy + UzVz
Scalar Product.
• Example
• If vector P=[1, 5, 0], vector Q=[4, 2, 0].
Compute P . Q
• answer
• P . Q = 1(4) + 5(2) + 0(0)
•
= 14
Scalar Product properties
• U.V = |U||V|cos 
• angle between two vectors
–  = cos
–
–1
(U.V)
|U||V|
• Example
• Find the angle between vector
b=(3, 2) and vector c = (-2, 3)
U

V
Scalar Product properties
Solution
• b.c = (3, 2). (-2, 3)
•
3(-2) + 2(3) = 0
• |b| = 32 + 22 = 13 = 3.61
• |c| = (-2)2 + 32 = 13 = 3.61
•  = cos –1 ( 0/(3.61((3.61))
•  = cos –1 ( 0 ) = 90
Scalar Product properties
•
•
•
•
•
If U is perpendicular to V, U.V = 0
U.U = |U|2
U.V = V.U
U.(V+W) = U.V + U.W
(kU).V = U.(kV)
Vector Product
• Also called the cross product
• Defined only for 3 D vectors
• Produce a vector which is perpendicular to
both of the given vectors.
y
c=axb
c
b
z
a
x
Vector Product
• To find the direction of vector C, use righhand rules
C
B
B
z
A
x
z
C
A
x
Vector Product
• To find the direction of vector C, use righhand rules
BxA
AxB
C
x
x
z
z
A
B
A
B
C
exercise
• Find the direction of vector C, (keluar skrin atau
kedalam skrin)
AxB
B
Q
P
PxQ
A
O
M
MxN
N
LxO
L
Vector Product
• If vektor A = [Ax, Ay, Az], vektor B = [Bx, By, Bz]
• AxB= i
j k
i j
•
Ax Ay Az Ax Ay
•
Bx By Bz Bx By
= [ (AyBz-AzBy), (AzBx-AxBz), (AxBy-AyBx)]
Vector Product
P
•
•
•
•
•
•
•
•
Example
Q
If P=[1, 5, 0], Q=[4, 2, 0]. Compute P x Q
Solution
PxQ= i
j k
i j
1 5 0
1 5
4
2 0
4 2
= [ (5.(0)-0.(5)), (0.(4)-1.(0)), (1.(2)-5.(4))]
= [ 0, 0, -18]
Vector Product
• Properties
• U x V = |U||V|n sin  where n = unit vector
perpendicular to both U and V
• U x V = -V x U
• U x (V + W) = U x V+ U x W
• If U is parallel to V, U x V = 0
• UxU=0
• kU x V = U x kV