Basic mathematics for geometric modeling. Coordinate Reference Frames Cartesian Coordinate (2D)...
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Transcript of Basic mathematics for geometric modeling. Coordinate Reference Frames Cartesian Coordinate (2D)...
Relationship : polar & cartesian
x
PY
x
y
r
r
x
y
P
Use trigonometric, polar cartesianx = r cos , y = r sin
Cartesian polar
r = x2 + y2, = tan-1 (y/x)
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
• distance and direction
• Does not have a fixed location in space
• Sometime called “displacement”.
VECTOR
VECTOR
• Can define a vector as the difference between two point positions.
x
y
P
Q
x1 x2
y1
y2 V
V = Q – P = (x2 – x1, y2 – y1) = (Vx, Vy)Also can be expressed as V = Vxi + Vyj
Component form
VECTOR : magnitude & direction
• Calculate magnitude using the Pythagoras theorem distance– |V| = Vx2 + Vy2
• Direction– = tan-1 (Vy/Vx)
• 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
QV
• 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
VECTOR : magnitude & direction
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
• Vector Component– (Vx, Vy, Vz)
• Magnitude– |V| = Vx2 + Vy2 + Vz2
• Direction = cos-1(Vx/|V|), = cos-1(Vy/|V|), =cos-1(Vz/|V|)
• Unit vector• V = V = [Vx/|V|, Vy/|V|, Vz/|V|]
|V|
x
y
z
V
VxVz
Vy
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
• 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]
x
yV
U x
yV
U
U + V
Vector Addition
• 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]
P
Q
P
Q
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
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)
Scalar Product properties
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.
x
y
z a
b
cc = a x b
Vector Product• To find the direction of vector C, use righ-
hand rules
x
z A B
C
A x B
x
zC
A B
B x A
exercise• Find the direction of vector C, (keluar skrin atau
kedalam skrin)
A
BA x B
PQ
P x Q
MN
M x N
L
O
L x O
• If vektor A = [Ax, Ay, Az], vektor B = [Bx, By, Bz]
• A x B = i j k i j
• Ax Ay Az Ax Ay
• Bx By Bz Bx By
= [ (AyBz-AzBy), (AzBx-AxBz), (AxBy-AyBx)]
Vector Product
Vector Product
• Example
• If P=[1, 5, 0], Q=[4, 2, 0]. Compute P x Q
• Solution
• P x Q = 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]
P
Q