Basic mathematics for geometric modeling

Coordinate Reference Frames• Cartesian Coordinate (2D)

• Polar coordinate

x

y

(x, y)

r

Relationship : polar & cartesian

x

PY

x

yr

r

x

y

P

Use trigonometric, polar cartesianx = r cos , y = r sin

Cartesian polar

r = x2 + y2, = tan-1 (y/x)

3D cartesian coordinates

x

y

z

Right-handed 3D coordinate system

z

x

y

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

Cx

z A

B

C

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

B A x BP

QP x Q

MN

M x NL

OL 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

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 • U x U = 0• kU x V = U x kV