Vector

Scaler versus Vector

Scaler (向量 ): : described by magnitudeE.g. length, mass, time, speed, etc

Vector(矢量 ): described by both magnitude and directionE.g. velocity, force, acceleration, etc

Quiz: Temperature is a scaler/vector.

Representing Vector

Vector can be referred to as AB or a or a

Two vectors are equal if they have the same magnitude and direction

Magnitudes equal: |a| = |c| or a = c

Direction equal: they are parallel and pointing to the same direction

A

B

AB or a

CD or c

C

D

How about these? Are they equal?a

b

Opposite Vectors

magnitudes are equal, parallel but opposite in

sense These two vectors are not

equal Actually, they have the

relationb = -a

ab

Rectangular components of Vector

A vector a can be resolved into two rectangular components or x and y components

x-component: ax

y-component: ay

a = [ax, ay]

y

x

aay

ax

Ө

Addition of Vectors

V1

V2

V1

V2

V1 + V2

V1

V2

V1 + V2

Method 1Method 2

Subtraction of Vectors

V1

V2

-V2

-V2

V1

V1 - V2

Scaling of vectors (Multiply by a constant)

V1

V1

V1

2V1

0.5V1

-V1

Class work

Given the following vectors V1 and V2. Draw on the provided graph paper:

V1+V2

V1-V2

2V1

V1

V2

Class Work

For V1 given in the previous graph: X-component is _______ Y-component is _______ Magnitude is _______ Angle is _________

Rectangular Form and Polar Form For the previous V1

Rectangular Form [x, y]: [4, 2]

Polar Form r Ө : √20 26.57

magnitude angle

x-component

y-component

x

yAngle

yxMagnitude

1

22

tan

Polar Form Rectangular Form Since

Therefore:

Vx = |V| cos Ө

Vy = |V| sin Ө

|V|Vy

Vx

Ө

magnitude of vector V

V

V

V

V

y

x

sin

cos

Example

Find the x-y components of the following vectors A, B & C

Given : |A|=2, ӨA =135o

|B|=4, ӨB = 30o

|C|=2, ӨC = 45o

y

x

A

B

C

ӨA

ӨB

ӨC

Example (Cont’d)

For vector A, Ay = 2 x sin(135o) = 2, Ax = 2 x cos(135o) = -2

For vector B, Bx = 4 x cos(240o) = -4 x cos(60o) = -2,

By = 4 x sin(240o) = -4 x sin(60o) = -23

For vector C, Cy = 2 x sin(-315o) = 2 x sin(45o) = 2

Cx = 2 x cos(-315o) = 2 x cos(45o) = 2

Example

What are the rectangular coordinates of the point P with polar coordinates 8π/6

Solution: use x=r sin Ө and y=r cos Ө y=8sin(π/6)=8(1/2)=4 x=8cos(π/6)=8(3 /2)=43 Hence, the rectangular coordinates are

[43,4]

π/6

8y

x

Class work

Find the polar coordinates for the following vectors in rectangular coordinates.

V1 = [1,1] r=____ Ө=_______ V2=[-1,1] r=____ Ө=_______ V3=[-1,-1] r=____ Ө=_______ V4=[1,-1] r=____ Ө=_______

Class work

a = [6, -10], r=____ Ө=_______ b = [-6, -10], r=____ Ө=______ c = [-6, 10], r=____ Ө=______ d = [6, 6], r=____ Ө=_______

y

=’ x0

x

y

v

v1tan'

’’ x

y

0

'180

tan' 1

x

y

v

v

’’

x

y

0

)'180(

'180

tan' 1

x

y

v

v ’’

x

y

0

'

'360

tan' 1

x

y

v

v

oror

Unit Vector A vector of length 1 unit is called a unit vector

represents a unit vector in the direction of positive x-axis represents a unit vector in the direction of positive y-axis represents a unit vector in the direction of positive z-axis Example on 2-dimensional case:

v

vvv of vector unit ˆ:

-3i

i

5i

x

y

-2j4j

j

x

y

2i

k

j

i

ˆ

ˆ

ˆ

Representing a 2D vector by i and j

modulus or magnitude (length or strength) of vector

For a 3D vector represented by i, j, and kkcjbiav ˆˆˆ

y

x

|v|

a

b

i

j

22|| bav

222|| cbav

jbiav ˆˆ

Vector addition If Then

Example: Given

Then

,ˆˆˆ kajaiaa zyx

y

x

a

b

bx

by

aya+b

ax

kbjbibb zyxˆˆˆ

kbajbaibaba zzyyxxˆ)(ˆ)(ˆ)(

jirjip ˆ2ˆ,ˆ3ˆ2

ji

jirp

ˆ5ˆ3

ˆ)23(ˆ)12(

Vector subtraction Similarly, for

Then

Example Given

Then

kbajbaibaba zzyyxxˆ)(ˆ)(ˆ)(

y

x

a

b

by

by

ay

ax-b

y

x

aa+(-b)

,ˆˆˆ kajaiaa zyx kbjbibb zyx

ˆˆˆ

jirjip ˆ2ˆ,ˆ3ˆ2

ji

jirp

ˆˆ

ˆ)23(ˆ)12(

Exercise

A = 2i + 3j, B= i j

A+B =______________AB =_______________3A =_________________|A| = ______________the modulus of B =______

Example Given a=7i+2j and b=6i-5j, find a+b, a-b and modulus of a+b

Solution

jijiji ˆ3ˆ13)ˆ5ˆ6()ˆ2ˆ7(ba

jiji

jiji

ˆ7ˆˆ)52(ˆ)67(

)ˆ5ˆ6()ˆ2ˆ7(ba

178)3(13ba 22

Application of force system Find the x and y components of the resultant

forces acting on the particle in the diagram

Solution:

kNjijRiRR

kNRcomponenty

kNRcomponentx

yx

y

x

ˆ035.4ˆ332.1ˆˆ

035.4)15sin(430sin6:

332.1)15cos(430cos6:00

00

6kN 4kN

15 30

y

x

Scalar Product of Vectors

Scalar product, or dot product, of 2 vectors:

The result is a scalar

a

bcos|||| baba

=Angle between the 2 vectors

Example:

= _________

=___________

=__________

= ___________

ii

40 degrees20 degrees

2

1

a

b

ij

ji

ba

For x-y-z coordinates,

It can be shown that

zzyyxx babababa

kajaiaa zyxˆˆˆ

kbjbibb zyxˆˆˆ

Example: If and Find , and angle between two

vectors Solution:

jia ˆ6ˆ4

jib ˆ3ˆ3

ba

ab

66)3(43

6)3(634

ab

ba

Notice that a b = b a

0

22

22

3.101

196.01852

6

||||cos

18])3(3[||

52)64(||

ba

ba

b

a

Properties of scalar product1. Commutative:

2. Distributive:

3. For two vectors and , and a scalar C,

Example:

abba

cabacba

)(

)()()( bababa

CCC

a

b

Given A = [1,2], B=[2,-3], C=[-4,5]Find:

A(B+C) = _________AB + A C = _________3 A B = __________A (3B) = ___________

If two vectors are perpendicular to each other, then their scalar product is equal to zero.

i.e. if then Example

Given and

Show that and are mutually perpendicular

Solution:

ba

0ba

a

bjia ˆ4ˆ3

jib ˆ3ˆ4

ba

jijiba

01212

)ˆ3ˆ4()ˆ4ˆ3(

Vector Product of Vectors Vector product, or cross product,

Defined as

The result is a vector Magnitude = Pointing in the direction of is a unit vector perpendicular to

the plane containing and in a sense defined by the right-handed screw rule

ba

ebaba ˆsin||||

a

sin|||| ba

ee

b

sinbalength

bxa

a

b

e

• Right-Hand-Rule for Cross Product

a

b

a X b

Some properties of vector product If Ө=0o, then

If Ө=90o, then

It can be proven that

0ba

ebaba

abba

Properties of vector product NOT commutative: Distributive: For 3D basic unit vector:

Easy way: using right-hand rule

0ˆˆ,0ˆˆ,0ˆˆ kkjjii

jik

ikj

kji

ˆˆˆ

ˆˆˆ

ˆˆˆ

abba

)()()( cabacba

i

k j

+ve-ve

jki

ijk

kij

ˆˆˆ

ˆˆˆ

ˆˆˆ

For 3D vectors:

By determinant:

kbabajbabaibababa xyyxxzzxyzzyˆ)(ˆ)(ˆ)(

,ˆˆˆ kajaiaa zyx

kbjbibb zyxˆˆˆ

kbabajbabaibaba

bb

aak

bb

aaj

bb

aai

bbb

aaa

kji

ba

xyyxxzzxyzzy

yx

yx

zx

zx

zy

zy

zyx

zyx

ˆ)(ˆ)(ˆ)(

ˆˆˆ

ˆˆˆ

Example:

Simplify Solution:

)ˆˆ(ˆ jij

k

k

jjijjij

ˆ

0ˆ

ˆˆˆˆ)ˆˆ(ˆ

Example: Evaluate and calculate

if and

Solution:

ba

kjia ˆ5ˆ2ˆ3

kjib ˆ8ˆ4ˆ7

|| ba

kji

kji

kji

ba

ˆ26ˆ59ˆ4

ˆ]7)2(43[ˆ]75)8(3[ˆ]45)8)(2[(

847

523

ˆˆˆ

6.644173

)26()59()4(|| 222

ba