Engineering Engineering FundamentalsFundamentalsEngineering Engineering

FundamentalsFundamentals

Session 7 (3 hours)Session 7 (3 hours)

Unit Vector

• A vector of length 1 unit is called a unit vector

• i represents a unit vector in the direction of positive x-axis

• j represents a unit vector in the direction of positive y-axis

a

aaa ˆ : ofr unit vecto

Unit Vector Examples

-3i

i2i+j

5i

x

y

-2j

4jj

^

^^

^

^ ^

^

x

y

^ ^

2i

A Vector in terms of i and j

• A 2D vector can be written as r=ai+bj

• modulus or magnitude (length or strength) of vector

y

x

r

a

b

i

j

22|| barr

^ ^

^ ^

Addition of Vectors• If • Then

• E.g.

jaiaa yxˆˆ

y

x

a

b

by

by

aya+b

ax

jbibb yxˆˆ

jbaibaba yyxxˆ)(ˆ)(

ji

jirp

jirjip

ˆ5ˆ3

ˆ)23(ˆ)12(

ˆ2ˆ,ˆ3ˆ2

Subtraction of Vectors• Similarly, for

• Then

• E.g.

jbaibaba yyxxˆ)(ˆ)(

y

x

a

b

by

by

ay

ax

-b

y

x

aa+(-b)

?jaiaa yxˆˆ

jbibb yxˆˆ

ji

jirp

jirjip

ˆˆ

ˆ)23(ˆ)12(

ˆ2ˆ,ˆ3ˆ2

Exercise

• A = 2i + 3j, B= -i –j (bolded symbol denotes vectors)

• A+B=______________• A-B=_______________• 3A=_________________• |A| = ______________• the modulus of B______

Example• If a=7i+2j and b=6i-5j, find a+b, a-b an

d modulus of a+b (bolded symbols denotes vectors)

• Solutionjijijiba ˆ3ˆ13)ˆ5ˆ6()ˆ2ˆ7(

jijijijiba ˆ7ˆˆ)52(ˆ)67()ˆ5ˆ6()ˆ2ˆ7(

178)3(13 22 ba

^ ^ ^^

Example• Find the x and y components of the

resultant forces acting on the particle in the diagram

• Solution: (Hint: the phase angles of the vectors are -15 and 210 degrees.)

kNjijRiRR

kNRcomponenty

kNRcomponentx

xx

x

x

ˆ035.4ˆ332.1ˆˆ

035.4)15sin(4210sin6:

332.1)15cos(4210cos6:

00

00

6kN 4kN

15 30

y

x

Scalar Product of Vectors

• Scalar product, or dot product, of 2 vectors:

a

bcos|||| baba

Angle between the 2 vectors

How does the dot product behave when a and b are

perpendicular to one another ?

When a and b have the same direction?

Exercise• i.i = _________• i.j=___________• j.j=__________• a.b = ___________

40 degrees20 degrees

2

1

a

b

Scalar Product of Rectangular Vectors

• For x-y coordinates,

• It can be shown that

yyxx bababa

jaiaa yxˆˆ

jbibb yxˆˆ

Exercise• [3,5].[2,-1]=________• The dot product of –i + j and 2i-3j

is ________________• The scalar product of 5i and 2i + j

is _____________

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

Example (cont’d)

0

22

22

3.101

196.01852

6

||||cos

18])3(3[||

52)64(||

ba

ba

b

a

Scalar Product of 3D Rectangular Vectors

• Similarly, for x-y-z coordinates,

• Then

kajaiaa zyxˆˆˆ

kbjbibb zyxˆˆˆ

zzyyxx babababa

x

y

z

az

ax

ay

a

bz

bx

by

b

Exercise• Scalar product of 3i + 2j –k and

–i + j = _______________

Scalar Product Properties

• Properties of scalar product1. Commutative:2. Distributive:3. For two vectors and , and a scalar

k,

abba

cabacba

)(

)()()( bkabakbak

a

b

Exercise

• A = [1,2], B=[2,-3], C=[-4,5]• A.(B+C) = _________• A.B + A.C = _________• 3 A.B = __________• A. (3B) = ___________(bolded symbols denotes vectors)

Scalar Product of Vectors

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

• i.e. if then• E.g. Given and • Show that and are mutually

perpendicular• Solution:

ba

0 ba

a

b jia ˆ4ˆ3

jib ˆ3ˆ4

ba

jijiba

01212

)ˆ3ˆ4()ˆ4ˆ3(

Vector Product of Vectors

• Vector product, or cross product, denoted

• Defined as

• The vector product of two vectors and is a vector of modulus in the direction of where is a unit vector perpendicular to the plane containing and in a sense (forward/backward direction) defined by the right-handed screw rule

ba

ebaba ˆsin||||

sinbalength

bxa

a

b

a

b

sin|||| ba

e ea

b

e

Right-Hand-Rule for Cross Product

a

b

a X b

Vector Product of Vectors

• Note that• if Ө=0o, then• if Ө=90o, then• It can be proven that

...ba

...ba

abba

0eab ˆ

Vector Product of Vectors

• Properties of vector product1. NOT commutative:2. Distributive:3.

Easy way to memorize #3: use right-hand rule

0ˆˆ,0ˆˆ,0ˆˆ kkjjii

jkiijkkij

jikikjkji

ˆˆˆ,ˆˆˆ,ˆˆˆ

ˆˆˆ,ˆˆˆ,ˆˆˆ

abba

)()()( cabacba

i

k j

+ve-ve

Example• Simplify • Solution:

)ˆˆ(ˆ jij

k

k

jjijjij

ˆ

0ˆ

ˆˆˆˆ)ˆˆ(ˆ

Vector Product of Rectangular Vectors

• If • then

• E.g. Evaluate if and

• Hence calculate

kbabajbabaibababa xyyxxzzxyzzyˆ)(ˆ)(ˆ)(

kajaiaa zyxˆˆˆ

kbjbibb zyxˆˆˆ

ba

kjia ˆ5ˆ2ˆ3

kjib ˆ8ˆ4ˆ7

|| ba

Example• Solution:• We know

• Substitute

kbabajbabaibababa xyyxxzzxyzzyˆ)(ˆ)(ˆ)(

kjia ˆ5ˆ2ˆ3

kjib ˆ8ˆ4ˆ7

kji

kjiba

ˆ26ˆ59ˆ4

ˆ]7)2(43[ˆ]75)8(3[ˆ]45)8)(2[(

6.64

4173

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

ba

Concept Map

Vectors

Rectangular form

in terms of i and junit vector

Vector operations

vector + vector , vector - vector, scalar X vector

Dot product vector.vector

cross product v vector X vector

A = A/|A|in terms of

matrix [Vx, Vy]

Vx i + Vy ji = unit vector in

x direction

j = unit vector in y direction

k = unit vector in z direction

magnitude=1 2D, 3D

A.B=|A| |B| cos θ

A.B = Ax Bx + Ay By

results in scalar

results in vector

|AXB|

= |A| |B| sin θ

AyBz-AzBy i

-(AxBz-AzBx) j

+(AxBy-AyBx) k

A.B = Ax Bx + Ay By + Az Bz

Direction: right-hand

rule