BAB 3 BA501 Vector Dan Scalar

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ENGINEERING MATHEMATICS 4 VECTORBA501

CHAP 3 - Vector and Scalar Products

Vector Quantities

a quantity that has magnitude and direction

Examples of vector quantities are displacement, velocity, direction, momentum,

force, lift, weight and etc.

Scalar Quantities

a quantity that has magnitude only.

Typical examples of scalar quantities are time, speed, temperature, and volume. A

scalar quantity or parameter has no directional component, only magnitude. For

example, the units for time (minutes, days, hours, etc.) represent an amount of time

only and tell nothing of direction. Additional examples of scalar quantities are

density, mass, and energy.

Fundamental of vector

i) Vector notation

vector notation is how to present vector, such us :a) vector is usually given a bold letter, such as A

b) place a right-handed arrow over the letter to denote a vector

c) vector can be write in engineering notation and matrix notation

ii) Vector representation

Vectors can be graphically represented by directed line segments

example : vector AB = a

a

40

A

B




If two vectors are represented by two adjacent sides of a parallelogram, then the

diagonal of parallelogram through the common point represents the sum of the

two vectors in both magnitude and direction.

Q Q R

O P O P

OP + OQ = OR

Q Q R

O P O P

OP + OQ = QP Figure 2

iii) Method of component

Rules of vector components:

i) Components should be perpendicular is called the orthogonal

components.ii) The component s of the vector may be in any axis (x and y axis) we called

the horizontal or the vertical dimension.

iii) The direction of the components is look like head to tail, so that we canadd that vector.

iv) If we are adding those x and y vectors we can get the resultant vector.

The components of a vector are those vectors which, when added together,give the original vector.

N

43




E

The sum of the components of two vectors is equal to the sum of these two

vectors.

A2 A

A = A1 + A2

A1

A2 A

A1

A1, the component in an easterly direction, will have a magnitude

A1 = A cos .

A2, the component in a northerly direction, will have a magnitude

A2 = A sin

Substraction

Subtraction is considered an addition process with one modification that the secondvector (to be subtracted) is first reversed in direction and is then added to the firstvector.

B B

=

O A O A

OA + BA− = OB

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A B C

a


Characteristic of additional vectors.

Characteristic Resultant

i. Commutative law a + b = b + a

ii. Associative law (a + b )+ c = a + (b + c )

iii. Identity law a + 0 = a

iv. Inversion law a + (-a ) = 0

Multiplication vector with scalar Multiplication vector , a with scalar value , t produce vector t a where

magnitude |t|.

Characteristic Resultant

i. Commutative law ma = amexp: 2a = a 2

ii. Associative law m(na ) = (mn)a exp : 2x(3a ) = (2x3) a = 6a

iii. Distributive law (m+n)a = ma + na

exp : (2 + 3) a = 2a + 3a = 5a

iv. Distributive law m(a + b ) = ma + mb

exp : 2 ( a + b ) = 2a + 2b

Definition 1 :

Given 3 point A, B, and C. Point A, B and C is collinear if AC t AB = , t is scalar non zero

t a

45




example 1 : Find the scalar value λ , for equation below :λ OC = OA6 + BC 12 + AO4 + AB2 + OB10

46




a) MD b ) DL c) LM

4. Given position vector point A, B and C respectively are 4 a + 2b , 8a - 4b

, and 16a - 16b . Show that point A, B and C are collinear and find

ratio of AB : BC .

48




If a~ = b~

hence a~ • a~ = aa

~~ =2~

a = a2 ; ∴ a~ • a~ = a2

Refer to fig 2,a~and b~ are parallel but in the opposite direction,

hence θ = 180°;

∴ , ba~~ = kosba

~~ 180° = ba~~ (-1) = - ba

~~

3. Perpendicular vector

If a~ and b~ are perpendicular, hence θ = 90°.

∴ a~ • b~ = ba~~ cos 90° = ba

~~ 0 = 0

∴ a~ • b~ = 0

4. Angle between two vector.

Theorem;

; a~ • b~ = kosba~

~θ ⇒ kos θ =

ba

ba

~~

~•~

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b

~

a~




4 :Given vector a~and b~ respectively a~ = 4 , b

~

=3 and a~ • b~= 7. Find

the magnitude of (a~+ b~ ) and the angle between a

~ and b~ .

52




5 : Calculate the work done F · s given |F|, |s| and θ (the angle between the

force F and the displacement s) when

|F| = 4 N, |s| = 2 m, θ= 27o

53

Exercise : Calculate a~ • b~ if given | a~ | = 3 , | b~ | = 5 and angle between a

~

and b~ is 60o

ans =1

7




Vector in a Cartesian plane

Example

Fig 3

54

In a Cartesian plane, if point C is (x,y), hence position vector for C can

expressed in the form ;

C(x, y)

xi

yj

xO

OC = Xi + Y j or (Y

X) .

Magnitude for OC , OC = 22+ yx

Unit vector in the positive direction of OC = 22+

+

yx

yjxi

N

A(8, 6)

8

6

y

xO

From fig 3;

ON = 8 , NA = 6,

Hence ON = 8i and NA = 6 j




Operational Vector in Cartesian Plane

Fig 4

Addition and Subtraction of position vectors

Given OP = 1

~r = X1 i + Y1 j or (

1

1

Y

X

) ;

OQ = 2

~r = X2 i + Y2 j or (

2

2

Y

X

)

Hence 1

~r + 2

~r = X1 i + Y1 j + X2 i + Y2 j

= (X1 + X2) i + (Y1 + Y2) j ( assemble i and j )

1

~r – 2

~r = X1 i + Y1 j - X2 i - Y2 j

= (X1 - X2) i + (Y1 - Y2) j

Orin column vector,

1

~r + 2

~r = (

1

1

Y

X

) + (2

2

Y

X

) =

+

+

21

21

Y Y

X X

1

~r – 2

~r = (

1

1

Y

X

) - (2

2

Y

X

) =

−−

21

21

Y Y

X X

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Q(X2, Y

2)

1

~r

y

xO

P(X1, Y

1)

2

~r

In fig 4, P(X1,Y

1) and Q(X

2,Y

2) are

two point in Cartesian Plane

and O is origin.




Example

6 : Given a~ = 5i + 2 j and b

~ = 2i – 5 j , find;

a) a~ + b

~ b) a~ - 2 b

~

Then find magnitude for each vector.

57




7: Given that position vector A is a~= 2i + 3 j and Position vector B is

b~= i – 5 j . Find :

a) angle between vector AB and A

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Exercise: 1. Given 2 vector a~ = 2i + 3 j and b~ = i + 2 j . If a~ perpendicular to a

~+ λb~ ,Find th

scalar value of λ (ans, : λ =813− )

2.Given a~ =

−4

3and b

~=

−

1

1

a) Find a~+ b

~ (

−3

2)

b) Calculate |a~ | ( 5 unit)

c) If a~= cb ~2~

3 + , Express c~ as column vector (

−

2

13

3

)



y

qq

r S

y

x

γ

βS

α


Vector in Three Dimension in Cartesian Plane

59

O

z

x

p[

O

zDirection Cosines

If OS makes angle of α, β and γ with the coordinat

axes i, j, and k respectively, then

The direction cosines for OS vector are:

cos α, = OS

x

, cos β = OS

y

, cos γ = OS

z

{Magnitude OS = a~ = 222 z y x ++ }

α = angle between vector S and x - axes

β = angle between vector S and y - axes

γ = angle between vector S and z - axes

α, β and γ known as direction angle.

Coordinate S (p, q, r).

Position vector for S.

OS = a~ = xi + yj + zk =

z

y

x

= pi + q j + r k =

r

q

p




Example

8: If position vector of point A, B and C are a~= 2i + j + 2k, b~ = 4i + 5j + 3k and

c~ = i - 3j + 2k respectively, find

a) vector AB

b) direction cosines of AB

c) unit vector in direction of a~+ b~+ c~

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The Scalar Product of 3 Dimension

(produce scalar value)

Let A = a1i + b1 j + c1k

and B = a2i + b2 j + c2k

B A • = (a1i + b1 j + c1k ) • (a2i + b2 j + c2k )

= a1a2i ⋅i +a1b2i ⋅ j +a1c2i ⋅k + b1a2 j ⋅i + b1b2 j ⋅ j + b1c2 j ⋅k + c1a2k ⋅i

+ c1b2k ⋅ j + c1c2k

⋅k

However , i ⋅i = i ⋅i kos 00 = 1

⇒ i ⋅I = j ⋅ j = k ⋅k = 1

and, i ⋅ j = i ⋅ j kos 900 = 0

⇒ i ⋅ j = j ⋅k = k ⋅I = 0

∴

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B A • = a1a2 + b1b2 + c1c2

Angle between vector A and B

cos θ = B A

B A~~

~~•




The Vector Product of 3 Dimension

(produce vector value)

the vector product can be written in determinant forms as :

Let, A = a1i + b1 j + c1k

and B = a2i + b2 j + c2k

B A× =222

111

cba

cba

k ji

Unit vector perpendicular to B A×

u = B x A

B x A~~

~~

Area of parallelogram to vector A and B where A and B are side by side

63

B A× = k baba jcacaicbcb )()()( 122112211221 −+−−−

Angle between vector A and B

Sin θ = B A

B x A

~~

~~



A

B


Area of parallelogram, B A× = B A~~

sin θ

Volume of Parallelepiped

a parallelepiped is a three-dimensional figure formed by six parallelograms

An alternative method defines the vectors a = (a1, a2, a3), b = (b1, b2, b3) and

c = (c 1, c 2, c 3) to represent three edges that meet at one vertex. The volume

of the parallelepiped then equals the absolute value of the scalar triple

product a · (b × c ):

Volume of Parallelepiped is:

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t

Ө

|)(| cbaV ×•= = |)(| acb ×• = |)(| bac ×•

http://en.wikipedia.org/wiki/Parallelogram

http://en.wikipedia.org/wiki/Scalar_triple_product


http://en.wikipedia.org/wiki/Parallelogram






Example10 : Find the angle between vector a~and b

~usinga) scalar productb) vector product

Given :a~= 2i + 3 j +k b~= i – 2 j – 6k

65




11: Unit vector perpendicular to P and Q respectively are 3i – 2 j + 4k and2i + 3 j – k

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12. Find the volume of the parallelepiped with adjacent edges PQ , PR and

PS where P (3, 0, 1), Q(−1, 2, 5), R (5, 1,−1), S(0, 4, 2)

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Exercise : 1. Angle between two vector a~= i + λ j +2k and b~= 2i + 3 j + k is cos-1

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1. Find the value of λ. (ans λ = -1)

2. If A~ = 5i – 2 j + 3k . B~ = 3i + j – 2k andC ~

= i – 3j +4k . Find

a) A~

• ( B~

x C ~

) (ans-12)

b) A~

x ( B~

x C ~

) (ans 62i + 44 j -74k )

3. If p = 4i – 3 j + 5k . and q = 3i – 5 j – 2k . Find :

i. q p 32 ×

ii. )4(2 pq p +•

BAB 3 BA501 Vector Dan Scalar

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