physical quantities matriculation stpm

7/28/2019 Physical quantities Matriculation STPM

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1

PHYSICS CHAPTER 1

CHAPTER 1:CHAPTER 1:

Physical quantities andPhysical quantities and

measurementsmeasurements

(3 Hours)(3 Hours)

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PHYSICS CHAPTER 1

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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: StateState basic quantities and their respective SI units: lengthbasic quantities and their respective SI units: length

(m), time (s), mass (kg), electrical current (A), temperature(m), time (s), mass (kg), electrical current (A), temperature(K), amount of substance (mol) and luminosity (cd).(K), amount of substance (mol) and luminosity (cd).

StateState derived quantities and their respective units andderived quantities and their respective units andsymbols: velocity (m ssymbols: velocity (m s-1-1), acceleration (m s), acceleration (m s-2-2), work (J),), work (J),force (N), pressure (Pa), energy (J), power (W) andforce (N), pressure (Pa), energy (J), power (W) andfrequency (Hz).frequency (Hz).

StateState and convert units with common SI prefixes.and convert units with common SI prefixes.

Learning Outcome:

1.1 Physical Quantities and Units (1 hour)

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PHYSICS CHAPTER 1

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Physical quantityPhysical quantity is defined as a quantity which can be measuredquantity which can be measured

using measuring instrument.using measuring instrument.

It can be categorised into 2 types

Basic (base) quantityBasic (base) quantity

Derived quantityDerived quantity

Basic quantityBasic quantity is defined as a quantity which cannot be derivedquantity which cannot be derived

from any other physical quantity.from any other physical quantity.

Table 1.1 shows all the basic (base) quantities.

1.1 Physical Quantities and Units

Quantity Symbol SI Unit Symbol

Length l metre m

Mass m kilogram kg

Time t second s

Temperature T/ kelvin K

Electric current I ampere A

Amount of substance N mole mol

Luminous Intensity candela cdTable 1.1Table 1.1


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Derived quantity Symbol Formulae Unit

Velocity v s/t m s-1

Frequency f 1/T s-1 or Hz (hertz)

Acceleration a v/t m s-2

Pressure F/A N m-2 or Pa

(pascal)

Momentum p m v kg m s-1

Force F m a kg m s-2 or

N(newton)

Work W F s kg m2 s-2 or

J(joule)

Power P W/t

Js-1 or W (watt)

Table 1.2Table 1.2

Derived quantityDerived quantity is defined as a quantity that is obtained from thequantity that is obtained from the

combination of base quantities.combination of base quantities.

Table 1.2 shows some examples of derived quantity.


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UnitUnit is defined as a standard size of measurement of aa standard size of measurement of a

physical quantity.physical quantity. Examples :

1 second1 second is defined as the time required forthe time required for

9,192,631,770 vibrations of radiation emitted by a9,192,631,770 vibrations of radiation emitted by a

caesium-133 atom.caesium-133 atom.

1 kilogram1 kilogram is defined as the mass of a platinum-iridiumthe mass of a platinum-iridiumcylinder kept at International Bureau of Weights andcylinder kept at International Bureau of Weights and

Measures ParisMeasures Paris.

1 meter1 meter is defined as the length of the path travelled bythe length of the path travelled by

light in vacuum during a time interval oflight in vacuum during a time interval of

1299, 792, 458

s


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rad = 180o

1 rad =180

o

=57.296o

The unit of a basic quantity is called base unit.

additional unit for base unit: unit of plane angle - radian (rd)

unit of solid angle- steradian (sr)

The common system of units used today are S.I unit (S.I unit (SystemSystem

International/metric systemInternational/metric system)) and cgs unit - UK.

The unit of derived quantity called derived unit


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It is used for represent larger and smaller values.for represent larger and smaller values.

Table 1.3 shows all the unit prefixes.

Examples:

5740000 m = 5740 km = 5.74 Mm

0.00000233 s = 2.33 106 s = 2.33 s

Prefix Multiple Symbol

tera 1012 T

giga 109 G

mega 106

Mkilo 103 k

deci 101 d

centi 102 c

milli 103 m

micro 106

nano 109 n

pico 1012 p

1.1.1 Unit Prefixes

Table 1.3Table 1.3


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Table 1.4 shows the conversion factors between SI and British units for

length and mass only.

1.1.2 Conversion of Unit

Length Mass

1 m = 39.37 in = 3.281 ft 1 kg = 103 g

1 in = 2.54 cm 1 slug = 14.59 kg1 km = 0.621 mi 1 lb = 0.453 592 kg

1 mi = 5280 ft = 1.609 km 1 kg = 0.0685 slug

1 angstrom ( ) = 10 10 m

Table 1.4Table 1.4


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Solve the following problems of unit conversion.

a. 15 mm2 = ? m2 b. 65 km h1 = ? m s1

c. 450 g cm3 = ? kg m3 d. 29 cm = ? in

e. 12 mi h1 = ? m s1

Solution :Solution :

a. 15 mm2

= ? m2

b. 65 km h-1 = ? m s-1

11stst method :method : 65 km h1=

65103 m1 h

Example 1.1 :

1 mm 2=103m 2


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22ndnd method :method :

c. 450 g cm-3 = ? kg m-3

65 km h1=65 km1 h

450 g cm3=450 g1 cm 3 103 kg

1 g

1 cm 3

102

3

m

3


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d. 29 cm = ? in

e. 12 mi h-1 = ? m s-1

29 cm=29 cm

1

2 .54

in

1 cm 12 mi h1=

12 mi1 h

1 . 609 km1 mi

1000 m1 km

1 h3600 s


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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: DefineDefine scalar and vector quantities.scalar and vector quantities. PerformPerform vector addition and subtraction operationsvector addition and subtraction operations

graphically:graphically: commutative rulecommutative rule associative rule, andassociative rule, and distributive ruledistributive rule

Resolve vectorResolve vectorinto two perpendicular components (2-D) :into two perpendicular components (2-D) :

Components in the x and y axes.Components in the x and y axes.

Components in the unit vectors in CartesianComponents in the unit vectors in Cartesiancoordinate.coordinate.

Learning Outcome:

1.2 Scalars and Vectors (2 hours)

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At the end of this topic, students should be able to:At the end of this topic, students should be able to: Define and useDefine and use dot (scalar) product:dot (scalar) product:

and cross (vector) product:and cross (vector) product:

Direction determined by corkscrew method or right handDirection determined by corkscrew method or right handrule.rule.

Learning Outcome:

1.2 Scalars and Vectors (2 hours)

AB=A B cos=B A cos

AB=A B sin =B A sinwww.kms

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PHYSICS CHAPTER 1

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ScalarScalarquantity is defined as a quantity with magnitudequantity with magnitude only. e.g. mass, time, temperature, pressure, electric current,

work, energy and etc.

Mathematics operation: ordinary algebra

VectorVectorquantityis defined as a quantity with both magnitudequantity with both magnitude

& direction.& direction.

e.g. displacement, velocity, acceleration, force, momentum,

electric field, magnetic field and etc.

Mathematics operation: vector algebra

1.2 Scalars and Vectors


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Table 1.6 shows written form (notation) of vectors.

Notation of magnitude of vectors.

1.2.1 Vectors

Vector ALengthLength of an arrow magnitudemagnitude of vector A

displacement velocity acceleration

s v as av

v=v

a=a

s (bold) v (bold) a (bold)

DirectionDirection of arrow directiondirection of vector A

Table 1.6Table 1.6


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Can be represented by using:

a)a) Direction of compassDirection of compass, i.e east, west, north, south, north-east,

north-west, south-east and south-west

b)b) Angle with a reference lineAngle with a reference line

e.g. A boy throws a stone at a velocity of 20 m s-1, 50 above

horizontal.

1.2.2 Direction of Vectors

50

v

x

y

0


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c)c) CartesianCartesian coordinates

2-Dimension (2-D)

s=x , y =1 m, 5 m

s

y/m

x/m

5

10


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3-Dimension (3-D)

s

2

3

4

s=x , y , z=4, 3, 2 my/m

x/m

z/m

0


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d)d) PolarPolarcoordinates

e)e) DenotesDenotes with + or signs+ or signs.

F=30 N,150

F150

+

+-

-


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There are two methods involved in addition of vectors graphically i.e.

ParallelogramParallelogram TriangleTriangle

For example :

1.2.3 Addition of Vectors


BA

B

A

AB

O

AB

B

A

AB

O


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Triangle of vectors method:

a) Use a suitable scale to draw vector A.

b) From the head of vector A draw a line to represent the vector B.

c) Complete the triangle. Draw a line from the tail of vector A to the

head of vector B to represent the vectorA + B.

A

B=

B

A

Commutative RuleCommutative Rule

B

A

BAO


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If there are more than 2 vectors therefore

Use vector polygon and associative rule. E.g. PQR

RQP

R

Q

P PQ

P

Q

R=

P

Q

R Associative RuleAssociative Rule

PQ R


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Distributive Rule :

a.

b.

For example :

Proof of case a:Proof of case a: let = 2

AB = A BA= AA

, are real number

AB =2 AB

B

A

AB

O 2 AB


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2 AO

2 B

2 A2 B

2 AB =2 A2 B

A B=2 A2 B


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Proof of case b:Proof of case b: let = 2 and = 1

A

A= 21 A=3 A

3 A

AA=2 A1 A2 A A

3 A

=

21 A=2 A1 A


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For example :

1.2.4 Subtraction of Vectors


DC

O

C D

O

D

C D=C D C

D

C

D

C

DC

D


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Vectors subtraction can be used

to determine the velocity of one object relative to another object

i.e. to determine the relative velocity. to determine the change in velocity of a moving object.

1. Vector A has a magnitude of 8.00 units and 45 above the positive x

axis. Vector B also has a magnitude of 8.00 units and is directed alongthe negative x axis. Using graphical methods and suitable scale to

determine

a) b)

c) d)(Hint : use 1 cm = 2.00 units)

Exercise 1.2 :

AB AB

A2

B 2

A

B


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11stst methodmethod :

1.2.5 Resolving a Vector

R

Ry

Rx

0x

y

Rx

R=cos Rx= R cos

Ry

R=sin Ry= R sin

22ndnd methodmethod :

R

Ry

Rx

0x

y

Rx

R=sin Rx= R sin

Ry

R=cos R

y= R cos


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The magnitude of vectormagnitude of vectorR :

Direction of vectorDirection of vectorR :

VectorR in terms of unit vectors written as

R orR=Rx 2Ry 2

tan=R

y

Rx

or=tan1RyR

x

R=RxiR

yj


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A car moves at a velocity of 50 m s-1 in a direction north 30 east.

Calculate the component of the velocitya) due north. b) due east.


Example 1.6 :

N

EW

S

vN

vE

v30

60

a)

b)

vN= v cos30

orvN= v sin60

vE= v sin30

or

vE= v cos60


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A particle S experienced a force of 100 N as shown in figure above.

Determine the x-component and the y-component of the force.


Example 1.7 :

150

F

Sx

15030

F

Sx

y

Fy

Fx

Vector x-component y-component

Fx=Fcos30

orFFx=Fcos150

Fy=Fsin 30

Fy=Fsin150


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The figure above shows three forces F1, F2and F3 acted on a particle O.

Calculate the magnitude and direction of the resultant force on particle

O.

Example 1.8 : y

30o

O

F230N

F110N

30o x

F340N


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30o


O

y

x

F3

30

o

F3y

Fr= F=F1F2F3Fr= Fx Fy

Fx=F1xF2xF3x

Fy=F1yF2yF3y

F2x

F1

F2

60o

F2

F3x


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Vector x-component y-component

F1

F3

F2

F1x=0 NF

1y=F

1F

1y=10 N

F2x=30cos60

F2x=15 N F

2y=30sin60

F2y=26 N

F3x=40cos30

F3x=34 .6 N

F3y=40sin30

F3y=20 N

VectorVector

sumsum


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y

xO


The magnitude of the resultant force is

and

Its direction is 162162 from positive x-axis OR 18from positive x-axis OR 18 above negative x-axis.above negative x-axis.

Fr=Fx

2Fy2

=tan1FyFx

Fy

Fx

162

Fr

18

PHYSICS CHAPTER 1


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1. Vector has componentsAx = 1.30 cm,Ay = 2.25 cm; vector

has componentsBx = 4.10 cm,By = -3.75 cm. Determine

a) the components of the vector sum ,

b) the magnitude and direction of ,

c) the components of the vector ,

d) the magnitude and direction of . (Young & freedman,pg.35,no.1.42)

ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345; 2.80 cm, -6.00 cm;; 2.80 cm, -6.00 cm;6.62 cm, 2956.62 cm, 295

2. For the vectors and in Figure 1.2, use the method of vector

resolution to determine the magnitude and direction of

a) the vector sum ,

b) the vector sum ,

c) the vector difference ,

d) the vector difference .(Young & freedman,pg.35,no.1.39)

ANS. : 11.1 m sANS. : 11.1 m s-1-1, 77.6, 77.6; U think;; U think;28.5 m s28.5 m s-1-1, 202, 202; 28.5 m s; 28.5 m s-1-1, 22.2, 22.2

Exercise 1.3 :

AB

A

ABBABA

B

A B

AB

BAABBA

Figure 1.2Figure 1.2

y

x0

37.0

B 18 .0 m s-1

A 12 .0 m s-1

PHYSICS CHAPTER 1


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3. Vector points in the negativex direction. Vector points at an

angle of 30 above the positivex axis. Vector has a magnitude of

15 m and points in a direction 40 below the positivex axis. Giventhat , determine the magnitudes of and .(Walker,pg.78,no. 65)

ANS. : 28 m; 19 mANS. : 28 m; 19 m

4. Given three vectorsP,Q andR as shown in Figure 1.3.

Calculate the resultant vector ofP,Q andR.

ANS. : 49.4 m sANS. : 49.4 m s22; 70.1; 70.1 above + x-axisabove + x-axis

Exercise 1.3 :

C

A

BABC=0 A

B


y

x0

50R 10 m s2

P35 m s2 Q 24 m s2

PHYSICS CHAPTER 1


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notations

E.g. unit vectora a vector with a magnitude of 1 unit in the direction

of vectorA.

Unit vectors are dimensionless.

Unit vector for 3 dimension axes :

1.2.6 Unit Vectors

A

a

a ,b , c

a=

A

A=1

[ a ]=1

y axis j @ j bold i=j=k=1xaxis i@ i bold

z axisk @ kbold

PHYSICS CHAPTER 1


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Vector can be written in term of unit vectors as :

Magnitude of vector,

x

z

y

kji

r=rx iry jrzk

r=rx 2 ry 2rz 2

PHYSICS CHAPTER 1


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E.g. : s=4 i3 j2 k m

3 j

x/m

y/m

z/m

0

s4 i2 k

PHYSICS CHAPTER 1


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Two vectors are given as:

Calculate

a) the vector and its magnitude,

b) the vector and its magnitude,

c) the vector and its magnitude.


a)

The magnitude,

Example 1.9 :

ba

a=i2 j6 k m

ab

b=4 i3 j k m

abx=axbx=14=5 iab

y

=ay

by

=23=5 j

abz=azbz=61=7 k

2ab

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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b)

The magnitude,

c)

The magnitude,

bax=bxax=41=3 i

ba y=byay=3 2 =jba z=bzaz=16=5 k

2ab x=2axbx=2 1 4=6 i

2ab y=2ayby=2 2 3 =7 j2abz=2azbz=2 6 1=13 k

PHYSICS CHAPTER 1


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Scalar (dot) productScalar (dot) product

The physical meaning of the scalar productphysical meaning of the scalar productcan be explained by

considering two vectors and as shown in Figure 1.4a.

Figure 1.4b shows the projection of vector onto the direction of

vector .

Figure 1.4c shows the projection of vector onto the direction of

vector .

1.2.7 Multiplication of Vectors

A B

A

B

AB

Figure 1.4aFigure 1.4a

A

BA

B

B cos

Figure 1.4bFigure 1.4b

A

BAcosFigure 1.4cFigure 1.4c

( )BABBA

toparallelofcomponent=

( )ABABA

toparallelofcomponent=

PHYSICS CHAPTER 1


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From the Figure 1.4b, the scalar product can be defined as

meanwhile from the Figure 1.4c,

where

The scalar product is a scalar quantityscalar quantity.

The angle ranges from 0 to 180 . When

The scalar product obeys the commutative law of multiplicationcommutative law of multiplication i.e.

AB=A B cos

: angle between two vectors

BA=B A cos

090 scalar product is positivepositive

90180 scalar product is negativenegative

=90

scalar product is zerozero

AB=BA

PHYSICS CHAPTER 1


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Example of scalar product is work donework done by a constant force where the

expression is given by

The scalar product of the unit vectors are shown below :W=Fs=Fscos =s Fcos

x

z

y

kji

ii=i2cos0o=12 1=1

ii=jj=kk=1

jj=j2cos0o=12 1 =1

kk=k2 cos0o=1 2

1 =1

ij= 1 1 cos 9 0o=0ij=jk=ik=0

ik= 1 1 cos 90

o

=0

jk= 1 1 cos 9 0o=0

PHYSICS CHAPTER 1


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Calculate the and the angle between vectors and for the

following problems.a) b)


a)

The magnitude of the vectors:

The angle ,

Example 1.10 :AAB B

AB= 1 4 ii1 2 jj1 3 kk

A=ik A=4 i3 jkB= 4 i2 j3 k B=2 j3 k

AB=423

AB=AB cos

=cos1 ABAB =cos1

3

329

ANS.:ANS.:3; 99.43; 99.4

PHYSICS CHAPTER 1


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Referring to the vectors in Figure 1.5,a) determine the scalar product between them.

b) express the resultant vector ofCandD in unit vector.


a) The angle between vectors

Cand

Dis

Therefore

Example 1.11 :

=18025 19=174


y

x0

C1 m

D 2 m 19

25

CD=CD cos= 1 2cos174

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

49

b) Vectors CandD in unit vector are

and

Hence

C=Cx

iCy

j

= 1cos25 i1sin25 j

C D= 0.911.89 i0.420.65 j

D=2cos19 i2sin19 j

PHYSICS CHAPTER 1


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Vector (cross) productVector (cross) product

Consider two vectors :

In general, the vector product is defined as

and its magnitudemagnitude is given by

where

The angle ranges from 0 to 180 so the vector product always

positivepositive value. Vector product is a vector quantityvector quantity.

The direction of vector is determined by

B= p iq jrk

A=x iy jz k

AB=C

AB=C=ABsin=AB sin: angle between two vectors

RIGHT-HAND RULERIGHT-HAND RULE

C

PHYSICS CHAPTER 1


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51

For example:

How to use right hand rule :

Point the 4 fingers to the direction of the 1st

vector. Swept the 4 fingers from the 1st vector towards the 2nd vector.

The thumb shows the direction of the vector product.

Direction of the vector product always perpendicularDirection of the vector product always perpendicular

to the plane containing the vectors andto the plane containing the vectors and .

A

C

BA

B

C

AB=C

BA=C

ABBA but AB=BA

B C

A

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

52

The vector product of the unit vectors are shown below :

Example of vector product is a magnetic force on the straighta magnetic force on the straight

conductor carrying current places in magnetic fieldconductor carrying current places in magnetic field where the

expression is given by

x

z

y

k

j

i

jk=kj=iij=ji=k

ki=ik=j

ii=jj= k k=0ii=i2s in 0o=0jj=j 2s in 0o=0

k k=k2

s in 0o

=0

F=IlB F=IlB sin

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

53

The vector product can also be expressed in determinant form as

11stst method :method :

22ndnd method :method :

Note :Note :

The angle between two vectorsThe angle between two vectorscan only be determined by using

the scalar (dot) productscalar (dot) product.

AB=i

j

kx y z

p q r

AB=yrzq ixrzp j xqyp k

AB=yrzq izpxr j xqyp k

PHYSICS CHAPTER 1


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54

Given two vectors :

Determine

a) and its magnitude b)

c) the angle between vectors and .


a)

The magnitude,

Example 1.12 :

AB=i

j

k3 2 1

1 0 5

AB AB

AB=2 5 1 0 i3 5 1 1 j3 0 2 1 kAB=100 i 151 j 02 k

A=3 i2 j k

B=i5 k

A B

AB=10 2 16 2 2 2

PHYSICS CHAPTER 1


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55

Given two vectors :

Determine

a) and its magnitude b)

c) the angle between vectors and .


a)

The magnitude,

Example 1.12 :

AB=i

j

k3 2 1

1 0 5

AB AB

AB=2 5 1 0 i3 5 1 1 j3 0 2 1 kAB=100 i 151 j 02 k

A=3 i2 j k

B=i5 k

A B

AB=10 2 16 2 2 2

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

56

b)

c) The magnitude of vectors,

Using the scalar (dot) productscalar (dot) product formula,

AB= 3 i2 j ki0 j5 k

AB=2

AB= 3 1 ii2 0 jj1 5 kkAB=305

AB=AB cos

=cos1 ABAB =cos1 214 26

A=3 22 2 1 2=14

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

57

1. If vector and vector , determine

a) , b) , c) .ANS. :ANS. :

2. Three vectors are given as follow :

Calculate

a) , b) , c) .

ANS. :ANS. :

3. If vector and vector ,

determine

a) the direction of

b) the angle between and .

ANS. : U think, 92.8ANS. : U think, 92.8

Exercise 1.4 :

2 k ; 26 ; 46

a=3 i5 j b=2 i4 jab ab abb

a=3i3

j2

k ; {

b=

i4

j2

k and { c =2

i2

j

k

abc abc abc 21;9 ;5 i11 { j9 k

P=3 i2 jk Q=2

i4

j3

kPQ

P Q

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

THE END

Next ChapterCHAPTER 2 :

Kinematics of Linear Motion

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