1.0 Physical Quantities and Measurement

Chapter 01Chapter 01PHYSICPHYSIC
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1
CHAPTER 1:CHAPTER 1:
Physical quantities andPhysical quantities and
measurementsmeasurements
(3 Hours)(3 Hours)

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Overview:
Physical quantitiesPhysical quantities
and measurementsand measurements
Basic andBasic and
derivedderived
quantitiesquantities
Scalar andScalar and
vectorvector
quantitiesquantities
UnitUnit
conversionconversion
VectorsVectors
multiplicationmultiplication
Addition,Addition,
subtractionsubtraction
and vectorand vector
resolutionresolution
Scalar andScalar and
vectorvector
productsproducts

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At the end of this chapter, students should beAt the end of this chapter, students should be
able to:able to:
 StateState basic quantities and their respective SIbasic quantities and their respective SI
units: length (m), time (s), mass (kg), electricalunits: length (m), time (s), mass (kg), electrical
current (A), temperature (K), amount ofcurrent (A), temperature (K), amount of
substance (mol) and luminosity (cd).substance (mol) and luminosity (cd).
 StateState derived quantities and their respectivederived quantities and their respective
units and symbols: velocity (m sunits and symbols: velocity (m s-1-1
), acceleration), acceleration
(m s(m s-2-2
), work (J), force (N), pressure (Pa), energy), work (J), force (N), pressure (Pa), energy
(J), power (W) and frequency (Hz).(J), power (W) and frequency (Hz).
 ConvertConvert units to common SI prefixes.units to common SI prefixes.
Learning Outcome:
1.1 Physical Quantities and units (1 hour)

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 Physical quantityPhysical quantity is defined as a quantity which can be measured.quantity which can be measured.
 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 physical quantities.from any physical quantities.
 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/θ 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 quantityDerived quantity is defined as a quantity which can bequantity which can be
derived from basic quantity.derived from basic quantity.
 Table 1.2 shows some examples of derived quantity.
Derived quantity Symbol Formulae SI Unit
Velocity v s/t m s-1
Volume V l × w × t m3
Acceleration a v/t m s-2
Density ρ m/V kg m-3
Momentum p m × v kg m s-1
Force F m × a kg m s-2
@ N
Work W F × s kg m2
s-2
@ J
Power P W/t kg m2
s-3
@ W
Frequency f 1/T s-1
@ Hz
Pressure P F/A kg m-1
s-2
@ Pa
Table 1.2Table 1.2

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 UnitUnit is defined as a standard size of measurement ofa standard size of measurement of
physical quantities.physical quantities.
 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-iridium
cylinder 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
s
458,792,299
1
second
kilogram
meter

SS The unit of basic quantity is called base unit
 addition unit for base unit:
 unit of plane angle - radian (rd)
 The common system of units used today are S.I unit (S.I unit (SystemSystem
International/metric systemInternational/metric system)).
 The unit of derived quantity – called derived unit
7
o
o
o
57.296
180
rad1
180rad
==
=
π
π

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 It is used for presenting larger and smaller values.for presenting 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 × 10−6
s = 2.33 µs
Prefix Multiple Symbol
tera × 1012 T
giga × 109 G
mega × 106 M
kilo × 103 k
deci × 10−1 d
centi × 10−2 c
milli × 10−3 m
micro × 10−6
µ
nano × 10−9 n
pico × 10−12 p
1.1.1 Unit Prefixes
Table 1.3Table 1.3

SS Note:Note:
Line of prefix: m, s, N, A, g and etc…
T G M k d c m µ n p
1012
109
106
103
100
100
10−1
10−2 10−3
10−6
10−9
10−12
UnitSymbol :
Value :
How to use?How to use?
1 Ts = ? ps
1 Ts =1012−
minus the index - divisionminus the index - division
(−12)
ps
= 1024
ps
9

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Solve the following problems.
a. 45 mm2
= ? m2
b. 37 km h−1
= ? m s−1
c. 30 g cm−3
= ? kg m−3
d. 29 µm = ? m
e. 23 m h−1
= ? m s−1
Solution :Solution :
a.
b. 11stst
method :method :
3
1 37 10 m
37 km h
1 h
−  ×
=  ÷
 
Example 1.1 :
3
37 10 m
3600 s
 ×
=  ÷
 1 1
37 km h 10.3 m s− −
=
2 2
45 mm 45 1 mm= ×
( )
2
3
45 10 m−
= ×
2 6 2 5 2
45 mm 45 10 m or 4.5 10 m− −
= × ×
1.1.2 Conversion of Unit

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37 km 1000 m 1 h
1 h 1 km 3600 s
   
=  ÷ ÷ ÷
   
11
22ndnd
method :method :
c.
1 37 km
37 km h
1 h
−  
=  ÷
 
1 1
37 km h 10.3 m s− −
=
( )
3 3
3
33 2 3
30 g 10 kg 1 cm
30 g cm
1 cm 1 g 10 m
−
−
−
 
    ÷=  ÷ ÷  ÷  
 
3 4 3
30 g cm 3.0 10 kg m− −
= ×

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d.
e.
29 mµ
5
29 m 2.9 10 mµ −
= ×
1 23 m 1 h
23 m h
1 h 3600 s
−   
=  ÷ ÷
  
1 3 1
23 m h 6.39 10 m s− − −
= ×
6
29 10 m−
= ×
Note:Note:
 Unit conversion is important, but it’s also important to recogniserecognise when
it’s needed.
 Always use the SI unitAlways use the SI unit.

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Calculate the volume in SI unit of a wire of length 125 cm and
diameter 0.65 mm.
Given l = 1.25 m ; d =0.65×10−3
m
The radius of the wire is
The volume of the cylindrical wire is given by
3
0.325 10 mr −
= ×
Example 1.2 :
( ) ( )
2
3
0.325 10 1.25π −
= ×
7 3
4.15 10 mV −
= ×
2
V r lπ=
3
0.65 10
2 2
d
r
−
×
= =

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Given 1 (angstrom) Å = 10-10
m
1. Solve all the following problems.
2. Convert the following problems into SI unit.
a. 0.249 mm3
b. 5.87 cm2
c. 12 g km h-1
d. 9.78 g cm-3
e. 10 km h-1
f. 8.5 g cm h-2
Exercise 1.1 :
a. 35 cm =_________m f. 24 mm2
=___________m2
b. 20 km =_________m g. 0.03 m2
=____________ mm2
c. 11 Mm =_________m h. 19 km2
=_____________ m2
d. 23 µF =_________F i. 56 g cm-3
=_________ kg m-3
e. 0.24 Å =_________m j. 560 km h-1
=_________ m s-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:
 DefineDefine scalar and vector quantities.scalar and vector quantities.
 PerformPerform vector addition and subtraction graphically.vector addition and subtraction graphically.
 Resolve vectorResolve vector into two perpendicular components(x andinto two perpendicular components(x and
y axes)y axes)
 Ilustrate unit vectors in cartesion coordinate.Ilustrate unit vectors in cartesion coordinate.
 State the physical meaning of dot(scalar) product:State the physical meaning of dot(scalar) product:

 State the physical meaning of cross(vector) product:State the physical meaning of cross(vector) product:
 Direction of cross product is determined by corkscrewDirection of cross product is determined by corkscrew
method or right hand rule.method or right hand rule.
Learning Outcome:
1.2 Scalars and Vectors (2 hours)
( , , )i j k%% %
( cos ) ( cos )A B A B B Aθ θ• = =
r r
( sin ) ( sin )A B A B B Aθ θ× = =
r r

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 State the physical meaning of dot(scalar)State the physical meaning of dot(scalar)
product:product:
 State the physical meaning of cross(vector)State the physical meaning of cross(vector)
product:product:
Direction of cross product is determined byDirection of cross product is determined by
corkscrew method or right hand rule.corkscrew method or right hand rule.
Learning Outcome:
1.2 Scalars and Vectors (2 hours)
( ) ( )θABθBABA coscos ==•
rr
( ) ( )θABθBABA sinsin ==×
rr

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 ScalarScalar quantity is defined as a quantity with magnitudequantity with magnitude only.
 e.g. mass, time, temperature, pressure, electric current,
work, energy, power and etc.
 Mathematics operational : ordinary algebra
 VectorVector quantity is 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 operational : vector algebra
1.2 Scalars and Vectors

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 Table 1.4 shows written form (notation) of vectors.
 Notation of magnitude of vectors.
1.2.1 Vectors
Vector A
LengthLength of an arrow– magnitudemagnitude of vector A
displacement velocity acceleration
s
r
v
r
a
r
s av
vv =
r
aa =
r
s (bold) v (bold) a (bold)
DirectionDirection of arrow – directiondirection of vector A
Table 1.4Table 1.4

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 Two vectorsTwo vectors equal if both magnitude and directionmagnitude and direction are the same.same.
(shown in figure 1.1)
 If vector A is multiplied by a scalar quantity k
 Then, vector A is

if kk = +ve= +ve, the vector is in the same directionsame direction as vector A.

if kk == --veve, the vector is in the opposite directionopposite direction of vector A.
P
r
Q
r
QP
rr
=
Figure 1.1Figure 1.1
Ak
r
Ak
r
A
r
A
r
−

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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
r
x
y
0

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c)c) CartesianCartesian coordinates
 2-Dimension (2-D)
m)5m,1(),( == yxs
r
s
r
y/m
x/m
5
10

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 3-Dimension (3-D)
s
r
2
3
4
m2)3,4,(),,( == zyxs
r
y/m
x/m
z/m
0

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d)d) PolarPolar coordinates
e)e) DenotesDenotes with + or – signs+ or – signs.
( )
r
N,15030=F
F
r
150°
+
+-
-

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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
ParallelogramParallelogram TriangleTriangle
B
r
A
r
B
r
A
r
BA
rr
+
O
BA
rr
+
B
r
A
r
BA
rr
+
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 vector A + B.
ABBA
rrrr
+=+ Commutative RuleCommutative Rule
B
r
A
r
AB
rr
+
O

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 For example :
1.2.4 Subtraction of Vectors
ParallelogramParallelogram TriangleTriangle
D
rC
r
O
DC
rr
−
O
D
r
−
( )DCDC
rrrr
−+=−
C
r
D
r
−
DC
rr
−
C
r
D
r
−DC
rr
−

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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 along
the 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 :
BA
rr
+ BA
rr
−
B2A
rr
+ BA2
rr
−

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 11stst
methodmethod :
1.2.5 Resolving a Vector
R
r
yR
r
xR
r
θ
0
x
y
θ
R
Rx
cos= cosxR Rθ⇒ =
θ
R
Ry
sin= θRRy sin=⇒
 22ndnd
methodmethod :
R
r
yR
r
xR
r
φ
0
x
y
φsin=
R
Rx
φsinRRx =⇒
φcos=
R
Ry
φcosRRy =⇒
φ
AdjacentAdjacent
componentcomponent
OppositeOpposite
componentcomponent

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 The magnitude of vectormagnitude of vector R :

Direction of vectorDirection of vector R :
 Vector R in terms of unit vectors written as
( ) ( )22
or yx RRRR +=

x
y
R
R
θ =tan or






= −
x
y
R
R
θ 1
tan
jRiRR yx
ˆˆ +=


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A car moves at a velocity of 30 m s-1
in a direction north 60° west.
Calculate the component of the velocity
a) due north. b) due west.
Example 1.3 :
N
EW
S
Nv

v

a)
b)

60cosN vv =
1
N sm51 −
=v

60cos30=
or

30sinN vv =

30sin30=

60sinW vv =
1
W sm26 −
=v

60sin30=
or

30cosW vv =

30cos30=
60°
30°
Wv


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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 1.4 :
Vector x-component y-component

30cosFFx −=
N6.68−=xF

30cos100−=F


30sinFFy −=
N05−=yF

30sin100−=
210°
F

S
x
S
210°
F

x
y
30°
xF

yF


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The figure above shows three forces F1, F2 and F3 acted on a particle O.
Calculate the magnitude and direction of the resultant force on particle
O.
Example 1.5 :
y
O
60o
x
( )N101F

( )N302F

( )N403F


SS
x
y
O 3F

33
∑ ++== 321 FFFFFr

∑ ∑+= yxr FFF

xxxx FFFF 321

++=∑
yyyy FFFF 321

++=∑
2F

1F

60o
xF2

yF2


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Vector x-component (N) y-component (N)
1F

3F

2F

0 1F− 01−=

60cos30−
15−=

60sin30
62=
3F 40= 0
VectorVector
sumsum
( ) 40510 +−+=∑ xF
25=
( ) 02601 ++−=∑ yF
16=

SSSolution :Solution :
The magnitude of the resultant force is
and
Its direction is 32.632.6°° from positive x-axis ORfrom positive x-axis OR
35
y
x
O
( ) ( )22
∑∑ += yxr FFF
N.729=rF
( ) ( )22
1625 +=








=
∑
∑−
x
y
F
F
θ 1
tan

6.32
25
16
tan 1
=





= −
θ
∑ yF

rF


32.6
∑ xF


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1. Vector has components Ax = 1.30 cm, Ay = 2.25 cm; vector
has components Bx = 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 :
BA

+
A

BA

+
AB

−
AB

−
B

A

B

BA

+
AB

+
BA

−
AB

−
y
x
0
37.0°
( )-1
sm18.0B

( )-1
sm12.0A


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37
3. Vector points in the negative x direction. Vector points at an
angle of 30° above the positive x axis. Vector has a magnitude of
15 m and points in a direction 40° below the positive x axis. Given
that , determine the magnitudes of and .
(Walker,pg.78,no. 65)
ANS. : 28 m; 19 mANS. : 28 m; 19 m
4. Given three vectors P, Q and R as shown in Figure 1.3.
Calculate the resultant vector of P, Q and R.
ANS. : 49.4 m sANS. : 49.4 m s−−22
; 70.1; 70.1°° above + x-axisabove + x-axis
Exercise 1.3 :
C
A

B

0=++ CBA

A

B

y
x0
50°
( )2
sm10 −
R

( )2
sm35 −
P

( )2
sm24 −
Q


SS
38
 notations –
 E.g. unit vector a – a vector with a magnitude of 1 unit in thevector with a magnitude of 1 unit in the
direction of vectordirection of vector AA.
 Unit vectors have no unitno unit.
 Unit vector for 3 dimension axes :
1.2.6 Unit Vectors
A

aˆ
cba ˆ,ˆ,ˆ
1ˆ ==
A
A
a 

)(@ˆ⇒- boldjjaxisy 1ˆˆˆ === kji
)(@ˆ⇒- boldiiaxisx
)(@ˆ⇒- boldkkaxisz

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39
 Vector can be written in term of unit vectors as :
 Magnitude of vector,
x
z
y
kˆ
jˆ
iˆ
krjrirr zyx
ˆˆˆ ++=

( ) ( ) ( )2
z
2
y
2
x rrrr ++=

SS
40
 E.g. : ( )mˆ2ˆ3ˆ4 kjis ++=

( ) ( ) ( ) m5.39234
222
=++=s
jˆ3
x/m
y/m
z/m
0
s

iˆ4kˆ2

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41
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 1.6 :
ab

−
( )mˆ3ˆˆ2 kjia −−=

ba

+
( )mˆ4ˆ2ˆ kjib −+=

( ) ibaba xxx
ˆ312 =+=+=+

( ) jbaba yyy
ˆ21 =+−=+=+

( )mˆ7ˆˆ3 kjiba −+=+

( ) ( ) kbaba zzz
ˆ743 −=−+−=+=+

( ) ( ) ( ) m68.7713
222
=−++=+ ba
ba

+2

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42
b)
The magnitude,
c)
The magnitude,
( ) iabab xxx
ˆ21 −=−=−=−

( ) ( ) jabab yyy
ˆ312 =−−=−=−

( )mˆˆ3ˆ kjiab −+−=−

( ) ( ) kabab zzz
ˆ34 −=−−−=−=−

( ) ( ) ( ) m.613131
222
=−++−=− ab
( ) ( ) ibaba xxx
ˆ512222 =+=+=+

( ) ( ) ( ) jbaba yyy
ˆ021222 =+−=+=+

( )mˆ10ˆ52 kiba −=+

( ) ( ) ( ) kbaba zzz
ˆ1043222 −=−+−=+=+

( ) ( ) m11.21052
22
=−+=+ ba

SS
43
Scalar (dot) productScalar (dot) product
 The physical meaning of the scalar productphysical meaning of the scalar product can 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

θ
A

B

A
 B

Figure 1.4aFigure 1.4a
θ
A

B

A

B

θBcos
Figure 1.4bFigure 1.4b
θ
A

B

θAcos
Figure 1.4cFigure 1.4c
( )ABABA

toparallelofcomponent=•
( )BABBA

toparallelofcomponent=•

SS
44
 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 θ ranges from 0° to 180 °.
 When
 The scalar product obeys the commutative law of multiplicationcommutative law of multiplication i.e.
( )θBABA cos=•

vectorsobetween twangle:θ
( )θABAB cos=•


90θ0 << scalar product is positivepositive

180θ09 << scalar product is negativenegative

90θ = scalar product is zerozero
ABBA

•=•

SS
45
 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 :
( ) ( )θFsθsFsFW coscos ==•=

x
z
y
kˆ
jˆ
iˆ
( ) ( ) 111cosˆˆ 2
===• o2
0iii
1ˆˆˆˆˆˆ =•=•=• kkjjii
( ) ( ) 111cosˆˆ 2
===• o2
0jjj
( ) ( ) 111cosˆˆ 2
===• o2
0kkk
( )( ) 09cosˆˆ ==• o
011ji
0ˆˆˆˆˆˆ =•=•=• kikjji
( )( ) 09cosˆˆ ==• o
011ki
( )( ) 09cosˆˆ ==• o
011kj

SS
46
Calculate the and the angle θ between vectors and for the
following problems.
a) b)
a)
The magnitude of the vectors:
The angle θ ,
Example 1.7 :
A

BA

• B

( )( ) ( )( ) ( )( ) kkjjiiBA ˆˆ33ˆˆ12ˆˆ12 •−+•+•=•

( ) ( ) ( ) 17322
222
=++=A
kjiA ˆ3ˆ2ˆ2 ++=

kjiA ˆˆ3ˆ4 +−=

kjiB ˆ3ˆˆ −+=

kjB ˆ3ˆ2 +=

922 −+=• BA

5−=• BA

( ) ( ) ( ) 11311
222
=−++=B
θABBA cos=•






 −
=






 •
= −−
1117
5
coscos 11
AB
BA
θ


112=θ
ANS.:ANS.:−−3; 99.43; 99.4°°

SS
47
Referring to the vectors in Figure 1.5,
a) determine the scalar product between them.
b) express the resultant vector of C and D in unit vector.
a) The angle between vectors C and D is
Therefore
Example 1.8 :
2
m441.DC −=•

( ) 
1361925180 =+−=θ
y
x0
( )m1C

( )m2D

19°25°
θCDDC cos=•

( )( ) 
136cos21=

SS
48
b) Vectors C and D in unit vector are
and
Hence
jCiCC yx
ˆˆ +=

( ) ( )ji ˆ25sin1ˆ25cos1 
+−=
( )mˆ42.0ˆ910 ji.C +−=

( ) ( ) jiDC ˆ65.042.0ˆ89.191.0 +++−=+

( )mˆ07.1ˆ98.0 ji +=
( ) ( )jiD ˆ19sin2ˆ19cos2 

+=
( )mˆ65.0ˆ891 ji.D +=


SS
49
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 θ 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
krjqipB ˆˆˆ ++=

kzjyixA ˆˆˆ ++=

CBA

=×
θABθBACBA sinsin ===×

vectorsobetween twangle:θ
RIGHT-HAND RULERIGHT-HAND RULE
C

Note:Note:
The angle betweenThe angle between
two vectorstwo vectors can only
be determined by
using the scalar (dot)scalar (dot)
productproduct.

SS
50
 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

B
 A

B

C

CBA

=×
CAB

=×
ABBA

×≠× but ( )ABBA

×−=×
B
)(C

A


SS
51
 The vector product of the unit vectors are shown below :
 Example of vector product is a torque (moment of force) on a metrea torque (moment of force) on a metre
rulerule where the expression is given by
Vector form:
Magnitude form:
x
z
y
kˆ
jˆ
iˆ
ijkkj ˆˆˆˆˆ =×−=×
kijji ˆˆˆˆˆ =×−=×
jkiik ˆˆˆˆˆ =×−=×
0ˆˆˆˆˆˆ =×=×=× kkjjii
0inˆˆ ==× o2
0siii
0inˆˆ ==× o2
0sjjj
0inˆˆ ==× o2
0skkk
r Fτ = ×
 
sinrFθτ =

SS
52
Given two vectors :
The angle between both vectors is 155 degree.
Determine
a) the magnitude of and its direction.
b)
Solution :Solution : θ=155°
a) The magnitude of vectors,
Apply:
Example 1.9 :
BA

×
BA

•
jiA ˆˆ3 −−=

kjiB ˆ2ˆ2ˆ4 −+=

sinA B AB θ× =
 
( ) ( ) 1013
22
=−+−=A
( ) ( ) ( ) 24224
222
=−++=B
( )( )10 24 sin155= 
2
6.55 unitA B× =
 

SS
53
Direction: always perpendicular to plane containing vector A
and B.
b) ( ) ( )kjikjiBA ˆ2ˆ2ˆ4ˆ0ˆˆ3 −+•+−−=•

2
14 unitA B• = −
 
( )( ) ( )( ) ( )( ) kkjjii ˆˆ20ˆˆ21ˆˆ43 •−+•−+•−=
0212 +−−=

SS
54
1. If vector and vector , determine
a) b)
ANS. :ANS. :
2. Three vectors are given as follow :
Calculate
a) b)
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 :
26; 46
jia ˆ+ˆ= 53

jib ˆ+ˆ= 42

ba

• ( ) bba

•+
kjickjibkjia ˆˆ2ˆ2andˆ2ˆ4ˆ;ˆ2ˆ3ˆ3 ++=+−−=−+=

( )a b c• −
 
( )cba

+•
29; 9− −
kjiP ˆˆ2ˆ3 −+=

kjiQ ˆ3ˆ4ˆ2 ++−=

QP

×
P

Q


SS
55
THE END.
Next Chapter…
CHAPTER 2 :
Kinematics of Linear Motion

1.0 Physical Quantities and Measurement

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