Dimension

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1.3 Dimension
1
LESSON 2
1.3 Dimension
1.3 Dimension
2
Objectives
• Deduce the correct dimension for a certain derived
quantity and its equation of physics.

• Use dimensional analysis to check homogeneity and
construct equations.
1.3 Dimension
3
Applications of dimension
• To determine the dimension of a physical quantity

• To check the homogeneity of an equation

• To construct an equation with given quantities
1.3 Dimension
4
To determine the dimension of a physical quantity
The following procedure is recommended :
Step 1: Identify all the physical quantities
in the given equation.
Step 2: Write the dimension
of each known physical quantity


Step 3: Solve the equation to get the
dimension of the unknown
physical quantity.
1.3 Dimension
5
What are the dimensions of the constants a and b in the
gas equation below ?
( )
2
a
p V b RT
V
| |
+ ÷ =
|
\ .
p = pressure
V = volume
R = gas constant
T = temperature
Solution :
| |
2
.........(i)
a
p
V
(
=
(
¸ ¸
From (i) , [a] = [p][V]
2

= (ML
-1
T
-2
)(L
3
)
2

= ML
5
T
-2

From (ii) , [b] = [V]
= L
3

Example 1 : To determine the dimension of a physical
quantity.
| | | |
.......(ii) V b =
Step 1
Step 2
Step 3
1.3 Dimension
6
To check the homogeneity of an equation
The following procedure is recommended :
Step 1: Identify all the physical
quantities in the
equation given
Step 2: Write the dimension of
each known physical
quantity
Step 3: Equate the dimensions
of the quantities on both
sides of the equation
An equation is homogeneous
if the dimensions on both sides
of the equation are equal

1.3 Dimension
7
L.H.S : [v]
2
= L
2
T
-2

R.H.S : [u]
2
= L
2
T
-2
, [ 2 as] = L
2
T
-2
Show that v
2
= u
2
+ 2 as is homogeneous

Solution :
Both sides of this equation have the same dimension
¬ homogeneous
Example 1 : To check the homogeneity of an equation
v,u = velocity
a = acceleration
s = displacement
Step 1
dimensionless
Step 2
[v]
2
= [u]
2
= [ 2 as] = L
2
T
-2

Step 3
1.3 Dimension
8
Solution :
L.H.S : [v]
2
= L
2
T
-2

R.H.S : [u] = LT
-1
, [at] = (LT
-2
)(T) = LT
-1
Dimension of the L.H.S = R.H.S ¬ not homogeneous
Example 2 : To check the homogeneity of an equation
v
2
= u + at
v,u = velocity
a = acceleration
t = time
Step 1
Step 2
[v]
2
= [u]

L
2
T
-2
= LT
-1

Step 3
[v]
2
= [at]
L
2
T
-2
= LT
-1

and,
1.3 Dimension
9
1
2
1
1
1
L
1
and L
1

quantity. ess dimensionl is : R.H.S
L
1
: L.H.S
÷ ÷
÷
=
(
¸
(

¸

=
(
¸
(

¸

=
(
¸
(

¸

r r
n
f
Both sides of this equation have the same dimension ¬ homogeneous
Solution :
Example 3 : To check the homogeneity of an equation
|
|
.
|


\
|
+ ÷ =
2 1
1 1
) 1 (
1

r r
n
f
Show that
is dimensionally correct.
f = focal length
n = refractive index
r = radius
Step 1
Step 3
Step 2
-1
1 2
1 1 1
L
f r r
( ( (
= = =
( ( (
¸ ¸ ¸ ¸ ¸ ¸
1.3 Dimension
10
To construct an equation with given quantities
The following procedure is recommended :
Step 6 : Rewrite the equation with the correct value of indices
for each quantity.
Step 5 : Determine the value of the indices.
Step 4 : Equate the indices of the dimensions on both sides of
the equation.
Step 3 : Write the dimension of all the quantities on both
sides of the equation.
Step 2 : Identify all the dimensionless quantities.
Step 1 : Write general equation that relates all the quantities.
1.3 Dimension
11
Construct an expression for the period, t of a simple pendulum which shows
how t is related to the mass m of the pendulum bob, the length l of the string
and acceleration g due to gravity.
Solution :
Let t = k m
x
l
y
g
z
where k is a dimensionless constant ; x, y, z are unknowns.
Dimensions on both sides must be the same.
[t] = [k][m]
x
[l]
y
[g]
z

T = M
x
L
y
(LT
-2
)
z
Equating the indices of M, L, T on both sides ;
M : x = 0,
L : y + z = 0
T : -2z = 1
z = -½ ,
y = ½ ,
x = 0
Period, t = k m
0
l
½
g



g
l
k t =
Step 1
Step 3
Step 4
Step 2
Step 5
Step 6
Example 1 : To construct an equation with given quantities
1
or
M
0
L
0
T
1
= M
x
L
(y+z)
T
-2z
1.3 Dimension
12
Example 2 : To construct an equation with given quantities
The air resistance F on a vehicle depends on the velocity v of the vehicle, the
density µ of air and the cross-sectional area A of the vehicle. Derive an
expression which relates F to v, µ and A.
Solution :
F · v
x
µ
y
A
z

F = k v
x
µ
y
A
z

[F] = [v]
x
[µ]
y
[A]
z

| | | | | |
y
x z m
ma v A
V
(
=
(
¸ ¸
| || | | |
| |
| |
| |
y
x z
y
m
m a v A
V
=
2 1 2
3
( ) ( )
( )
y
x z
y
M
MLT LT L
L
÷ ÷
=
Equating the indices on both sides;
2 ( 3 2 ) y x y z x
MLT M L T
÷ ÷ + ÷
=
M ; y = 1
L ; x-3y+2z = 1
T ; -x = -2
x = 2
y = 1
z = 1
Hence,
F = k v
2
µ
1
A
1

F = k v
2
µA
dimensionless
1.3 Dimension
13
Conclusion
Applications of dimension :

• To determine the dimension of a physical quantity

• To check the homogeneity of an equation

• To construct an equation with given quantities
1.3 Dimension
14
[a] = [p][V]
2

= ML
5
T
-2

| |
2 F
V
A
(
=
(
¸ ¸
| |
| |
| |
2
F
V
A
=
| |
| |
| |
2
ma
V
A
=
| || |
| |
| |
2
m a
V
A
=
2
3 2
2
( )
MLT
L
L
÷
=
1 2 6
( )( ) ML T L
÷ ÷
=
1.3 Dimension
15
1.4 Scalar & Vector
LESSON 3
1.3 Dimension
16
• Define scalar and vector quantities, unit vector in
Cartesian coordinate.

• Vector addition operation and their rules and
visualize resultant vector graphically by applying
a) commutative
b) associative
c) distributive and rules


Objectives
1.3 Dimension
17
• A quantity that has both magnitude and direction.
• Examples: velocity, acceleration, force, and momentum.
• Vector notation : or A or
• A vector can be represented by an arrow ;

Scalar quantity
• A quantity that has magnitude but no direction.
• Examples : mass, work, speed, energy, and density.
Vector quantity
A
vector direction
A a
magnitude of A A A = or
1.3 Dimension
18
P
Q
P = Q and point in the P Q same dire tion c ¬ =
• Equality of two vectors
Two vectors are equal if they have equal lengths and
point in the same direction.
P
• Negative of a vector
The negative of a vector is vector having the same
length but opposite direction.
-P
• Multiplying a vector by a scalar
m P = mP ; = mP

magnitude
direct same direction ion of P
×
÷
 positive scalar quantity :
 negative scalar quantity :
(-m) P = -mP ; = mP

magnitude
directi opposite direction on of P
×
÷
1.3 Dimension
19
A = E
• equal vectors :
• negative of a vector :
D = -A = - E
• parallel vectors :
• anti parallel vectors :
A ; E ; F ; G
A and D ; A and H ; F and H ; D and E ;
E and H ; F and D ; D and G ; G and H
F = A 2
1
2
A
G = A =
2
Example :
1.3 Dimension
20
Quiz :
1. Define scalar quantity
2. Define vector quantity
1.3 Dimension
21
Vector Addition & Subtraction
Vector addition
• Vector addition obeys commutative, associative and distributive
laws
• A resultant vector is a single vector which produces the same
effect ( in both magnitude and direction) as the vector sum of
two or more vectors.
• 2 methods of vector addition :
 graphical method - head to tail / tip to tail
- parallelogram
 calculation /algebraically (component method)
1.3 Dimension
22
A
A

B

B +
C =
C

 Additional graphical : head to tail
Example 1
1.3 Dimension
23
A

B

C

D

D C B A R
    
+ + + =
 Additional graphical : head to tail
Example 2
Geometric construction for
summing four vectors.
is the resultant vector R
1.3 Dimension
24
A

A

B

B

C

C=A+B
 Additional graphical : parallelogram
Example
The resultant vector , is the
diagonal of the parallelogram.
C
- The resultant of two vectors acting at any angle may be represented
by the diagonal of a parallelogram.
1.3 Dimension
25
A

A

B

B

C

Commutative law :
Example
Vector addition is commutative
C A B B A
    
= + = +
1.3 Dimension
26
Associative law :
Vector addition is associative
Example
) C B ( A
  
+ +
C B
 
+
A

B

C

C ) B A (
  
+ +
B A
 
+
A

B

C

) C B ( A
  
+ + C ) B A (
  
+ +
=
1.3 Dimension
27
Distributive law :
A

B

B A
 
+
A m

B m

) B A m(
 
+
B m A m ) B A m(
   
+ = +
Vector addition is distributive
Example
1.3 Dimension
28
Vector subtraction :
A - B = A + (-B) = C
A

B

C

B -

• The vector is equal in magnitude to vector
and points in the opposite direction.
-B B
• To subtract from , apply the rule of vector
addition to the combination of and
B A
-B A
1.3 Dimension
29
Vector addition & subtraction
Example 1
A
B
Draw the vectors :
a) A + B
b) A - B
c) 2A + B
1.3 Dimension
30
Vector addition & subtraction
Example 1
A
B
Draw the vectors :
a) A + B
b) A - B
c) 2A + B
Solution
A
B
A+B
a)
b)
A
-B
A - B
1.3 Dimension
31
c)
2A + B
2A
B
2A + B
1.3 Dimension
32
Components of a vector & Unit Vectors
Components of a vector :
)u
A

y
x
can be resolved into its components that are A

A vector
perpendicular to each other.
i) In 2 – D
A
x
A
y
Vector A is resolved into a -component -component nd x y
(vector components)
A = A + A
x y
A = = A cos or cos
x x
A A u u
A = A sin or sin
y y
A A u u =
1
tan
y
x
A
A
u
÷
=
Direction : Magnitude :
2 2
x y
A A A = +
(scalar
components)
1.3 Dimension
33
A =
A
x
A
y
)o
|
¸
A

y
x
z
ii) In 3 – D
Vector A can be resolved into 3 c , omponents : a compo nd nents x y z
A
x
A
y
A
z
A = = A cos
x x
A o
A = A cos
y y
A | =
A = A cos
z z
A ¸ =
A
z
Magnitudes :
A
x
+ A
y
+ A
z
(vector
components)
(scalar
components)
1.3 Dimension
34
1.3 Dimension
35
A
x
A
y
A
z
A
x
y
z
1.3 Dimension
36
1.3 Dimension
37
REMEMBER !!!
x
y
A
x
positive
A
x
positive
A
y
positive
A
y
positive
A
x
negative
A
x
negative
A
y
negative
A
y
negative
The signs of the components of a vector (eg: vector )
depend on the quadrant in which the vector is located :
A
1.3 Dimension
38
o
Addition of vectors using components (¬ Cartesian coordinates)
x
y
A
B
C
|
¸
A = + = cos sin
x y
A A A A o o +
B = + = cos sin
x y
B B B B | | +
C = + = cos sin
x y
C C C C ¸ ¸ +
Vectors x-component y-component
A
B
C
Let R is the resultant vector,
cos cos cos
x x x x
R A B C A B C o | ¸ = + + = + +
sin sin sin
y y y y
R A B C A B C o | ¸ = + + = + +
2 2
Magnitude, R
x y
R R = +
-1
Direction , = tan
y
x
R
R
u
cos A o
sin A o
cos B | sin B |
cos C ¸ sin C ¸
R
y

R
x

R
u
x
y
1.3 Dimension
39
Forces,F x-component y-component
37
o

Addition of vectors using components (¬ Cartesan coordinates)
x
y
45
o

Let R is the resultant vector,
(80 71 95 150)N = -94 N
x x
R F = = + ÷ ÷
¿
2 2
Magnitude, R ( 94) (71) 118 N F = = ÷ + =
-1 o
71
Direction , = tan 37
94
u = ÷
÷
Example 1: Four coplanar forces act on a body at point O. Find their resultant.
100 N
110 N
160 N
20
o

O
80 N
30
o

(0 71 55 55)N = 71 N
y y
R F = = + + ÷
¿
71 N
-94 N
R
143
o

(or, 143
o
from positive x-axis )
Solution :
80 N
80 cos 0
o
= 80 N
100 N
80 sin 0
o
= 0
100 cos 45
o
= 71 N
100 sin 45
o
= 71 N
110 N -110 cos 30
o
= -95 N 110 sin 30
o
= 55 N
160 N -160 cos 20
o
= -150 N -160 sin 20
o
= -55 N
x
y
1.3 Dimension
40
• A dimensionless vector
• Have magnitude of 1, with no units.
• Show direction
• Examples :
Unit Vectors :
ˆ
i
ˆ
j
ˆ
k
ˆ
i
ˆ
j
ˆ
k
: unit vector in the +ve x- direction
: unit vector in the +ve y-direction
: unit vector in the +ve z-direction
1.3 Dimension
41
ˆ ˆ
A i j
x y
A A = +
)u
A

y
x
Components of a vector in the form of unit vector :
i) In 2 – D
ii) In 3 – D
ˆ ˆ ˆ
A i j k
x y z
A A A = + +
A

y
x
z
ˆ
j
y
A
ˆ
j
y
A
ˆ
i
x
A
ˆ
i
x
A
ˆ
k
z
A
Example :
ˆ ˆ
A 2i 3j = +
Example :
ˆ ˆ ˆ
A 3i 4j + 2k = +
2
2
0
0
3
3
4
1.3 Dimension
42
1.3 Dimension
43
Example 1:
ˆ ˆ ˆ ˆ ˆ ˆ
A (6i 3j k) and B (4i 5j 8k) = + ÷ = ÷ + Given two vectors
B A   + 2
F
i
n
d

t
h
e

m
a
g
n
i
t
u
d
e

o
f

t
h
e

d
i
s
p
l
a
c
e
m
e
n
t

Solution :
C = A + B
ˆ ˆ ˆ ˆ ˆ ˆ
(6i 3j k) (4i -5j+8k)
ˆ ˆ ˆ
(6 4)i + (3-5)j + (-1+8)k
ˆ ˆ ˆ
10i - 2j + 7k
= + ÷ +
(
= +
¸ ¸
=
Addition of vectors using components (¬ unit vectors)
Find C A B and its magnitude. = +
2 2 2
Magnitude, C 10 ( 2) 7 12.4 = + ÷ + =
1.3 Dimension
44
Example 2:
ˆ ˆ ˆ ˆ ˆ
P = (2i + 3j) and Q (3i - 2j + 3k) = Given two vectors
Solution :
R = 2P - Q
ˆ ˆ ˆ ˆ ˆ ˆ
=2(2i + 3j + 0k) - (3i - 2j + 3k)
ˆ ˆ ˆ ˆ ˆ ˆ
= (4i + 6j + 0k) - (3i - 2j + 3k)
ˆ ˆ ˆ
= (4 - 3)i + (6 + 2)j + (0 - 3)k
ˆ ˆ ˆ
= i + 8j - 3k
(
¸ ¸
(
¸ ¸
Find R 2P Q and its magnitude. = ÷
2 2 2
Magnitude, R 1 8 ( 3) 8.6 = + + ÷ =
Addition of vectors using components (¬ unit vectors)
1.3 Dimension
45
Example 3:
ˆ ˆ ˆ ˆ
A (6i 3j) and B (4i 5j) = + = ÷ Given two vectors
B A   + 2
F
i
n
d

t
h
e

m
a
g
n
i
t
u
d
e

o
f

t
h
e

d
i
s
p
l
a
c
e
m
e
n
t

Solution :
C = A + B
ˆ ˆ ˆ ˆ
(6i 3j) (4i -5j)
ˆ ˆ
(6 4)i + (3-5)j
ˆ ˆ
10i - 2j
= + +
(
= +
¸ ¸
=
Addition of vectors using components (¬ unit vectors)
Find C A B and its magnitude. = +
2 2
Magnitude, C 10 ( 2) 10.2 = + ÷ =
Direction ;
1
1
0
tan
( 2)
= tan
10
= -11.3
y
x
C
C
u
÷
÷
=
÷
x
y
-2
10
1.3 Dimension
46
- graphical
ˆ ˆ ˆ
A = i + j + k
x y z
A A A
x y
ˆ ˆ
A = i + j A A
Conclusion
head to tail
parallelogram
• Scalar quantity – has magnitude
• Vector quantity – has magnitude and direction
• Unit vector – magnitude ‘1’ ; has no unit ; shows direction
• Component of a vector in 2-D :
• Component of a vector in 3-D :
• Vector addition :
- by calculation ( component method)
1.3 Dimension
47
1.4 Scalar & Vector
LESSON 4
1.3 Dimension
48
Objective:




• Understand the operation , use and properties of
dot (scalar) product and cross (vector) product of
two vectors
1.3 Dimension
49
u : angle between and (ranges from 0
o
to 180
o
)
Scalar (Dot) Product
A B A B cos u - =
• Definition :
where ;
B cos u : component of parallel to the direction of
B
u
A
• Scalar quantity

AB
B A
cos
1 -
|
|
.
|


\
|
-
=
 
u
A
A
B
A : magnitude of
A B
or
1.3 Dimension
50
Dot Product of unit vectors
Unit vectors , and are
perpendicular to each
ˆ
ˆ
o h
ˆ
t er
i j k
Using equation A B A B cos u - =
1 0 cos ) 1 )( 1 (
ˆ ˆ
ˆ ˆ ˆ ˆ
o
= = - = - = - k k j j i i
0 90 cos ) 1 )( 1 (
ˆ
ˆ ˆ ˆ ˆ ˆ
o
= = - = - = - k j j i j i
i
ˆ
j
ˆ
k
ˆ
1.3 Dimension
51
Dot Product of unit vectors
Given
ˆ
ˆ ˆ
x y z
A Ai A j A k = + + and
ˆ
ˆ ˆ
x y z
B B i B j B k = + +
ˆ ˆ
ˆ ˆ ˆ ˆ
( ) ( )
x y z x y z
A B A i A j A k B i B j B k - = + + - + +
x x y y z z
A B A B A B = + +
ˆ
ˆ ˆ ˆ ˆ ˆ
(
x x x y x z
Ai B i Ai B j Ai B k = - + - + - +
ˆ
ˆ ˆ ˆ ˆ ˆ

y x y y y z
A j B i A j B j A j B k - + - + - +
ˆ ˆ ˆ ˆ
ˆ ˆ
)
z x z y z z
A k B i A k B j A k B k - + - + -
ˆ
ˆ ˆ ˆ ˆ ˆ
( ) ( ) ( )
x x x y x z
i i i A B A B A j B i k = - + - + - +
ˆ
ˆ ˆ ˆ ˆ ˆ
( ) ( ) ( )
y x y y y z
j i j j j k A B A B A B - + - + - +
ˆ ˆ ˆ ˆ
ˆ ˆ
( ) ( ) ( )
z x z y z z
A B A B A B k i k j k k - + - + -
(1) (0) (0)
x x x y x z
A B A B A B = + + +
(0) (1) (0)
y x y y y z
A B A B A B + + +
(0) (0) (1)
z x z y z z
A B A B A B + +
1.3 Dimension
52
Example 1 : Dot product of unit vectors
Given,
and
k j i A
ˆ
ˆ
3
ˆ
2 + + =

k j i B
ˆ
ˆ
2
ˆ
4 ÷ + ÷ =

Find a) the magnitude of B and
 
A
b) the scalar product B
 
- A
c) the angle between B and
 
A
Solution :
21 ) 1 ( 2 ) 4 ( B B Magnitude
14 1 3 2 A A Magnitude a)
2 2 2
2 2 2
= ÷ + + ÷ = =
= + + = =
 
 
ˆ ˆ
ˆ ˆ ˆ ˆ
b) (2 3 ) ( 4 2 )
(2)( 4) (3)(2) (1)( 1)
3
A i j k i j k = + + - ÷ + ÷
= ÷ + + ÷
= ÷
o
100
21 14
3
cos c)
1 -
= |
.
|

\
| ÷
= u
1.3 Dimension
53
Example 2 – Dot product of unit vectors
are the force acting on an object and its displacement. j i x j i F
ˆ
2
ˆ
4 and
ˆ
3
ˆ
2 + = + =

Find a) the magnitude of x F


and
b) the magnitude of work done
c) the angle between x F


and
Solution :
20 2 4 Magnitude
13 3 2 F F Magnitude a)
2 2
2 2
= + = =
= + = =
x x
 
 
J 14
) 2 )( 3 ( ) 4 )( 2 (
)
ˆ
2
ˆ
4 ( )
ˆ
3
ˆ
2 ( Work b)
=
+ =
+ - + = - = j i j i x F


o
7 . 29
20 13
14
cos c)
1 -
=
|
|
.
|


\
|
= u
1.3 Dimension
54
The direction of the new vector is perpendicular to both
and can be determined by the right hand
rule
Vector (cross) Product
A B A B sin u × =
• Magnitude :
where ;
• Direction :
B and
 
A
• Vector quantity
B
u
A
B sin u : component of perpendicular to the direction of
u : angle between and
A
A B
A : magnitude of
A B
1.3 Dimension
55
Vector (cross) Product
- Right hand screw rule
• The thumb points in the
direction of the vector product
B A
 
×
• Sweep your fingers through
angle u from
A

• Put your forefingers parallel to
to B A
u
1.3 Dimension
56
(Questions on determining the direction of the
vector product using the right hand rule)
1.3 Dimension
57
Cross Product of unit vectors
o
ˆ ˆ
ˆ ˆ ˆ ˆ
(1)(1)sin 0 0 i i j j k k × = × = × = =
ˆ ˆ
ˆ ˆ ˆ ˆ
; k i j i k j × = × = ÷
A B A B sin u × =
From equation
Hence ,
(Counter-
clockwise)
ˆ ˆ
ˆ ˆ ˆ ˆ
; i j k j i k × = × = ÷
ˆ ˆ
ˆ ˆ ˆ ˆ
; j k i k j i × = × = ÷
1.3 Dimension
58
Cross Product of unit vectors
Given
ˆ
ˆ ˆ
x y z
A Ai A j A k = + + and
ˆ
ˆ ˆ
x y z
B B i B j B k = + +
ˆ ˆ
ˆ ˆ ˆ ˆ
( ) ( )
x y z x y z
A B A i A j A k B i B j B k × = + + × + +
ˆ
ˆ ˆ ˆ ˆ ˆ
(
x x x y x z
A i B i A i B j A i B k = × + × + × +
ˆ
ˆ ˆ ˆ ˆ ˆ

y x y y y z
A j B i A j B j A j B k × + × + × +
ˆ ˆ ˆ ˆ
ˆ ˆ
)
z x z y z z
A k B i A k B j A k B k × + × + ×
ˆ
ˆ ˆ ˆ ˆ ˆ
( ) ( ) ( )
x x x y x z
i i i A B A B A j B i k = × + × + × +
ˆ
ˆ ˆ ˆ ˆ ˆ
( ) ( ) ( )
y x y y y z
j i j j j k A B A B A B × + × + × +
ˆ ˆ ˆ ˆ
ˆ ˆ
( ) ( ) ( )
z x z y z z
A B A B A B k i k j k k × + × + ×
ˆ
ˆ
(0) ( ) ( )
x x x y x z
A B A B A B k j = + + ÷ +
ˆ
ˆ
( ) (0) ( )
y x y y y z
A B A B A B k i ÷ + + +
ˆ ˆ
( ) ( ) (0)
z x z y z z
A B A B A j B i + ÷ +
ˆ
ˆ ˆ
( ) ( ) ( )
y z z y x z z x x y y x
A B A B i A B A B j A B A B k = ÷ ÷ ÷ + ÷
1.3 Dimension
59
Cross Product of unit vectors
k A j A i A A
z y x
ˆ
ˆ ˆ
+ + =

Given,
k B j B i B B
z y x
ˆ
ˆ ˆ
+ + =

x y z
x y z
ˆ
ˆ ˆ

A A A
B B B
i j k
A B × =
B A B A of magnitude The
   
× = ×
( ) ( ) ( )
y z z y x z z x x y y x
ˆ
ˆ ˆ
A B A B - A B A B A B A B i j k = ÷ ÷ + ÷
y z
y z
A A
ˆ

B B
i = +
x y
x y
A A
ˆ

B B
k +
x z
x z
A A
ˆ

B B
j ÷
1.3 Dimension
60
Example 1 :
Given,
and
ˆ
ˆ ˆ
2 3 A i j k = + ÷
ˆ
ˆ ˆ
3 2 2 B i j k = + +
Find the magnitude of B
 
× A
Solution :

234 1 ) 13 ( 8 B A , Magnitude
2 2 2
= + ÷ + = ×
 
-1
2
cos 82.6
14 17
o
u
| |
= =
|
\ .
| | | | | |
k j i
k j i
k j i
B A
ˆ
ˆ
13
ˆ
8
ˆ
(1)(3) (2)(2)
ˆ
) 3 )( 3 ( (2)(2) -
ˆ
) 2 )( 3 ( (1)(2)
2 2 3
3 - 1 2
ˆ

ˆ

ˆ

+ ÷ =
÷ + ÷ ÷ ÷ ÷ + =
= ×
 
From
u cos B A B A = -
 
B and A between angle the and
 
where A B 2 and A 14
B 17
- = =
=
1.3 Dimension
61
Example 2 :
Solution :
1.3 Dimension
62
Conclusion
• Scalar (dot) product - scalar quantity
A B A B cos u - =
• Vector (cross) product - vector quantity
A B A B sin u × =
1.3 Dimension
63
Significant Figures
The rules of significant figures:
1. Any figures that is non-zero, are considered as a
significant figure.
2. Zeros at the beginning of a number are not significant
Example: 0.254 ----------------- 3 s.f
3. Zeros within a number are significant.
Example: 104.6 m ------------- 4 s.f
4. Zeros at the end of a number after the decimal point
are significant.
Example: 27050.0 ------------- 6 s.f

1.3 Dimension
64
Significant Figures
5. Zeros at the end of a whole number without a decimal point
may or may not be significant.
It depends on how that particular number was obtained, using
what kind of instrument, and the uncertainty involved.
Example: 500m ------------------- could be 1 or 3 sf.
Convert the unit:
500m = 0.5km (would you say it has 1 sf ? )
500m = 50 000cm (would you say it has 1 or 5 sf ? )
How to solve this problem ?

1.3 Dimension
65
Significant figures – Addition and
Subtraction processes
The rule:
The final result of an addition and/or subtraction should have
the same number of significant figures as the quantity with the
least number of decimal places used in the calculation.
Example:
23.1 + 45 + 0.68 + 100 = 169

Example:
23.5 + 0.567 + 0.85 =

24.9
1.3 Dimension
66
Significant figures – Multiplication and
division processes

The rule:
The final result of an multiplication and/or division should
have the same number of significant figures as the quantity
with the least number of significant figures used in the
calculation.
Example:
0.586 x 3.4 = 1.9924
= 2.0
Example:
13.90 / 0.580 = 23.9655

= 24.0

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