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Engineering
Fundamentals
Session 7 (3 hours)
Unit Vector
• A vector of length 1 unit is called a unit vector
a
unit vecto r of a : aˆ 
a
• i represents a unit vector in the direction of
positive x-axis
• j represents a unit vector in the direction of
positive y-axis
Unit Vector Examples
-3i^
y
^i
y
^^
2i+j
2i^
-2j^
^j
4j^
5i^
x
x
^
^
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
 
r | r | a 2  b 2
y
b
r
j
i
a
x
Addition of Vectors

• If a  axiˆ  a y ˆj
• Then

b  bxiˆ  by ˆj
 
a  b  (ax  bx )iˆ  (a y  by ) ˆj
• E.g.

 ˆ
ˆ
ˆ
p  2i  3 j , r  i  2 ˆj
 
p  r  (2  1)iˆ  (3  2) ˆj
 3iˆ  5 ˆj
y
b
by
ay
a
ax
a+b
by
x
Subtraction of Vectors
• Similarly, for


a  a xiˆ  a y ˆj b  bxiˆ  by ˆj
• Then
 
a  b  (ax  bx )iˆ  (a y  by ) ˆj
y
b
by
ay
a
• E.g.

 ˆ
ˆ
ˆ
p  2i  3 j , r  i  2 ˆj
 
p  r  (2  1)iˆ  (3  2) ˆj
 iˆ  ˆj
?
ax
by
x
y
-b
a+(-b)
a
x
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
and modulus of a+b (bolded symbols
denotes vectors)
• Solution
 
a  b  (7iˆ  2 ˆj )  (6iˆ  5 ˆj )  13iˆ  3 ˆj
 
a  b  (7iˆ  2 ˆj )  (6iˆ  5 ˆj )  (7  6)iˆ  (2  5) ˆj  iˆ  7 ˆ
 
a b 
13
2

 (3) 2  178
Example
• Find the x and y components of the
resultant forces acting on the particle in
y
6kN
the diagram
4kN
15
• Solution:
30
(Hint: the phase angles of the vectors are -15 and 210
degrees.)
x  component : Rx  6 cos 2100  4 cos( 150 )  1.332kN
y  component : Rx  6 sin 210 0  4 sin( 150 )  4.035kN

R  Rx iˆ  Rx ˆj  1.332iˆ  4.035 ˆjkN
x
Scalar Product of Vectors
• Scalar product, or dot
product, of 2 vectors:
   
a  b | a || b | cos 
How does the dot product behave
when a and b are perpendicular to
one another ?
When a and b have the same
direction?

a


b
Angle between the 2 vectors
Exercise
•
•
•
•
i.i = _________
i.j=___________
j.j=__________
a.b = ___________
a
2
40 degrees
1
20 degrees
b
Scalar Product of
Rectangular Vectors
• For x-y coordinates,

a  a xiˆ  a y ˆj

b  bxiˆ  by ˆj
• It can be shown that


a  b  axbx  a y by
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 a  4iˆ  6 ˆj and b  3iˆ  3 ˆj
 


• Find a  b , b  a and angle between
two vectors
• Solution:
a  b  4  3  6  (3)  6
b  a  3  4  (3)  6  6
Notice that a.b = b.a
Example (cont’d)

| a | (4 2  6 2 )  52

2
2
| b | [3  (3) ]  18
 
a b
6
cos     
 0.196
| a || b |
52 18
  101.30
Scalar Product of 3D
Rectangular Vectors
z
• Similarly, for x-y-z
coordinates,

az
ˆ  a y ˆj  az kˆ
a

a
i
x

b  bxiˆ  by ˆj  bz kˆ
bz
b
• Then


a  b  axbx  a y by  az bz
by
bx
ax
x
a
ay
y
Exercise
• Scalar product of 3i + 2j –k
and
–i + j = _______________
Scalar Product Properties
•
1.
2.
3.
Properties of scalar product
   
Commutative: a  b  b  a 



 
Distributive: a  (b  c )  a  b  a  c


For two vectors a and b , and a scalar k,

 
  
k (a  b )  (ka )  b  a  (kb )
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 a  b then a b  0

ˆ
ˆ
ˆ
• E.g. Given a  3iˆ 
and
b

4
i

3
j
4
j


• Show that a and b are mutually
perpendicular 

• Solution: a  b  (3iˆ  4 ˆj )  (4iˆ  3 ˆj )
 12  12  0
 
a  b
Vector Product of
Vectors ê
• Vector product,
 or cross product,

denoted a  b
• Defined as
 
   
a  b | a || b | sin  eˆ
 
length  a b sin 
axb



• The vector product of two vectors a and b
is a vector of modulus | a || b | sin  in the
direction of ê where ê is a unit vector

perpendicular
to
the
plane
containing
a

and b in a sense (forward/backward
direction) defined by the right-handed
screw rule

b

a
Right-Hand-Rule for
Cross Product
aXb
b
a
Vector Product of
Vectors
•
•
•
•
Note that
 
o
if Ө=0 , then a b  ... 0

o
if Ө=90 , then a b  ... abeˆ
 
 
It can be proven that a  b  b  a
Vector Product of
Vectors
• Properties of vector product
 
 
a  b  b  a
1. NOT commutative:



 
 
2. Distributive: a  (b  c )  (a  b )  (a  c )
3. iˆ  iˆ  0, ˆj  ˆj  0, kˆ  kˆ  0
iˆ  ˆj  kˆ, ˆj  kˆ  iˆ, kˆ  iˆ  ˆj
ˆj  iˆ  kˆ, kˆ  ˆj  iˆ, iˆ  kˆ   ˆj
Easy way to memorize #3: use right-hand
rule
+ve
-ve
i
k
j
Example
• Simplify
• Solution:
ˆj  (iˆ  ˆj )
ˆj  (iˆ  ˆj )  ˆj  iˆ  ˆj  ˆj
 kˆ  0
 kˆ
Vector Product of
Rectangular Vectors

b  bxiˆ  by ˆj  bz kˆ
• If
• then
 
a  b  (a y bz  az by )iˆ  (axbz  az bx ) ˆj  (axby  a y bx )kˆ

a  axiˆ  a y ˆj  az kˆ

 
• E.g. Evaluate a  b if a  3iˆ  2 ˆj  5kˆ and


• b  7iˆ  4 ˆj  8kˆ Hence calculate | a  b |
Example
• Solution:
• We
know


a  b  (a y bz  az by )iˆ  (axbz  az bx ) ˆj  (axby  a y bx )kˆ


• Substitute a  3iˆ  2 ˆj  5kˆ b  7iˆ  4 ˆj  8kˆ
 
a  b  [( 2)( 8)  5  4]iˆ  [3(8)  5  7] ˆj  [3  4  (2)  7]kˆ
 4iˆ  59 ˆj  26kˆ
 
| a  b | (4) 2  (59) 2  (26) 2
 4173
 64.6
Concept Map
A = A/|A|
unit vector
i = unit vector in x
direction
j = unit vector in y
direction
k = unit vector in z
direction
Vectors
magnitude=1
in terms of
matrix
[Vx, Vy]
Rectangular
form
in terms of i and j
Vx i + Vy j
2D, 3D
cross product v
vector X vector
Vector
operations
AyBz-AzBy i
vector + vector ,
vector - vector,
scalar X vector
A.B=|A|
Dot product
vector.vector
|B| cos θ
-(AxBz-AzBx) j
results in
vector
Direction
right-han
rule
+(AxBy-AyBx) k
results in
scalar
A .B = A x B x + A y B y
A .B = A x B x + A y B y + A z B z
|AXB|
= |A| |B| sin θ