1. Being able to predict the path of a thunderstorm is essential
for minimizing the damage it does to lives and property. If a
thunderstorm is moving at 20 km h in a direction 53° north of
east, how far north does the thunderstorm move in 1 h?
PHYSICAL QUANTITIES: VECTORS

2. LEARNING GOALS
• The difference between scalars and vectors, and how to add and
subtract vectors graphically.
• What the components of a vector are, and how to use them in
calculations.
• What unit vectors are, and how to use them with components to
describe vectors.
• Explain the two ways of multiplying vectors vector product and
scalar product

3. Vectors and Vector Addition

4. Question: How to add and subtract vectors graphically.

8. Question: What the components of a vector are and how to use them
in calculations.

9. Finding a vector’s magnitude and direction from its
components
We can find the magnitude and direction if we know the components. By
applying the
Pythagorean theorem to we can find the magnitude of vector
Direction
If is measured from the positive x-axis, and a positive angle is measured
toward the
positive y-axis as in Fig., then

2
2
y
x A
A
A 












x
y
x
y
A
A
A
A
1
tan
tan



10. Question: How to use components to calculate the vector
sum (resultant) of two or more vectors.
Figure 1.21 shows two vectors and
and their vector sum along with the x-
and y-components of all three vectors. We
can see from the diagram that
A B
R
y
y
y
x
x
x
B
A
R
B
A
R





11. Figure shows this result for the case in
which the components
and are all positive.
y
x
y
x B
B
A
A ,
,
,

12. We can extend this procedure to find the sum of
any number of vectors.
We can introduce a z-axis perpendicular to
the xy-plane; then in general a vector has
components in the three coordinate
directions. Its magnitude A is
2
2
2
z
y
x A
A
A
A 


......
.......












y
y
y
y
y
y
x
x
x
x
x
x
E
D
C
B
A
R
E
D
C
B
A
R

13. Question: How to multiply a vector by a scalar
Multiplying a vector by a scalar.
For example, Eq. (1.9) says that each component
of the vector is twice as great as the
corresponding component of the vector so
is in the same direction as but has twice the
magnitude. Each component of the vector is
three times as great as the corresponding
component of the vector but has the opposite
A
2
A A
2
A
A
3
y
y
x
x
cA
D
cA
D


Component of A
c
D 

15. Unit Vectors
A unit vector is a vector that has a magnitude of 1,
with no units. Its only purpose is to point—that is, to
describe a direction in space.
In an x-y coordinate system we can
define a unit vector that points in
the direction of the positive x-axis.
iˆ
a unit vector that points in the
direction of the positive y-axis.
ĵ j
A
A
i
A
A
y
y
x
x
ˆ
ˆ




Question: What are unit vectors and how to use
them with components to describe vectors.

16. Similarly, we can write a vector in terms of its
components as
Using unit vectors, we can express the
vector sum of two vectors and as
j
A
A
i
A
A
y
y
x
x
ˆ
ˆ




j
A
i
A
A y
x
ˆ
ˆ 


j
B
i
B
B y
x
ˆ
ˆ 


j
A
i
A
A y
x
ˆ
ˆ 


B
A
R





j
R
i
R
j
B
A
i
B
A
j
B
i
B
j
A
i
A
y
x
y
y
x
x
y
x
y
x
ˆ
ˆ
ˆ
)
(
ˆ
)
(
)
ˆ
ˆ
(
)
ˆ
ˆ
(











17. If the vectors do not all lie in the xy-
plane, then we need a third
component.
We introduce a third unit vector that
points in the direction of the positive
-axis (Fig. 1.24). Then
k
R
j
R
i
R
k
B
A
j
B
A
i
B
A
R
k
B
j
B
i
B
B
k
A
j
A
i
A
A
z
y
x
z
z
y
y
x
x
z
y
x
Z
y
x
ˆ
ˆ
ˆ
ˆ
)
(
ˆ
)
(
ˆ
)
(
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ


















18. Question: Explain the two ways to multiply
vectors: the scalar (dot) product and the vector
(cross) product.

19. A















B















The scalar product of two vectors and is denoted by
. Because of this notation, the scalar product is also called the
dot product. Although and are vectors, the quantity is a
scalar. To define the scalar product, the two vectors and are
drawn with their tails at the same point (Fig). The angle ϕ between
their directions ranges from 0° to 180°. Figure bellow shows the
projection of the vector
B















.
A B















A















.
A B















Scalar Product
1.25 Calculating the scalar product of
two vectors,
. cos
A B AB 

















20. . cos
A B A B 

  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
Definition of cross the scalar
product or dot product.
into the direction of ; this projection is the component of in
the direction of and is equal to . . Then
to be the magnitude of multi- plied by the component of
in the direction of which is expressed as an equation as
A















.
A B















A















B















A















B















A















cos
B 

22. The vector product of two vectors and also called the
cross product, is denoted by . As the name suggests, the
vector product is itself a vector.
A















B















A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To define the vector product , two vectors and with their tails
at the same point (Fig. 1.29a) are drawn. The two vectors then lie in
a plane. The vector product is a vector quantity with a direction
perpendicular to this plane (that is, perpendicular to both and
and a magnitude equal to
A















B















A















B















sin
AB 
That is,

sin
A B AB 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Where is unit


Vector product

24. There are always two directions perpendicular to a given plane, one
on each side of the plane. The direction of is chosen as
follows.
A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
If we imagine to rotate vector about the perpendicular line
until it is aligned with choosing the smaller of the two possible
angles between and . The fingers of our right hand is
curled around the perpendicular line so that the fingertips point in
the direction of rotation; the thumb will then point in the direction
of . Figure 1.29a shows this right-hand Rule and describes a
second way to think about this rule.
A















B















A















B















A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

25. Similarly, the direction of can be determined by rotating
into as in Fig. 1.29b. The result is a vector that is
opposite to the vector The vector product is not
commutative. In fact, for any two vectors and
B A

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
A















B















A B B A
  
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
B















A
















26.  180
The angle is measured from toward and the smaller of the
two possible angles is taken, so is in the ranges from 0° to 180°
Then and is never negative. It is also
noted that when and are parallel or antiparallel, or or
The vector product of two parallel or antiparallel vectors is
always zero.
 A















B















A















B















0
  
sin
AB 
180
  
In particular, the vector product of any vector
with itself is zero.
0
sin 



27. B















Similar to the scalar product, a geometrical interpretation for the
magnitude of the vector product can be given as shown in Fig.
1.30a,
is the component of vector that is perpendicular to the
direction of vector . The magnitude of equals the
magnitude of multiplied by the component of
perpendicular to
sin
B  B















A















A















A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
A















B
















28. Figure 1.30b shows that the magnitude of also equals
the magnitude of multiplied by the component of
perpendicular to .
A B

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
A















B















B
















30. Question: show that the scalar product
can be positive, negative and zero

31. we can define to be the magnitude of
multiplied by the component of in the direction of as in
Fig. 1.25c. Hence
.
A B















B















A















B















 
. cos cos
A B B A AB
 
 















The scalar product is a scalar quantity, not a vector, and it may be
positive, negative, or zero. When is between 0° and and
then the scalar product is positive.
 90
cos 0
 

32. When is between 90° and so that the
component of in the direction of is negative, and is
negative (Fig. 1.26b). Finally, when
(Fig. 1.26c). The scalar product of two perpendicular vectors is
always zero.
 180 cos 0
 
B















A















.
A B















90 ,
   . cos90 0
A B AB
 
















33. For any two vectors and and
This means that the scalar product obeys the commutative law of
multiplication; the order of the two vectors does not matter.
A















,
B















cos cos
AB BA
 


34. 1.26 The scalar product
can be positive, negative, or zero,
depending on the angle between
and
A















B
















35. If
If
If

36. Quick Quiz
1. The magnitudes of two vectors and
are A=12 units and B=8 units.
Which pair of numbers represents the
largest and smallest possible values for the
magnitude of the resultant vector (a) 14.4
units, 4 units (b) 12 units, 8 units (c) 20
units, 4 units (d) none of these answers

37. A car travels 20.0 km due north and then 35.0 km in a direction 60.0°
west of north as shown in Figure. Find the magnitude and direction of
the car’s resultant displacement.
The law of cosines to find R:

38. A car travels 20.0 km due north and then 35.0 km in a direction 60.0°
west of north as shown in Figure 3.11a. Find the magnitude and
direction of the car’s resultant displacement.
to find the direction of measured from the northerly direction the law of
sines can be used
The resultant displacement of the car is 48.2 km in a direction 38.9° west of north

39. Figure illustrates typical proportions of male (m) and female
(f) anatomies.
The displacements and from the soles of the feet to the
navel have magnitudes of 104 cm and 84.0 cm, respectively.
The displacements and from the navel to outstretched
fingertips have magnitudes of 100 cm and 86.0 cm,
respectively. Find the vector sum of these displacements for
both people

40. The vector product of any vector
with itself is zero, so
Using Eqs. (1.22)
and (1.23) and the right-hand rule,
we find
0
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 




 k
k
j
j
i
i
j
k
i
i
k
i
j
k
k
j
k
i
j
j
i
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ















Question: Express the cross product of and in terms of
their components and the corresponding unit vectors

41.    
k
B
i
A
j
B
i
A
i
B
i
A
k
B
j
A
j
B
j
A
i
B
j
A
k
B
i
A
j
B
i
A
i
B
i
A
k
B
j
B
i
B
k
A
j
A
i
A
B
A
z
z
y
z
x
z
z
y
y
y
x
y
z
x
y
x
x
x
z
y
x
z
y
x
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ



























     k
B
A
B
A
j
B
A
B
A
i
B
A
B
A
B
A x
y
y
x
z
x
x
z
y
z
z
y
ˆ
ˆ
ˆ 








B
A
C





Thus the component of
x
y
y
x
z
z
x
x
z
y
y
z
z
y
x
B
A
B
A
C
B
A
B
A
C
B
A
B
A
C







42. The vector product can also be expressed in determinant
form as
z
y
x
z
y
x
B
B
B
A
A
A
k
j
i
B
A
ˆ
ˆ
ˆ





43.    
k
B
k
A
j
B
k
A
i
B
k
A
k
B
j
A
j
B
j
A
i
B
j
A
k
B
i
A
j
B
i
A
i
B
i
A
k
B
j
B
i
B
k
A
j
A
i
A
B
A
z
z
y
z
x
z
z
y
y
y
x
y
z
x
y
x
x
x
z
y
x
z
y
x
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
.

















0
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ
1
ˆ
.
ˆ
ˆ
.
ˆ
ˆ
.
ˆ






k
j
k
i
j
i
k
k
j
j
i
i
0
90
cos
1
0
cos




Because
z
z
y
y
x
x B
A
B
A
B
A
B
A 




.
since all have magnitude 1 and are
perpendicular to each other
k
j
i ˆ
,
ˆ
,
ˆ
Question: Express the dot product of and in terms of
their components and the corresponding unit vectors

45. Dislocated Shoulder: A
patient with a dislocated
shoulder is put into a traction
apparatus as shown in Fig.
P1.73. The pulls and have
equal magnitudes and must
combine to produce an
outward traction force of
5.60 N on the patient’s arm.
How large should these pulls
Biosciences problems

46. A sailor in a small sailboat encounters shifting winds. She
sails 2.00 km east, then 3.50 km southeast, and then an
additional distance in an unknown direction. Her final position
is 5.80 km directly east of the starting point (Fig. P1.72). Find
the magnitude and direction of the third leg of the journey.
Draw the vector addition diagram and show that it is in
qualitative agreement with
Your numerical solution.

47. Three horizontal ropes pull on a
large stone stuck in the ground,
producing the vector forces and
shown in Fig. Find the
magnitude and direction of a
fourth force on the stone that
will make the vector sum of the
four forces zero.

48. Question: Three horizontal ropes pull on a large stone stuck in the
ground, producing the vector forces , and and shown in Figure.
Find the magnitude and direction of a fourth force on the stone that
will make the vector sum of the four forces zero.

49. The angles of the vectors are measured from +x –axis toward +y –
axis and these are A= 300
, B=(900
+ 300
)=1200
and C=(180o
+530
)=2330
Now Using these equation we will find the component of these forces
along +x axis and +y Axis.

50. Force Angle X Component
(N)
Y Component
(N)
A=100N 30 =
100cos300
= 86.60
= 100sin30o
= 50
B=80N 120 = 80cos1200
= -40
= 80sin1200
= 69.28
C=40N 233 = 40cos233o
= -24.07
= 40sin233o
= -31.95
x
A
x
B
x
C
y
A
y
B
y
C
x
x
x
x C
B
A
R 

 y
y
y
y C
B
A
R 


= 22.53 N =
87.34 N
x
R y
R

51. 2
2
y
x R
R
R 

2
2
)
34
.
87
(
)
53
.
22
( 

N
2
.
90

)
(
tan 1
y
x
R
R



)
53
.
22
34
.
87
(
tan 1


0
64
.
75

x

X

Y

Y

0
30
0
30
0
53
R
A
B
C
0
64
.
75
Fourth Force
The direction of the fourth is 75.640
with – X axis.

52. Problem: A patient with a dislocated shoulder is put into a
traction apparatus as shown in Figure. The pulls and have
equal magnitudes and must combine to produce an outward
traction force of 5.60 N on the patient’s arm. How large should
these pulls be?
A

B


53. 0
32
0
32
)
64
cos
1
(
2
64
cos
2
2
64
cos
2
cos
2
0
2
0
2
2
0
2
2
2
2










A
A
A
AA
A
A
AB
B
A
R 
A

B

R

From this parallelogram we can write the
resultant

54. 3
.
3
89
.
10
)
44
.
1
(
2
36
.
31
)
64
cos
1
(
2
)
6
.
5
(
)
64
cos
1
(
)
6
.
5
(
2
)
64
cos
1
(
2
)
6
.
5
(
)
64
cos
1
(
2
0
2
2
0
2
2
0
2
2
0
2
2












A
A
A
A
A
A
A
R
As A=B, So the pulls are 3.3 N

55. Vector has magnitude 6 units and is in the direction of the
+X-axis. Vector has magnitude 4 units and lies in the xy-
plane, making an angle of 30° with the +X-axis. Find the
vector product
    12
30
sin
4
6
sin 
 

AB
By the right-hand rule, the direction of is along the +z axis.
(the direction of the unit vector

56. 3
2
30
cos
4
6




x
x
B
A
2
30
sin
4
0




y
y
B
A
0
0


z
z
B
A
     
     
     
k
C
C
C
C
z
y
x
ˆ
12
12
3
2
0
2
6
0
0
6
3
2
0
0
2
0
0
0












x
y
y
x
z
z
x
x
z
y
y
z
z
y
x
B
A
B
A
C
B
A
B
A
C
B
A
B
A
C






Note:

57. Find the angle between and
⃗
𝐴 .⃗
𝐵= 𝐴𝐵𝑐𝑜𝑠𝜃 ⟹𝑐𝑜𝑠 𝜃=
⃗
𝐴. ⃗
𝐵
𝐴𝐵
=
4
21
=0.1905𝑎𝑛𝑑𝜃=79°
𝐴=√2
2
+2
2
+(−1)
2
=3
𝐵=√6
2
+(− 3)
2
+(2 )
2
=7
⃗
𝐴 .⃗
𝐵=(2) (6)+(2)(− 3)+(−1)(2)=4

58. Then
Determine the value of a so that and are
perpendicular
and are perpendicular if

59. What is the angle between the vector
and

60. x
y
y
x
z
z
x
x
z
y
y
z
z
y
x
B
A
B
A
C
B
A
B
A
C
B
A
B
A
C






m
B
m
A
4
.
2
60
.
3




B
A
C





07
.
2
210
cos
4
.
2
23
.
1
70
cos
6
.
3






x
x
B
A
2
.
1
210
sin
4
.
2
38
.
3
70
sin
6
.
3






y
y
B
A
0
0


z
z
B
A
     
     
     
k
C
C
C
C
z
y
x
ˆ
524
.
5
524
.
5
07
.
2
38
.
3
2
.
1
23
.
1
0
0
23
.
1
07
.
2
0
0
2
.
1
0
0
38
.
3
















B
A
C 




62. Determine a unit vector perpendicular to the plane and
⃗
𝐶=𝑐1
^
𝑖+𝑐2
^
𝑗 +𝑐3
^
𝑘
or
or
⃗
𝐶
𝐶
=
𝑐3 (1
2
^
𝑖−
1
3
^
𝑗 + ^
𝑘)
√𝑐3
2
[(1
2 )
2
+(−
1
3 )
2
+(1)2
]
=±(3
7
^
𝑖−
2
7
^
𝑗+
6
7
Let vector be perpendicular to the plane of and Then is perpendicular to
and also to . Hence
Solving (i) and (ii) we get

63. Determine a unit vector perpendicular to the plane
and
Let vector be perpendicular to the plane of and
Then is perpendicular to and also to .
Hence or and
or
Solving (i) and (ii) we get
Then a unit vector in the direction of is

64. Determine a unit vector perpendicular
to the plane