Chapter 3 - Vectors
I.
Definition
II.
Arithmetic operations involving vectors
A) Addition and subtraction
- Graphical method
- Analytical method  Vector components
B) Multiplication
I. Definition
Vector quantity: quantity with a magnitude and a direction. It can be
represented by a vector.
Examples: displacement, velocity, acceleration.
Same displacement
Displacement  does not describe the object’s path.
Scalar quantity: quantity with magnitude, no direction.
Examples: temperature, pressure
Vector Notation
• When handwritten, use an arrow:

A
• When printed, will be in bold print: A
• When dealing with just the magnitude of a vector
in print: |A|
• The magnitude of the vector has physical units
• The magnitude of a vector is always a positive
number
Coordinate Systems
• Used to describe the position of a point in
space
• Coordinate system consists of
– a fixed reference point called the origin
– specific axes with scales and labels
– instructions on how to label a point relative to
the origin and the axes
Cartesian Coordinate System
• Also called
rectangular
coordinate system
• x- and y- axes
intersect at the origin
• Points are labeled
(x,y)
Polar Coordinate System
– Origin and reference
line are noted
– Point is distance r
from the origin in the
direction of angle ,
ccw from reference
line
– Points are labeled (r,)
(r,θ)
Polar to Cartesian Coordinates
•
• Based on
forming a right
triangle from r
and 
• x = r cos 
• y = r sin 
Cartesian to Polar Coordinates
• r is the hypotenuse and 
an angle
y
tan  
x
r  x2  y2
 must be ccw from
positive x axis for these
equations to be valid
•
Review of angle reference system
90º
0º<θ1<90º
90º<θ2<180º
θ2
θ1
180º
θ3
180º<θ3<270º
0º
Origin of angle reference system
θ4
270º<θ4<360º
270º
Angle origin
θ4=300º=-60º
Example
• The Cartesian coordinates of
a point in the xy plane are
(x,y) = (-3.50, -2.50) m, as
shown in the figure. Find the
polar coordinates of this
point.
• Solution:
Example
• The Cartesian coordinates of
a point in the xy plane are
(x,y) = (-3.50, -2.50) m, as
shown in the figure. Find the
polar coordinates of this
point.
• Solution:
r  x 2  y 2  ( 3.50 m)2  ( 2.50 m) 2  4.30 m
y 2.50 m
tan   
 0.714
x 3.50 m
  216
Two points in a plane have polar coordinates (2.50 m,
30.0°) and (3.80 m, 120.0°). Determine (a) the
Cartesian coordinates of these points and (b) the
distance between them.
Two points in a plane have polar coordinates (2.50 m,
30.0°) and (3.80 m, 120.0°). Determine (a) the
Cartesian coordinates of these points and (b) the
distance between them.
(a)
x  rcos
y  rsin
x1   2.50 m  cos30.0 y1   2.50 m  sin 30.0
 x1,y1   2.17,1.25 m
x2   3.80 m  cos120
y2   3.80 m  sin120
 x2 ,y2    1.90, 3.29 m
(b)
d  ( x)2  ( y)2  16.6  4.16  4.55 m
A skater glides along a circular path of radius 5.00 m. If he
coasts around one half of the circle, find (a) the magnitude
of the displacement vector and (b) how far the person
skated. (c) What is the magnitude of the displacement if he
skates all the way around the circle?
A skater glides along a circular path of radius 5.00 m. If he
coasts around one half of the circle, find (a) the magnitude
of the displacement vector and (b) how far the person
skated. (c) What is the magnitude of the displacement if he
skates all the way around the circle?
C
B
5.00 m
d
(a)
A
(b)
(c)
d   10.0ˆ
i  10.0 m
1
s  2 r  5  15.7 m
2
d  0
II.
Arithmetic operations involving vectors
  
s  a b

a
Vector addition:
- Geometrical method

a

b
Rules:
   
a b b a
(commutativ e law)
     
(a  b )  c  a  (b  c )
(associativ e law)

b
  
s  a b
Vector subtraction:
   

d  a  b  a  (b )
Vector component: projection of the vector on an axis.
a x  a cos 
a y  a sin 
a  a x2  a 2y
tan 
ay
ax

Scalar components of a
Vector magnitude
Vector direction
Unit vector:
Vector with magnitude 1.
No dimensions, no units.
iˆ, ˆj, kˆ  unit vectors in positive direction of x, y, z axes

a  axiˆ  a y ˆj
Vector component
Vector addition:
- Analytical method: adding vectors by components.
  
r  a  b  (ax  bx )iˆ  (a y  by ) ˆj
A man pushing a mop across a floor causes it to undergo
two displacements. The first has a magnitude of 150 cm and
makes an angle of 120° with the positive x axis. The
resultant displacement has a magnitude of 140 cm and is
directed at an angle of 35.0° to the positive x axis. Find the
magnitude and direction of the second displacement.
A man pushing a mop across a floor causes it to undergo two
displacements. The first has a magnitude of 150 cm and makes an angle
of 120° with the positive x axis. The resultant displacement has a
magnitude of 140 cm and is directed at an angle of 35.0° to the positive x
axis. Find the magnitude and direction of the second displacement.
B  R A
A x  150 cos120  75.0 cm
A y  150sin120  130 cm
R x  140cos35.0  115 cm
R y  140sin 35.0  80.3 cm


B  115   75  ˆ
i  80.3  130 ˆ
j 190ˆ
i 49.7ˆ
j cm
B  1902  49.72  196 cm
 49.7 
 14.7 .

 190 
  tan1  
θ = 3600 – 14.70 = 345.30
As it passes over Grand Bahama Island, the eye of a hurricane
is moving in a direction 60.0 north of west with a speed of
41.0 km/h. Three hours later, the course of the hurricane
suddenly shifts due north, and its speed slows to 25.0 km/h.
How far from Grand Bahama is the eye 4.50 h after it passes
over the island?
As it passes over Grand Bahama Island, the eye of a hurricane is
moving in a direction 60.0 north of west with a speed of 41.0 km/h.
Three hours later, the course of the hurricane suddenly shifts due north,
and its speed slows to 25.0 km/h. How far from Grand Bahama is the
eye 4.50 h after it passes over the island?
d1  41.0km / h  3.00h  123.0km
d 2  25.0km / h  1.50h  37.5km
at
at
600
N of W
N
iˆ  East
ˆj  North
dtotal= 41.0

 
 
km
km
km 




ˆ
cos60.0  3.00 h  ˆ
i   41.0
sin 60.0  3.00 h ˆ
j  25.0
1.50 h ˆ
j 61.5 km  i






h
h
h 
144 km ˆ
j
2
2
Magnitude =
 61.5 km 
 144 km   157 km
Vectors & Physics:
-The relationships among vectors do not depend on the location of the origin of
the coordinate system or on the orientation of the axes.
- The laws of physics are independent of the choice of coordinate system.
a  a x2  a 2y  a'2x  a'2y
   '
Multiplying vectors:
- Vector by a scalar:


f  sa
- Vector by a vector:
Scalar product = scalar quantity
(dot product)
 
a  b  ab cos   axbx  a yby  azbz
Rule:
   
a b  b a
 
a  b  ab if
  0
(cos   1)
 
a  b  0 if
  90
(cos   0)
     
i  i  j  j  k  k  11 cos 0  1
           
i  j  j  i  i  k  k  i  j  k  k  j  11 cos 90  0
 
a b
cos    
Angle between two vectors:
ab
Multiplying vectors:
- Vector by a vector
Vector product = vector
(cross product)
  
a  b  c  (a y bz  by a z )iˆ  (a z bx  bz a x ) ˆj  (a xby  bx a y )kˆ
c  ab sin 
Magnitude
 
a  b  0 if
 
a  b  ab if
  0
  90
(sin   0)
(sin   1)
Vector product
Direction  right hand rule
Rule:
 
 
b  a  ( a  b )

 
c perpendicu lar to plane containing a, b
1) Place a and b tail to tail without altering their orientations.
2) c will be along a line perpendicular to the plane that contains a and b
where they meet.
3) Sweep a into b through the small angle between them.
Right-handed coordinate system
z
k
i
j
y
x
Left-handed coordinate system
z
k
j
y
i
x
iˆ ˆj kˆ
 
a  b  a x a y a z  a y bz iˆ  a xby kˆ  a z bx ˆj  a y bx kˆ  a z by iˆ  a xbz ˆj
bx by bz
 (a y bz  a z by )iˆ  (a z bx  axbz ) ˆj  (a xby  a y bx )kˆ
     
i  i  j  j  k  k  11 sin 0  0
      
i i  j  j  k k  0

 
 
i  j  ( j  i )  k
  
 
j  k  ( k  j )  i
 
 

k  i   (i  k )  j
42: If B is added to C = 3i + 4j, the result is a vector in the
positive direction of the y axis, with a magnitude equal to
that of C. What is the magnitude of B?
If B is added to C = 3i + 4j, the result is a vector in the
positive direction of the y axis, with a magnitude equal to
that of C. What is the magnitude of B?
42:
  

B  C  B  (3iˆ  4 ˆj )  D  Dˆj
C  D  32  4 2  5


B  (3iˆ  4 ˆj )  5 ˆj  B  3iˆ  ˆj  B  9  1  3.2
50: A fire ant goes through three displacements along level
ground: d1 for 0.4m SW, d2=0.5m E, d3=0.6m at 60º North of
East. Let the positive x direction be East and the positive y
direction be North. (a) What are the x and y components of d1,
d2 and d3? (b) What are the x and the y components, the
magnitude and the direction of the ant’s net displacement?
(c) If the ant is to return directly to the starting point, how far
and in what direction should it move?
N
D
45º
d4 d3
d1
d2
E
50: A fire ant goes through three displacements along level
ground: d1 for 0.4m SW, d2 0.5m E, d3=0.6m at 60º North of
East. Let the positive x direction be East and the positive y
direction be North. (a) What are the x and y components of d1,
d2 and d3? (b) What are the x and the y components, the
magnitude and the direction of the ant’s net displacement?
(c) If the ant is to return directly to the starting point, how far and
in what direction should it move?
(a)
d1x  0.4 cos 45  0.28m
d1 y  0.4 sin 45  0.28m
d 2 x  0.5m
N
D
45º
d4 d3
d1
d2
E
d2 y  0
d 3 x  0.6 cos 60  0.30m
d 3 y  0.6 sin 60  0.52m
50: A fire ant goes through three displacements along level
ground: d1 for 0.4m SW, d2 0.5m E, d3=0.6m at 60º North of
East. Let the positive x direction be East and the positive y
direction be North. (a) What are the x and y components of d1,
d2 and d3? (b) What are the x and the y components, the
magnitude and the direction of the ant’s net displacement?
(c) If the ant is to return directly to the starting point, how far
and in what direction should it move?
(b)

 
d 4  d1  d 2  (0.28iˆ  0.28 ˆj )  0.5iˆ  (0.22iˆ  0.28 ˆj )m
  
D  d 4  d 3  (0.22iˆ  0.28 ˆj )  (0.3iˆ  0.52 ˆj )  (0.52iˆ  0.24 ˆj )m
D  0.52 2  0.24 2  0.57 m
N
D
45º
E
 0.24 

  24.8 North of East
 0.52 
  tan 1 
d4 d3
d1
d2
(c)
Return vector  negative of net
displacement, D=0.57m, directed 25º
South of West
53:

d1  4iˆ  5 ˆj  6kˆ

d 2  iˆ  2 ˆj  3kˆ

d  4iˆ  3 ˆj  2kˆ
3
Find:
   
(a) r  d1  d 2  d 3 ?

(b) Angle between r and  z ?


(c) Component of d1 along d 2 ?


 
(d ) Component of d1 perpendicu lar to d 2 and in plane of d1 , d 2 ?
53:

d1  4iˆ  5 ˆj  6kˆ

d 2  iˆ  2 ˆj  3kˆ

d  4iˆ  3 ˆj  2kˆ
3
   
(a) r  d1  d 2  d 3 ?

(b) Angle between r and  z ?


(c) Component of d1 along d 2 ?


 
(d ) Component of d1 perpendicu lar to d 2 and in plane of d1 , d 2 ?
   
(a) r  d1  d 2  d3  (4iˆ  5 ˆj  6kˆ)  (iˆ  2 ˆj  3kˆ)  (4iˆ  3 ˆj  2kˆ)  9iˆ  6 ˆj  7kˆ
d1perp

 7 

(b) r  kˆ  r 1 cos   7    cos 1 
  123
 12.88 
r  92  62  7 2  12.88m
d1
θ d1//
d
 2
 
d d
(c) d1  d 2  4  10  18  12  d1d 2 cos   cos   1 2
d1d 2
 
d1  d 2  12
d1//  d1 cos   d1

 3.2m
d1d 2
3.74
d 2  12  2 2  32  3.74m
(d ) d1  d12//  d12perp  d1 perp  8.77 2  3.2 2  8.16m
d1  42  52  62  8.77m
30:

If d1  3iˆ  2 ˆj  4kˆ

d 2  5iˆ  2 ˆj  kˆ
 


(d1  d 2 )  (d1  4d 2 ) ?
30:

If d1  3iˆ  2 ˆj  4kˆ

d 2  5iˆ  2 ˆj  kˆ
 


(d1  d 2 )  (d1  4d 2 ) ?
 
 

(d1  d 2 )  a  contained in (d1 , d 2 ) plane


 

 
(d1  4d 2 )  4(d1  d 2 )  4b  perpendicu lar to (d1 , d 2 ) plane


 

a perpendicu lar to b  cos 90  0  4a  b  0
54:
Vectors A and B lie in an xy plane. A has a
ˆ
magnitude 8.00 and angle 130º; B has components
Bx= -7.72, By= -9.20. What are
y
the angles between the
A
130º
negative direction of the y axis
and (a) the direction of A,
x
B
(b) the direction of A x B,
(c) the direction of A x (B+3k)?
54:
Vectors A and B lie in an xy plane. A has a magnitude 8.00 and angle 130º; B has
components Bx= -7.72, By= -9.20. What are the angles between the negative direction of
ˆ
the y axis and (a) the direction of A, (b) the direction of AxB, (c) the direction of Ax(B+3k)?

(a ) Angle between  y and A  90  50  140
y
 


ˆ
ˆ
(b) Angle  y , ( A  B)  C  angle  j , k because C perpendicu lar
 
plane ( A, B)  ( xy)  90
A
130º
x
B
 

ˆ
(c) Direction A  ( B  3k )  D

A  (8 cos 1300 )iˆ  (8 sin 1300 ) ˆj  5.14iˆ  6.13 ˆj
 
E  B  3kˆ  7.72iˆ  9.2 ˆj  3kˆ
iˆ
  
D  A  E   5.14
ˆj
6.13
kˆ
0  18.39iˆ  15.42 ˆj  94.61kˆ
 7.72  9.20 3
D  18.39 2  15.42 2  94.612  97.61

 ˆj  D   ˆj  (18.39iˆ  15.42 ˆj  94.61kˆ)  15.42

  ˆj  D    15.42 


cos   
    99

 1  D   97.61 
39: A wheel with a radius of 45 cm rolls without sleeping along
a horizontal floor. At time t1 the dot P painted on the rim of the
wheel is at the point of contact between the wheel and the
floor. At a later time t2, the wheel has rolled through one-half of
a revolution. What are (a) the magnitude and (b) the angle
(relative to the floor) of the displacement P during this interval?
39: A wheel with a radius of 45 cm rolls without sleeping along a horizontal
floor. At time t1 the dot P painted on the rim of the wheel is at the point of
contact between the wheel and the floor. At a later ytime t2, the wheel has
rolled through one-half of a revolution.
What are (a) the magnitude and
(b) the angle (relative to the floor) of
d
the displacement P during this interval?
Vertical displacement:
Horizontal displacement:
2 R  0.9m
1
(2R)  1.41m
2

d  (1.41m)iˆ  (0.9m) ˆj

d  1.412  0.92  1.68m
 2R 
tan   
    32.5
 R 
x
Vector a has a magnitude of 5.0 m and is directed East.
Vector b has a magnitude of 4.0 m and is directed 35º West
of North. What are (a) the magnitude and direction of (a+b)?
(b) What are the magnitude and direction of (b-a)? (c) Draw a
vector diagram for each combination.
Vector a has a magnitude of 5.0 m and is directed East.
Vector b has a magnitude of 4.0 m and is directed 35º West
of North. What are (a) the magnitude and direction of (a+b)?
(b) What are the magnitude and direction of (b-a)? (c) Draw a
vector diagram for each combination.

a  5iˆ

b  4 sin 35 iˆ  4 cos 35 ˆj  2.29iˆ  3.28 ˆj
 
(a ) a  b  2.71iˆ  3.28 ˆj
 
a  b  2.712  3.282  4.25m
 3.28 

tan   
    50.43
 2.71 
-a
b
  

(b) b  a  b  (a )  7.29iˆ  3.28 ˆj
 
b  a  7.29 2  3.282  8m
N
125º
b-a
a
W
a+b
E
 3.28 

tan    
    24.2
 7.29 
or 180  (24.2 )  155.8
S
180  155.8  24.2 North of West