Chapter 3 - Vectors
I.
Definition
II. Arithmetic operations involving vectors
A) Addition and subtraction
- Graphical method
- Analytical method  Vector components
B) Multiplication
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
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
II.
Arithmetic operations involving vectors
Vector addition:
  
s  a b
- Geometrical method

a

b
  
s  a b
Rules:
   
a b b a
(commutative law)
     
(a  b )  c  a  (b  c )
(3.1)
(associative law)
(3.2)
Vector subtraction:
   

d  a  b  a  (b )
(3.3)
Vector component: projection of the vector on an axis.
a x  a cos 
(3.4)
a y  a sin 
a  a x2  a 2y

Scalar components of a
Vector magnitude
(3.5)
tan  
ay
ax
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  a x iˆ  a y ˆj
(3.6)
Vector component
Vector addition:
- Analytical method: adding vectors by components.
  
r  a  b  (a x  bx )iˆ  (a y  by ) ˆj
(3.7)
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
(3.8)
   '
Multiplying vectors:
- Vector by a scalar:


f  sa
- Vector by a vector:
Scalar product = scalar quantity
(dot product)
 
a  b  ab cos   a x bx  a y by  a z bz
(3.9)
Rule:
   
a b  b  a
 
a  b  ab  cos   1 (  0 )
 
a  b  0  cos   0 (  90 )
(3.10)
     
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    
ab
Angle between two vectors:
Multiplying vectors:
- Vector by a vector
Vector product = vector
(cross product)
  
a  b  c  (a y bz  by a z )iˆ  (bz a x  a z bx ) ˆj  (a x by  bx a y )kˆ
c  ab sin 
Magnitude
 
a  b  0  sin   0 (  0 )
 
a  b  ab  sin   1 (  90 )
Vector product
Direction  right hand rule
Rule:
 
 
b  a  ( a  b )
(3.12)

 
c perpendicular 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 smallest angle between them.
Right-handed coordinate system
z
k
i
j
y
x
Left-handed coordinate system
z
k
j
y
i
x
     
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
ˆ the result is a vector in the positive direction of the y axis, with a
P1: If B is added to C = 3iˆ + 4j,
magnitude equal to that of C. What is the magnitude of B?
Method 2
Method 1
Isosceles triangle
  

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
θ
C
D
tan   (3 / 4)    36.9
 
  B / 2
sin   
 B  2 D sin    3.2
D
2
2
B
P2: 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
N
D
45º
d4 d3
d1
E
d1 y  0.4 sin 45  0.28m
d 2 x  0.5m
d2 y  0
d 3 x  0.6 cos 60  0.30m
d2
d 3 y  0.6 sin 60  0.52m
(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  0.57m
 0.24 

  24.8 North of East
 0.52 
  tan 1 
(c) Return vector  negative of net displacement,
D=0.57m, directed 25º South of West
P2

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 perpendicular to d 2 and in plane of d1 , d 2 ?
   
(a ) r  d1  d 2  d 3  (4iˆ  5 ˆj  6kˆ)  (iˆ  2 ˆj  3kˆ)  (4iˆ  3 ˆj  2kˆ)  9iˆ  6 ˆj  7kˆ

 7 

(b) r  kˆ  r 1  cos   7    cos 1 
  123
 12.88 
d1perp
r  9 2  6 2  7 2  12.88m
 
 
d1  d 2
(c) d1  d 2  4  10  18  12  d1d 2 cos   cos  
d1d 2
 
d d
 12
d1//  d1 cos   d1 1 2 
 3.2m
d1d 2
3.74
d1
θ d1//
d2
d 2  12  2 2  32  3.74m
(d ) d1  d12//  d12perp  d1 perp  8.77 2  3.2 2  8.16m
d1  4 2  52  6 2  8.77 m
P3

If d1  3iˆ  2 ˆj  4kˆ

d 2  5iˆ  2 ˆj  kˆ
Tip:
 


(d1  d 2 )  (d1  4d 2 ) ?
Think before calculate !!!
 
 

(d1  d 2 )  a  contained in (d1 , d 2 ) plane


 

 
(d1  4d 2 )  4(d1  d 2 )  4b  perpendicular to (d1 , d 2 ) plane


 
a perpendicular to b  cos 90  0  4a  b  0
P4:
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)?
y
A

(a) Angle between  y and A  90  50  140
130º
x
B
 


(b) Angle  y , ( A  B)  C  angle  ˆj , kˆ because C perpendicular
 
plane ( A, B)  ( xy )  90
 

(c) Direction A  ( B  3kˆ)  D
 
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   
  97.61     99
1
D



P5: 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?
y
Vertical displacement: 2 R  0.9m
Horizontal displacement:
1
(2R)  1.41m
2
d

r  (1.41m)iˆ  (0.9m) ˆj

r  1.412  0.9 2  1.68m
x
 2R 

tan   
    32.5
 R 
P6: 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
b
N
125º
b-a
a
W
a+b
E

a  5iˆ

b  4 sin 35 iˆ  4 cos 35 ˆj  2.29iˆ  3.28 ˆj
  

 
(b) b  a  b  ( a )  7.29iˆ  3.28 ˆj
(a) a  b  2.71iˆ  3.28 ˆj
 
 
b  a  7.29 2  3.282  8m
a  b  2.712  3.282  4.25m
 3.28 

tan   
    50.43
 2.71 
 3.28 

tan    
    24.2
7
.
29


or 180  (24.2 )  155.8
S
180  155.8  24.2 North of West