AP Physics Chapter 3
Vector
1
AP Physics




2
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Lecture
Q&A
Vector and Scalar

Vector:
–
–
–

Scalar:
–
–
–
3
Magnitude: How large, how fast, …
Direction: In what direction (moving or pointing)
Representation depends on frame of reference
Magnitude only
No direction
Representation does not depend on frame of
reference
Examples of vector and scalar

Vectors:
–

Scalars:
–
4
Position, displacement, velocity, acceleration, force,
momentum, …
Mass, temperature, distance, speed, energy,
charge, …
Vector symbol

Vector: bold or an arrow on top
– Typed: v and V or v and V
– Handwritten: v and V

Scalar: regular
–

5
v or V
v stands for the magnitude of vector v.
Adding Vectors

Graphical
–
–

6
Head-to-Tail (Triangular)
Parallelogram
Analytical (by components)
Graphical representation of vector: Arrow
An arrow is used to graphically represent a vector.

–
The length of the arrow represents the magnitude of the vector.
b
head
a
tail
–
–
7
The direction of the arrow represents the direction of the vector.
When comparing the magnitudes of vectors, we ignore
directions.
 Vector a is smaller than vector b because a is shorter than b.
Equivalent Vectors

Two vectors are identical and equivalent if they both
have the same magnitude and are in the same
direction.
–
They do not have to start from the same point. (Their tails
don’t have to be at the same point.)
A
B
–
8
C
A, B and C are all equivalent vectors.
Negative of Vector


Vector -A has the same magnitude as vector A but
points in the opposite direction.
If vector A and B have the same magnitude but point in
opposite directions, then A = -B, and B = -A
A
-A
-A
9
Adding Vectors: Head-to-Tail

Head-to-Tail method:
Example: A + B
–
–
–
A
B
Draw vector A
Draw vector B starting
from the head of A
The vector drawn from the
tail of A to the head of B is
the sum of A + B.
B
A
10
Make sure arrows are
parallel and of same length.
A+B=B+A
B
A
A+B
A
How about
B + A?
B
B
A
11
What can we
conclude?
A+B+C
B
A
C
C
B
A
Resultant vector:
12
from tail of first to head of last.
A-B=A+(-B)
A
B
A
–
-B
–
–
13
Draw vector A
Draw vector -B from head
of A.
The vector drawn from the
tail of A to head of –B is
then A – B.
What Are the Relationships?
b
a
bc a
14
c
c
b
a
abc  0
Magnitude of sum A + B = C
A and B in same
direction
max. c

15
A and B in opposite
direction
min. c
|A – B|  C  A + B
A and B at some
angle
Adding Vectors: Parallelogram

A+B
–
–
–
Draw the two vectors from
the same point
Construct a parallelogram
with these two vectors as
two adjacent sides
The sum is the diagonal
vector starting from the
same tail point.
B
A
Advantage:
No need to measure length.
16
A
B
N
35o
W
Example

17
E
S
Vector a has a magnitude of 5.0 units and is
directed east. Vector b is directed 35o west of
north and has a magnitude of 4.0 units.
Construct vector diagrams for calculating a + b
and b – a. Estimate the magnitudes and
directions of a + b and b – a from your
diagram.
Solution
N
-a
b
35o
a
Using ruler and protractor, we find:
a+b: 4.3 unit, 50o North of East
b-a: 8.0 unit, 66o West of North
18
E
66o
50o

W
S
Vector Components

Drop perpendicular lines from
the head of vector a to the
coordinate axes, the
components of vector a can
be found:

 ax  a cos 

a y  a sin 


19

 is the angle between the
vector and the +x axis.

ax and ay are scalars.
y
a
ay

ax
x
Finding components of a vector


20
Resolving the vector
Decomposing the vector
y
ay
Vector magnitude
and direction


ax
x
The magnitude and direction of a vector can be
found if the components (ax and ay) are given:
 magnitude: a  a 2  a 2
2
2
2

a

a

a
x
y

x
y
z


ay
a
1 y
 tan  
 direction:   tan
a

ax
x

 is the angle from the +x axis to the vector.
21
a
(for 3-D)
Example

22
A ship sets out to sail to a point 120 km due
north. Before the voyage, an unexpected
storm blows the ship to a point 100 km due
east of its starting point. How far, and in what
direction, must it now sail to reach its original
destination?

Solution
A ship sets out to sail to a point 120 km due
north. Before the voyage, an unexpected storm
blows the ship to a point 100 km due east of its
starting point. How far, and in what direction,
must it now sail to reach its original destination?
c  a 2  b2  1202  1002  156km
b
1 100
 tan
 39.8o
  tan
a
120
N
1

23
It must sail 156 km at 39.8o
West of North to reach its
original destination.

c

b = 100
E
Practice: What are the magnitude and
direction of vector a  4iˆ  5 ˆj ?
y
ax  4, a y  5,
5
a  ?,   ?
a
ax  a y  4  5  6.4
2
2
ay
2
2
5
 tan
 51o
  tan
ax
4
1
24
1

4
x
Practice: What are the components of a vector
that has a magnitude of 12 units and makes an
angle of 126o with the positive x direction?
y
a  12,   126o
ax  ?, a y  ?
126o
x
ax  a cos   12cos126  7.1
o
a y  a sin   12sin126o  9.7
25
Unit vectors

26
Unit vector: magnitude of exactly 1 and points
in a particular direction.
– x direction: i or iˆ
–
y direction: j or ĵ
–
z direction: k or k̂
Vector components and expression

Any vector can be written in its components
and the unit vectors:
 a  ax iˆ  a y ˆj  az kˆ   ax , a y , az 

b  bx iˆ  by ˆj  bz kˆ
27
Terminology


28
axi is the vector component of a.
ax is the (scalar) component of a.
Example

Express the
following vector in
component and unit
vector form.
y
ay
a=12.0 units
=30o
ax
a x  a cos  12.0cos30o  10.4
a y  a sin   12.0sin 30o  6.00
 a  a x iˆ  a y ˆj  10.4iˆ  6.00 ˆj
29
x
y
ry
Adding Vectors
by Components

ay
b
by
a
r
ax
rx
When adding vectors by components, we add
components in a direction separately from other
components.
r  a  b
x
x
x
r  ab


  ry

rz


2-D
 a y  by
 az  bz
3-D
Component form:
r  a  b   ax  bx  iˆ  ay  by ˆj   az  bz  kˆ
30
bx

s  a  b   a  b  iˆ   a
x
x
y

 b  ˆj   a
y
z
 bz  kˆ
x
Example

The minute hand of a wall clock measures 10
cm from axis to tip. What is the displacement
vector of its tip (a) from a quarter after the hour
to half past, (b) in the next half hour, and (c) in
the next hour?
•
•
31
•
Solution
or magnitude and direction form.
a)
rA B  rB  rA  (0, 10)  (10,0)
 (0  10, 10  0)  ( 10, 10)cm
Try b) and c)
b)
rB C  rC  rB  (0,10)  (0, 10)
 (0,20)cm
c)
32 r  rC  rC  0
y
C (0,10)
rBC
x
O
A (10, 0)
rAB
B (0,-10)
Practice
a)

Two vectors are given by a = 4i – 3j + k and
b = -i + j + 4k. Find:
a)
a+b
b)
a–b
c)
a vector c such that a – b + c = 0
 
a  b  4iˆ  3 ˆj  kˆ  iˆ  ˆj  4kˆ

 (4  1)iˆ  ( 3  1) ˆj  (1  4)kˆ  3iˆ  2 ˆj  5kˆ
b)

 
a  b  4iˆ  3 ˆj  kˆ  iˆ  ˆj  4kˆ

 (4  1)iˆ  (3  1) ˆj  (1  4)kˆ  5iˆ  4 ˆj  3kˆ
c)
33
a  b  c  0  c  b  a  ( a  b )  5iˆ  4 ˆj  3kˆ
Practice: 54-19
The two vectors a and b in Fig. 3-29 have equal magnitudes of
10.0 m and the angels are 1 = 30o and 2 = 105o. Find the (a) x
and (b) y components of their vector sum r, (c) the magnitude of r,
and (d) the angle r makes with the positive direction of the x axis.
a )  b)
 a x  a cos  10cos30  8.66

o
a

a
sin


10sin
30
5
 y
o

bx  10 cos135  7.07

o
b

10
sin
135
 7.07

 y
o
34
 rx  ax  bx  8.66  7.07  1.59

 ry  a y  by  5  7.07  12.07
a x  a cos

a y  a sin 
y
b
2
r
1  a
b=105o+30o=135o
a
x
54-19 (Continued)
c)
r
rx 2  ry 2  1.592  12.072  12.2
d)
 r  tan
35
1
ry
12.07
 tan
 82.5o
rx
1.59
1
Vector Multiplication
a  2iˆ  4 ˆj  kˆ
 4a   4  2  iˆ   4  4  ˆj   4   1  kˆ  8iˆ  16 ˆj  4kˆ

More:




36
Scalar (aka dot or inner) product: a  b
Vector (aka cross) product: a  b
We cannot write ab if a and b are vectors.
But we still can write 2a since 2 is a scalar.
: angle between vector and +x axis
Scalar product:
ab
: angle between two vectors
a  b  ab cos 




a

 (phi) is the angle between vector a and b.
 is always between 0o and 180o. (0o    180o)
The scalar product is a scalar  It has no
direction.
What if the two vectors are perpendicular to each
other?
  90o  a  b  0
37
b
Physical meaning of ab
ba is the projection of b onto a.
ba  b cos 
b
 a  b  ab cos   aba

ba
a
Also
ab is the project of a onto b.
ab  a cos 
 a  b  ab cos   abb
38
ab

b
a
Properties of scalar product
1.
2.
3.
4.
5.
6.
ab = ba
ii = jj = kk = 1
ij = ji = jk = kj = ki = ik = 0
aa = a2
ab = 0 if a  b
ab = axbx+ayby+azbz


a  b  axiˆ  a y ˆj  az kˆ  bxiˆ  by ˆj  bz kˆ
39

Angle between two vectors
When we know magnitudes and :
a  b  ab cos 
When we know the components:
a  b  axbx  ayby  az bz
Put together:
ab cos   ax bx  a y by  az bz
 cos  
40
ax bx  a y by  az bz
ab
   cos
1
ax bx  a y by  az bz
ab
Example:
a. Determine the components and magnitude of r = a – b
+ c if a = 5.0i + 4.0j – 6.0k, b = -2.0i + 2.0j + 3.0k, and c
= 4.0i + 3.0 j + 2.0k.
b. Calculate the angle between r and the positive z axis.
a)
r  ab c
 (5.0iˆ  4.0 ˆj  6.0kˆ)  ( 2.0iˆ  2.0 ˆj  3.0kˆ)  (4.0iˆ  3.0 ˆj  2.0kˆ)
 (5.0  2.0  4.0)iˆ  ( 4.0  2.0  3.0) ˆj  ( 6.0  3.0  2.0) kˆ
z
 11.0iˆ  5.0 ˆj  7.0kˆ
C•
O
b)
OA  11.0  5.0  12.1
2
AB  7.0
OAB  90o
AB 7.0

 0.579
OA 12.1
 AOB  tan 1 0.579  30.0o
tan AOB 
41
 COB  AOB  90o  120.0o
5
y
2
11
A
x
-7
B
a  b  ab cos 
Another approach
We are looking for the angle between r and any vector in the z direction.
Let’s choose the unit vector in the z direction, k
r  11.0iˆ  5.0 ˆj  7.0kˆ , and kˆ  0iˆ  0 ˆj  1kˆ
r  kˆ  rx kx  ry k y  rz kz  11 0  5  0   7  1  7
r  112  52   7   13.9
2
k 1
r  kˆ  rk cos 
42
r  kˆ
7
 cos  

 0.5    cos 1  0.5   120o
rk 13.9  1
Practice: 55-43
For the vectors in Fig. 3-35, with a = 4, b =3, and c =5,
calculate (a) a  b , (b) a  c , and (c) b  c .
B
a)
a b  0  a  b
b)
b
3
  tan 1  tan 1  36.87o
a
4
   180o  36.87o  143.17o
c


A
5
4
a
a  c  ac cos   4  5cos143.17o  16
b)
A  tan 1
43
a
4
 tan 1  53.13o  B  180o  53.13o  127o
b
3
b  c  bc cos B  3  5cos127o  9
3b
Approach 2
y
a  4iˆ  0 ˆj
b  0iˆ  3 j
c  4iˆ  3 ˆj
b
c
-4
a
a  b  4   0   0   3  0
a  c  4   4   0   3  16
b  c  0   4   3   3  9
44
b
c
-3
x
Approach 3
a)
a b  0  a  b
b)
5
4
a
abc  0
 c  a  b




a  c  a  a  b  a  a  a  b  a 2  16
c)
b  c  b  a  b   a  b  b  b  b 2  9
45
c
3b
Approach 4
d = -c
5
3b
4
a
b)
da
a  c  a   c   a  d  ad a  aa  16
c)
 
b  c   b  c  e  c  ece    b  b  9
46
e = -b
c
ce
A B
Vector Product: c = a  b
A


Magnitude of c is: c  ab sin 
c is a vector, and it has a direction given by
the right-hand-rule (RHR):
–
–
–
–
47
Place the vectors a and b so that their tails are at the
same point.
Extend your right arm and fingers in the direction of a.
Rotate your hands along your arm so that you can flap
your fingers toward b through the smaller angle
between a and b. Then
Your outstretched thumb points in the direction of c.
Properties of cross product
b  a = - (a  b)
a  b is  a, and a  b is  b
aa=0
iˆ  ˆj  kˆ

 ˆj  kˆ  iˆ
ˆ ˆ ˆ
k  i  j
48

 
a  b  axiˆ  a y ˆj  az kˆ  bxiˆ  by ˆj  bz kˆ
 
a  b  a y bz  az by iˆ  az bx  axbz  ˆj  axby  a y bx kˆ

Practice:
 
a  b  a y bz  az by iˆ  az bx  axbz  ˆj  axby  a y bx kˆ
Three vectors are given by a = 3.0i + 3.0j – 2.0k, b = -1.0i
–4.0j + 2.0k, and c = 2.0i + 2.0j + 1.0k. Find (a) a • (b 
c), (b) a • (b + c), and (c) a  (b + c).
a)

 
b  c  1.0iˆ  4.0 ˆj  2.0kˆ  2.0iˆ  2.0 ˆj  1.0kˆ
  (4.0)(1.0)  (2.0)(2.0)  iˆ

  (2.0)(2.0)  ( 1.0)(1.0)  ˆj
  (1.0)(2.0)  ( 4.0)(2.0)  kˆ  8.0iˆ  5.0 ˆj  6.0kˆ
a  (b  c )
 (3.0iˆ  3.0 ˆj  2.0kˆ)  (8.0iˆ  5.0 ˆj  6.0kˆ)
 (3.0)(8.0)  (3.0)(5.0)  (2.0)(6.0)  21
49
Pg52-53P (2)
b)

 
b  c  1.0iˆ  4.0 ˆj  2.0kˆ  2.0iˆ  2.0 ˆj  1.0kˆ

 (1.0  2.0)iˆ  (4.0  2.0) ˆj  (2.0  1.0) kˆ
 1.0iˆ  2.0 ˆj  3.0kˆ
a  (b  c )
 (3.0iˆ  3.0 ˆj  2.0kˆ)  (1.0iˆ  2.0 ˆj  3.0kˆ)
 (3.0)(1.0)  (3.0)(2.0)  (2.0)(3.0)  9
50
Pg52-53P (3)
c)
 
b  c  1.0iˆ  2.0 ˆj  3.0kˆ
a  (b  c )  (3.0iˆ  3.0 ˆj  2.0kˆ)  (1.0iˆ  2.0 ˆj  3.0kˆ)
  (3.0)(3.0)  (2.0)(2.0)  iˆ
  (2.0)(1.0)  (3.0)(3.0)  ˆj
  (3.0)(2.0)  (3.0)(1.0)  kˆ
 5.0iˆ  11.0 ˆj  9.0kˆ
51
Law of cosine
c  a  b  2ab cos C
2
2
2
c
b
C
a
52
Law of Sine
a
b
c


sin A sin B sin C
or
C
a
b
B
A
c
53
sin A sin B sin C


a
b
c