PH 221-1D Spring 2013
Vectors
Lecture 4,5
Chapter 3
(Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)
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Chapter 3
Vectors
In Physics we have parameters that can be completely described by a
number and are known as “scalars” .Temperature, and mass are such
parameters.
Other physical parameters require additional information about direction and
are known as “vectors” . Examples of vectors are displacement, velocity
and acceleration.
In this chapter we learn the basic mathematical language to describe
vectors. In particular we will learn the following:
Geometric vector addition and subtraction
Resolving a vector into its components
The notion of a unit vector
Add and subtract vectors by components
Multiplication of a vector by a scalar
The scalar (dot) product of two vectors
The vector (cross) product of two vectors
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An example of a vector is the displacement vector which
describes the change in position of an object as it moves
from point A to point B. This is represented by an arrow
that points from point A to point B. The length of the
arrow is proportional to the displacement magnitude.
The direction of the arrow indicated the displacement
direction.
The three arrows from A to B, from A' to B', and from A''
to B'', have the same magnitude and direction. A vector
can be shifted without changing its value if its length and
direction are not changed.
In books vectors are written in two ways:
Method 1:

a
Method 2:
a
(using an arrow above)
(using bold face print)
The magnitude of the vector is indicated by italic print: a
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Geometric vector Addition
  
s  a b
Tail‐to‐head method
Parallelogram method

Sketch vector a using an appropriate scale

Sketch vector b using the same scale


Place the tail of b at the tip of a


The vector s starts from the tail of a

and terminates at the tip of b
   
Vector addition is commutative a  b  b  a
When there are more than two vectors, we can
group them in any order as we add them
     
( a  b )  c  a  (b  c ) (associative law)


Negative  b of a given vector b


b has the same magnitude as b
but opposite direction
4
Geometric vector Subtraction
  
d  a b
   

We write: d  a  b  a  b


From vector b we find  b


Then we add b to vector a
 
 
We thus reduce vector subtraction to
vector addition which we know how to do
Note: We can add and subtract vectors using the method of components.
For many applications this is a more convenient method
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A component of a vector along an axis is the projection
of the vector on this axis. For example a x is the

projection of a along the x-axis. The component a x
is defined by drawing straight lines from the tail

and tip of the vector a which are perpendicular to
the x-axis.
C
A
From triangle ABC the x- and y-components

of vector a are given by the equations:
B
a x  a cos 
,
a y  a sin  Component Notation
If we know a x and a y we can determine a and  .
From triangle ABC we have:
a  a a
2
x
2
y
, tan  
ay
ax
Magnitude-angle Notation
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
Problem 1. The x component of vector A is -25.0 m and the y component

is +40.0 m. (a) What is the magnitude of A ? (b) What is the angle between

the direction of A and the positive direction of x?

A vector a can be represented in the magnitude-angle notation (a, ), where
a  ax2  a y2
is the magnitude and
 ay 
  tan  
 ax 

is the angle a makes with the positive x axis.
1
(a) Given Ax = 25.0 m and Ay = 40.0 m, A  ( 25.0 m) 2  (40.0 m) 2  47.2 m
(b) Recalling that tan  = tan ( + 180°), tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°.
Noting that the vector is in the third quadrant (by the signs of its x and y components) we
see that 122° is the correct answer.
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Unit Vectors
A unit vector is defined as vector that has magnitude
equal to 1 and points in a particular direction.
Its sole purpose is
to point in a particular direction. The unit vectors
along the x, y, and z axes are labeled
iˆ , ˆj, and kˆ, respectively.
Unit vectors are used to express other vectors


For example vectors a and b can be written as:


a  a xiˆ  a y ˆj and b  bxiˆ  by ˆj
The quantities a xiˆ and a y ˆj are called

the vector components of vector a. The
quantities ax and ay are scalars, called the

scalar components of a or simply components.
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y

r

b

a
Adding Vectors by Components
x
O


We are given two vectors a  a xiˆ  a y ˆj and b  bxiˆ  by ˆj
We want to calculate the vector sum
  
r  a  b  a xiˆ  a y ˆj  bxiˆ  by ˆj  ( a x  bx )iˆ  ( a y  by ) ˆj  rxiˆ  ry ˆj
The components rx and ry are given by the equations:
rx  a x  bx and ry  a y  by
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P ro b le m 8 . A c a r is d riv e n e a st fo r a d ista n c e o f 5 0 k m , th e n n o rth fo r 3 0 k m ,
a n d th e n in a d ire c tio n 3 0  e a st o f n o rth fo r 2 5 k m . S k e tc h th e v e c to r d ia g ra m
a n d d e te rm in e (a ) th e m a g n itu d e a n d (b ) th e a n g le o f th e c a r's to ta l
d isp la c e m e n t fro m its sta rtin g p o in t.

r
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Problem 13. Two vectors are given by

a  (4.0m )iˆ  (3.0m) ˆj  (1.0m )kˆ

b  ( 1.0m)iˆ  (1.0m) ˆj  (4.0m)kˆ
 
In unit-vector notation, find (a) a  b
 

  
(b) a  b , and (c) a third vector c such that a  b  c  0.
All distances in this solution are understood to be in meters.
 
ˆ m.
(a) a  b [4.0  (1.0)] ˆi  [(3.0) 1.0]ˆj  (1.0  4.0)kˆ  (3.0iˆ  2.0jˆ  5.0k)
 
ˆ m.
(b) a b [4.0  (1.0)]iˆ  [(3.0) 1.0]jˆ  (1.0  4.0)kˆ  (5.0 ˆi  4.0jˆ 3.0k)
  
  
(c) The requirement a  b  c  0 leads to c  b  a, which we note is the opposite of

ˆ m.
what we found in part (b). Thus, c  (5.0iˆ  4.0 ˆj  3.0k)
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
Problem 26. Vector A, which is directed along an x axis, is to be

added to vector B, which has a magnitude of 7.0 m. The sum is a third
vector that is directed along the y axis, with a magnitude that is 3.0 times


that of A. What is the magnitude of A ?
As a vector addition problem, we express the situation (described in the problem



^
^
^
^
statement) as A + B = (3A) j , where A = A i and B = 7.0 m. Since i  j we may
use the Pythagorean theorem to express B in terms of the magnitudes of the other two
vectors:
B = (3A)2 + A2

A=
1
B = 2.2 m .
10
12
y

d

b

a
Subtracting Vectors by Components
x
O


ˆ
ˆ
We are given two vectors a  ax i  a y j and b  bx iˆ  by ˆj
We want to calculate the vector difference
  
d  a  b  d x iˆ  d y ˆj

The components d x and d y of d are given by the equations:
d x  ax  bx and d y  a y  by
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Multiplying a Vector by a Scalar



Multiplication of a vector a by a scalar s r esults in a new vector b  sa
The magnitude b of the new vector is given by: b  | s | a


If s  0 vector b has the same direction as vector a


If s  0 vector b has a direction opposite to that of vector a
The Scalar Product of two Vectors

 

The scalar product a  b of two vectors a and b is given by:
 
a  b =ab cos 
The scalar product of two vectors is also
known as the "dot" product.
When two vectors are in unit-vector notation, we write their dot
 
product as a  b  ( a iˆ  a ˆj  a kˆ)  (b iˆ  b ˆj  b kˆ )
x
y
z
x
y
z
which we can expand acording to the distributive law:
each vector component of the first vector is to be dotted
with each vector component of the second vector.
By doing so the scalar product
of two vectors is given by the equation:
 
a  b =a x bx  a y by  a z bz
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The Vector Product of two Vectors

  

The vector product c  a  b of the vectors a and b

is a vector c

The magnitude of c is given by the equation:
c  ab sin 

The direction of c is perpendicular to the plane P defined


by the vectors a and b

The sense of the vector c is given by the right hand rule:


a. Place the vectors a and b tail to tail

b. Rotate a in the plane P along the shortest angle

so that it coincides with b
c. Rotate the fingers of the right hand in the same direction

d. The thumb of the right hand gives the sense of c
The cumulative does not apply
to a vector product.
The vector product of two vectors is also known as
the "cross" product
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The Vector Product
In unit-vector notation, we write
 
a  b  ( a xiˆ  a y ˆj  a z kˆ)  (bxiˆ  by ˆj  bz kˆ )
which can be expanded according to the distibutive law,
each component of the first vector is to be crossed with
each component of the second vector.
 
a  b  ( a xiˆ  a y ˆj  a z kˆ)  (bxiˆ  by ˆj  bz kˆ ) 
 ( a xiˆ  bxiˆ)  ( a xiˆ  by ˆj )  ( a xiˆ  bz kˆ)  ( a y ˆj  bxiˆ)  ( a y ˆj  by ˆj )  ( a y ˆj  bz kˆ) 
 ( a z kˆ  bxiˆ)  ( a z kˆ  by ˆj )  ( a z kˆ  bz kˆ) 
To check whether any xyz coordinate
 0  a x by kˆ  a x bz ˆj  a y bx kˆ  0  a y bziˆ  a z bx ˆj  a z by iˆ  0 
system is a right-handed
coordinate system, use the right-hand
  
rule for the cross product i  j  k
 (a y bz  by a z )iˆ  ( a z bx  bz a x ) ˆj  ( a x by  bx a y )kˆ
A determinant can also be used
iˆ
 
 
a  b  b  a  a x
ˆj
ay
bx
by
kˆ
ay
a z  iˆ
by
bz
az
a
 ˆj x
bz
bx
az ˆ a x
k
bx
bz
ay

by
 (a y bz  by a z )iˆ  ( a z bx  bz a x ) ˆj  ( a x by  bx a y )kˆ
with this system. If your fingets sweep


i (positive direction of x) into j
(positive direction of y) with the
outstretched thump pointing in the
positive direction of z, then
the system is right-handed.
Note: The order of the two vectors in the cross product is important
 
 
b a   a b


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P r o b le m 4 0 . F o r th e fo llo w in g th r e e v e c to r s , w h a t is



3C  ( 2 A  B ) ?

A  2 .0 0 iˆ  3 .0 0 ˆj  4 .0 0 kˆ

B   3 .0 0 iˆ  4 .0 0 ˆj  2 .0 0 kˆ

C  7 .0 0 iˆ  8 .0 0 ˆj
ˆj kˆ
iˆ
Using the fact that
a y az
a az ˆ ax
 
 ˆj x
k
a  b  a x a y az  iˆ
by bz
bx
bx bz
ˆi  ˆj  k,
ˆ ˆj  kˆ  ˆi, kˆ  ˆi  ˆj
bx by bz
or
we obtain
 ( a y bz  by az )iˆ  ( az bx  bz a x ) ˆj  ( a x by  bx a y )kˆ
ay

by


ˆ
2 A  B  2 2.00iˆ  3.00ˆj  4.00kˆ  3.00iˆ  4.00ˆj  2.00kˆ  44.0iˆ  16.0jˆ  34.0k.



 
a  b =a x bx  a y by  az bz
Next, making use of
ˆi  ˆi = ˆj  ˆj = kˆ  kˆ = 1
ˆi  ˆj = ˆj  kˆ = kˆ  ˆi = 0
or
we have

 
3C  2 A  B  3 7.00 ˆi  8.00ˆj  44.0 ˆi  16.0ˆj  34.0 kˆ

 


 3[(7.00) (44.0)+( 8.00) (16.0)  (0) (34.0)]  540.
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