VECTORS
Table of Contents
RESOURCES ............................................................................................................................ 3
Textbook: ............................................................................................................................... 3
Pre-knowledge ................................................................................................................... 3
Geometric vectors .............................................................................................................. 3
Resources consulted ............................................................................................................... 3
Notes ...................................................................................................................................... 3
Internet ................................................................................................................................... 3
Download ........................................................................................................................... 3
More on moments .............................................................................................................. 3
Direction of the cross product ............................................................................................ 3
History of vectors ............................................................................................................... 4
myTUTor: Interactive demos ................................................................................................ 4
PRACTICE ................................................................................................................................ 4
OUTCOME ................................................................................................................................ 5
ASSESSMENT STANDARDS ................................................................................................. 5
MOTIVATION .......................................................................................................................... 5
1.
DEFINITIONS AND NOTATIONS ................................................................................. 6
Geometric vectors .................................................................................................................. 6
Algebraic vectors ................................................................................................................... 7
Vectors in three dimensions ................................................................................................. 10
ACTIVITY 1.1 ................................................................................................................. 10
Calculations with algebraic vectors ..................................................................................... 11
The magnitude of a vector ............................................................................................... 11
ACTIVITY 1.2 ................................................................................................................. 11
Vector addition and scalar multiplication ........................................................................ 11
ACTIVITY 1.3 ................................................................................................................. 12
ACTIVITY 1.4 ................................................................................................................. 12
2.
UNIT VECTORS ............................................................................................................. 13
Definition ............................................................................................................................. 13
Standard unit vectors............................................................................................................ 13
Normalizing vectors ............................................................................................................. 14
More examples ..................................................................................................................... 16
ACTIVITY 2 .................................................................................................................... 17
Vectors
© Tshwane University of Technology: EL Voges
Page 1 of 35
3.
Scalar product .................................................................................................................. 17
Definition ............................................................................................................................. 17
Properties of the dot product ................................................................................................ 18
Geometrical interpretation ................................................................................................... 18
Special cases ........................................................................................................................ 19
Angle between vectors ......................................................................................................... 19
Parallel and orthogonal vectors ............................................................................................ 20
Unit vectors and components ............................................................................................... 21
Work .................................................................................................................................... 22
ACTIVITY 3 .................................................................................................................... 23
4.
Vector product ................................................................................................................. 24
Definition ............................................................................................................................. 24
Properties ............................................................................................................................. 24
Geometrical interpretation ................................................................................................... 25
The right-hand rule .......................................................................................................... 25
Moment ................................................................................................................................ 27
What is "the moment of a force"? .................................................................................... 28
Analysis of the implications of the formula ..................................................................... 28
ACTIVITY 4 .................................................................................................................... 30
Answers.................................................................................................................................... 31
Activity 1.1 ...................................................................................................................... 31
Activity 1.2 ...................................................................................................................... 31
Activity 1.3 ...................................................................................................................... 31
Activity 1.4 ...................................................................................................................... 31
Activity 2 ......................................................................................................................... 31
Activity 3 ......................................................................................................................... 32
Activity 4 ......................................................................................................................... 32
QUESTIONS FROM PREVIOUS ASSESSMENTS .............................................................. 33
FROM EXAM PAPERS ...................................................................................................... 33
QUESTIONS FROM AN ASSIGNMENT ......................................................................... 33
QUESTIONS FROM A SEMESTER TEST ....................................................................... 33
VECTORS: Tutorial ................................................................................................................ 34
Vectors
© Tshwane University of Technology: EL Voges
Page 2 of 35
RESOURCES
Textbook:
Washington, Allyn J. 2011. Basic Technical Mathematics with Calculus. 9th ed. SI Version.
Pearson Canada: Toronto
Pre-knowledge
pp. 273 – 284: sine and cosine rules
Geometric vectors
pp. 245 – 273
Resources consulted
The following books were consulted during the compilation of these notes. It may be used for
additional exercises.

Singh, Kuldeep. 2011. Engineering Mathematics through Applications. 2nd ed.

Bird, OJ. 2010. Higher Engineering Mathematics. 6th ed.
The following web sites were accessed for ideas.




http://en.wikipedia.org/wiki/Cross_product
http://www.netcomuk.co.uk/~jenolive/vect8.html
http://en.wikipedia.org/wiki/Dot_product
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
Notes
This set of notes on algebraic vectors and some applications.
Internet
Download
You need Mathematica Player (CDF Player), a free interactive tool available from
http://www.wolfram.com/cdf-player/
Once you have this Player you will have access to lots of interactive demonstrations in
a wide variety of subjects.
More on moments
If you are interested in more advanced examples of moment visit the following web
site:
http://www.engin.brown.edu/courses/en3/notes/Statics/moments/moments.htm
There are also a few demo files on myTUTor which might assist you with
understanding certain concepts.
Direction of the cross product
The following URL will take you to a web site with three short video clips on the
direction of a cross product:
Vectors
© Tshwane University of Technology: EL Voges
Page 3 of 35
http://www.ecourses.ou.edu/cgibin/ebook.cgi?doc=&topic=st&chap_sec=04.3&page=theory
History of vectors
Follow the hyperlinks below should you be interested in the history of vectors.
http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html
myTUTor: Interactive demos
There are several interactive demos on myTUTor which are included with the permission of
Wolfram; they form part of the Wolfram Demonstrations project. You will need Mathematica
Player to "play". Go to the folder "Vectors", "Demos" and use the hyperlinks to access the
demos.

Remember, you need Mathematica Player, which is not available on TUT computers.

Cross Product Of Vector
o Change the vectors u and v to see how the cross product w changes.
Simple Lever
o Change the weights and positions and watch the movement of the lever
(beyond the scope of this course).
A Lever With Four Weights
o Move the blocks around; see how to place blocks to balance a beam (beyond
the scope of this course)
Mechanical Work
o You can animate – let the object move and watch how the work increases as
the distance increases.
Dot Product
o Change one vector and see how the work done changes.
Vector Projection
o Change the vectors w (red) and v (green) and note the change in the
component of w (yellow) in the direction of v. The yellow vector is called the
projection of w onto v.
More on Vector Projections
o This demo will help you understand the concept of the projection of one
vector onto another.
Vectors In 3D
o Change the vectors, rotate the diagram, zoom in, include negative vectors
and the vector sum.







PRACTICE


Do all the exercises in this document.
Do as much as possible from your textbook, Exercise 9.1, pp. 258 – 259; Exercise 9.2,
pp. 261 – 263; Exercise 9.3, p. 267; Exercise 9.4, pp. 270 – 272.
Vectors
© Tshwane University of Technology: EL Voges
Page 4 of 35
OUTCOME
Apply the knowledge of geometric and algebraic vectors to solve a wide variety of problems.
ASSESSMENT STANDARDS
1.
2.
Use the correct vector notation;
Give a geometric interpretation of vectors addition, subtraction and multiplication
with a scalar;
Write a vector given in terms of its initial and terminal points as a vector in the
standard position;
Represent a vector in component form using the standard unit vectors i, j and k;
Calculate the magnitude a of given vector;
Determine the unit vector in a specified direction;
Determine the component of a vector in a specified direction;
Determine the resultant of vectors;
Determine the angle between two vectors;
Calculate the two types of vector products, that is, the dot product and the cross
product;
Use the dot product to calculate the work done W  F r by a force F to move an
3.
4.
5.
6.
7.
8.
9.
10.
11.
object from A to B where r  AB ;
Calculate the moment M  r  F of a force F applied at a point B about a point A
12.
where r  AB using the cross product; and
13.
Use vectors to solve other basic problems.
MOTIVATION
The theory around vectors developed over time and is now an integral part of the life of an
engineer and scientist. You are already familiar with geometric vectors and their
applications. For example, Example 4 on page 269 (Washington, 2011) uses vectors to
calculate the force of friction between a block and a plank. Here are two more examples of
vectors:

When a charged particle move with a velocity v in a magnetic field H, then the
particle experiences a force perpendicular to both v and H which is proportional to
the so-called cross product v  H .

The work done by a force F to move an object from A to B is given by the scalar
product F r AB where r AB is the position vector from A to B.
In this set of notes the concept of algebraic vectors is introduced. Calculations with algebraic
vectors and a few applications are discussed.
Vectors
© Tshwane University of Technology: EL Voges
Page 5 of 35
1.
DEFINITIONS AND NOTATIONS
You have now worked through the theory, examples and exercises in Washington (2011).
Let's confirm a few definitions and notations we'll use in this Mathematics course.
Geometric vectors
From Washington (2011) and previous courses you know that a scalar is a physical
quantity specified by its magnitude and unit. A vector is described by specifying its
magnitude, direction and unit. You represented a vector by a directed line segment, or arrow,
in a plane. Then you used the arrows to determine the resultant of two or more vectors.
These “pictures” of vectors are called geometric vectors. In mathematics we often omit the
“physical” and the unit in the definitions of vectors and scalars.

A scalar is a real number or a quantity that has magnitude. For example, a
temperature of 31°C, a speed of 15 m/s and a mass of 3 kg are all scalars.

A vector is a quantity that has both direction and magnitude. A velocity of 15 m/s
due east and a force of 30 N vertically downwards are examples of vectors.
Geometrically a vector is represented by a directed line segment. The length of line
segment AB represents the magnitude of vector AB in Figure 1. The angle  between
the line segment and the horizontal line gives the direction of the vector.
Figure 1 A geometric representation
of a vector

A vector is not a directed line segment; it is just represented by the line
segment.
We often denote a vector by a single letter. In Figure 1 we used a to represent AB . The
point A is called the initial point or tail or end of vector AB . B is the terminal point or
head or tip of AB .

Most textbooks use boldface when using a single letter to represent a vector. Thus,
most textbooks will use a to represent AB . However, it is not really possible to use
boldface when writing. We'll use the underscore instead, that is, we'll write a.
 DON'T forget the underscore!
Vectors
© Tshwane University of Technology: EL Voges
Page 6 of 35
The magnitude or length or modulus of vector a  AB is denoted by a  AB .

Don't confuse a and a ! a represents the length of vector a, while a is the absolute
value of scalar a.

Some textbooks write a to indicate the norm of a vector.

The zero vector 0 has magnitude 0 and may have any direction.
 Remember, 0  0 .
Vectors are free in the sense that we may move the arrow representing a vector from one
position to another provided that its magnitude and direction remain the same. We say the
vector is translated, that is, displaced without rotation. You've done this when you
determined, for example, the resultant of two vectors a and b. The initial point of b in Figure
2 was positioned so that it coincided with the end point of a.
Figure 2 Free vectors
We'll use the concept "free vectors" to define algebraic vectors.
Algebraic vectors
It is not always feasible to plot vectors according to scale to determine the resultant. To do
the calculation without the sketch we place the geometric vector AB in a rectangular
coordinate system as shown in Figure 3. Since AB is free we translate it to the position OP .
Note that OP  AB . We say OP is in the standard position since its initial point is at the
origin. The vector OP is the standard vector to AB .
Figure 3 A vector in the standard position
Vectors
© Tshwane University of Technology: EL Voges
Page 7 of 35

OP is the standard vector to all vectors with the same magnitude and direction.
Given the coordinates of the initial and terminal points of a geometric vector, how do we
determine the corresponding standard vector? Consider the vector AB with initial point
A( xa ; ya ) and terminal point B( xb ; yb ) shown in Figure 4.
Figure 4 The coordinates of the tip of a standard vector
We know, using our knowledge of analytic geometry, that the magnitude of the horizontal
and vertical components of AB are ( xb  xa ) and ( yb  y a ) respectively. The respective
components of OP are ( x p  0)  x p and ( y p  0)  y p .
Since OP  AB their horizontal and vertical component will be equal. Thus, the terminal
point of OP is given by
( x p , y p )  ( xb  xa ; yb  ya ) .
Since the initial point of OP is at the origin it may be represented by the coordinates of its
terminal point, that is, OP  ( x p ; y p ) .

For every vector AB in a rectangular coordinate system there always exists a point
P( x p ; y p ) such that OP  AB . The point P( x p ; y p ) therefore completely describes
the vector AB except for its actual position.

Conversely, given any point P( x p ; y p ) in a rectangular coordinate system, the directed
line segment joining O and P forms the geometric vector OP .
Vectors
© Tshwane University of Technology: EL Voges
Page 8 of 35

Note the possibility of confusion: is ( x p ; y p ) the coordinates of the point P or does it
represent the vector OP ? In this course we'll use triangular brackets for vector
representation. Thus,
OP  x p ; y p
represents the vector OP , while ( x p ; y p ) refers to the coordinates of the point P.
Example 1.1
Given the geometric vector AB with A(3 ; 4) and B(7 ; –1), determine the standard vector
OP to AB .
COMMENT
Use the formula ( x p ; y p )  ( xb  xa ; yb  ya ) . See Figure 5.
SOLUTION
Substitution of the given coordinates yields
( x p ; y p )  (7  3 ; 1  4)  (4 ;  5) .
Thus, OP  4 ;  5 . 
Figure 5 The vectors in Example 1.1
The vector a ; b is called an algebraic vector. It is often also called the component form
of the vector. The real numbers a and b are called the scalar components of a ; b . Here a
is the x-component or horizontal component, and b is the y-component or vertical component.
Two algebraic vectors a ; b and c ; d are equal if and only if a  c and b  d . The zero
vector is 0  0 ; 0 .
Vectors
© Tshwane University of Technology: EL Voges
Page 9 of 35
Vectors in three dimensions
Geometric vectors are limited to space that we can visualize, that is, in two or three
dimensions (2D or 3D). Algebraic vectors don't have this restriction. For example, 1; 2 ,
3 ; 7 ; 0 and 1;  1; 2 ; 3 ;1 are 2D, 3D and 5D algebraic vectors. In this course discussions
will be limited to 2D vectors in a plane and 3D vectors in space.
Figure 6 The 3D Cartesian coordinate system
In three dimension, the xy plane is horizontal and the third axis, the z-axis, is perpendicular to
this plane as shown in Figure 6.The axes Ox, Oy and Oz, in that order, are right-handed in the
sense that the rotation of a right-hand screw from Ox to Oy will advance in the direction of
Oz; a rotation from Oy to Oz will advance in the direction of Ox, and a rotation from Oz to Ox
will advance in the direction of Oy.
ACTIVITY 1.1
1.
Sketch the following vectors on one system of axes: 2 ;  3 , 0 ; 5 , 4 ; 0 ,
2 ;  3 and 4 ;1 .
2.1
2.3
Represent each geometric vector AB , with A and B as indicated, as an algebraic vector
in the standard form a ; b .
A(–1 ; 7), B(1 ; –1)
2.2
A(0 ; 0), B(3 ; 5)
A(3 ; –2), B(0 ; –5)
2.4
A(–1 ; 0), B(0 ; 1)
3.
Determine the initial point of the vector RS  5 ;  1 if the terminal point is (4 ; 7).
4.
4.1
4.2
4.3
Consider the algebraic vectors 1; 4 and 4 ;1 .
Sketch these vectors on one system of axes.
Use the parallelogram rule to determine the sum 1; 4  4 ;1 geometrically.
Deduce a method to calculate the sum of algebraic vectors without sketching them. 
2.
Vectors
© Tshwane University of Technology: EL Voges
Page 10 of 35
Calculations with algebraic vectors
The magnitude of a vector
The magnitude (length/norm/modulus) of v  a ; b , denoted by v (or v in some books),
is given by
v  a 2  b2 .
Geometrically, v  a 2  b2 is the length of the standard geometric vector OP associated
with the algebraic vector a ; b .

v is a vector and v is a scalar.

v  0 for any vector v. v  0 if and only of v  0 .

The modulus of the three-dimensional vector OP  a ; b ; c is given by
OP  a 2  b2  c 2 .
EXAMPLE 1.2
If A is the point with coordinates (2 ; 3) and B is the point (5 ; 1), determine AB and AB .
SOLUTION
AB  5  2,1  3  3 ;  2
AB  32  (2)2  13 
ACTIVITY 1.2
Calculate the magnitude of each of the following vectors.
4; 3
4; 3; 2
1.
2.
3.
4.
AB if A is the point (1 ; –2 ; 3) and B (1 ; –6 ; –6)
CD where C(4 ; 5 ; 6) and D(13 ; 2 ; 7)

Vector addition and scalar multiplication
If u  a ; b ; c and v  d ; e ; f are two vectors and k is a scalar, then
u  v  ad ;be;c f
and
ku  ka ; kb ; kc .
Vectors
© Tshwane University of Technology: EL Voges
Page 11 of 35
EXAMPLE 1.3
Given u  4 ;  3 and v  2 ; 3 , then
u  v  4  2 ; 3 3  6 ; 0
2u  2(4) ;  2(3)  8 ; 6
2u  3v  2(4) ; 2(3)  3(2) ; 3(3)
 8; 6  6;9
 86; 69
 2 ;  15 
ACTIVITY 1.3
1.
Calculate 3a, a – b and a  b if:
1.1
a  1;1 and b  2 ; 3
1.3
a  b and b  3 ; 4
2.
2.1
Determine 4a  2b and 3a  5b if:
a  1;  3 and b  1;1
1.2
a  1; 3 and b  5a
2.2
a  1;  1 and b  3 ; 4 
ACTIVITY 1.4
1.
Given the geometric vector AB with A(3 ; 4) and B(7 ; –1), calculate the vector OP
so that OP  AB where O is the point (0 ; 0).
2.
2.1
3.
3.1
Represent each geometric vector AB , with A and B as indicated, as an algebraic
vector of the form x ; y .
A(–1 ; 7), B(1 ; –1)
2.2
A(3 ; –2), B(0 ; –5)
Determine the magnitude and direction of each of the following vectors in the
standard position. Round off to three decimal places where necessary and give the
angles in radians with   ( ;  ] .
1;  4
2;3
3.2
4.1
1;  2, 2
2 ; 5, 6
3.4
Points P, Q and R have coordinates (9 ; 1 ; 0), (8 ; –3 ; 5) and (5 ; 5 ; 7) respectively.
Write down
4.2
PQ and QR
PQ and QR .
5.
5.1
Calculate the norm (magnitude) of each of the following vectors.
4; 3; 2
7 ;  3 ;1
5.2
6.
Determine 2a, a – b and a  b if a  0 ;  1;1 and b  1;1; 0 .
3.3
4.
Vectors
© Tshwane University of Technology: EL Voges
Page 12 of 35
7.1
7.3
7.5
Given that a  1; 3 , b  1; 2 , c  3 ; 6 ; 2 and d  6 ; 4 ;  1 , determine the
following:
a+b
7.2
a–b
a + 2b
7.4
d–c
6c – d
8.
8.1
Determine the values of x, y and z if
( x  y  z) i  ( y  z  1) j  ( x  z )k  0 .
8.2
( x  y) i  ( y  2 z ) j  ( x  z ) k  2 i  k
2.
UNIT VECTORS
7.

Definition
A unit vector is a vector with unit length, that is, the magnitude of a unit vector is 1. It is
usually used to indicate direction.
Example 2.1
Which of the following are unit vectors? Motivate your answer.
a)
c
3 1
;
2 2
b)
d
1 2 3
; ;
3 3 3
COMMENTS
A vector is a unit vector if its magnitude is 1.
SOLUTION
2
a)
 3   1 2
3 1
c  
 1
    
4 4
 2  2
The length of c is 1 and c is therefore a unit vector.
2
b)
2
2
1 4 3
8
1  2  3 
d        
  
 
9 9 9
9
 3  3  3 
d is not a unit vector since d  1 . 
Standard unit vectors
We define three unit vectors i, j and k in the direction of the x-, y- and z-axes respectively,
shown in Figure 7, as follows:
 i  1; 0 ; 0 , a unit vector in the direction of the x-axis
Vectors
© Tshwane University of Technology: EL Voges
Page 13 of 35

j  0 ;1; 0 , a unit vector in the direction of the y-axis

k  0 ; 0 ;1 , a unit vector in the direction of the z-axis
Figure 7 The three standard unit
vectors
Every vector in space is a linear combination of the standard unit functions. For example, the
vector a  2 ; 3 ;  1 may also be written as a  2 i  3 j  k , since
2 i  3 j  k  2 1; 0 ; 0  3 0 ;1; 0 1 0 ; 0 ;1  2 ; 3 ; 1  a
Example 2.2
Let P(1 ; –1) and Q(–2 ; 3) be two points in a plane. Write the vector a  PQ as a linear
combination of i and j.
COMMENTS
First calculate PQ and then express a in terms of the unit vectors.
SOLUTION
PQ  2  1; 3  (1)  3 ; 4
 a  PQ  3 ; 4  3 i  4 j 

Note the use of underscores to indicate vectors. Textbooks often use boldface but we
can't write in boldface.

In other courses these unit vectors might be denoted by e1, e2 and e3 respectively.
Normalizing vectors
We quite often use unit vectors to indicate the direction of a vector. To create a unit vector in
a specific direction, we say we normalize the vector by dividing the vector by its length.
Thus, the unit vector in the direction of a vector a, shown in Figure 8, is
a
aˆ  .
a
Vectors
© Tshwane University of Technology: EL Voges
Page 14 of 35
Figure 8 A unit vector in the direction of a

Note the use of the "hat", ^, to indicate the unit vector associated with the vector a.
Example 2.3
Determine a unit vector in the direction of the vector a = 3i - 5j + 4k.
COMMENT
a
2
2
2
We'll use the formula aˆ 
where a  a1  a2  a3 .
a
SOLUTION
Let â denote the unit vector in the direction of a. Then
3; 5; 4
3; 5; 4
aˆ 

50
32  (5) 2  42
3
5
4
i
j
k
50
50
50
Write the answer in terms of the standard unit vectors because that is the
format used in the question.
 aˆ 
When the magnitude of a vector a in the direction of another vector b is given, we calculate
the vector a using the formula
a  a bˆ
where b̂ is the unit vector in the direction of b.
Example 2.4
Calculate the vector b with magnitude 6 in the direction of a vector a = i + 2j - 2k.
COMMENTS
Magnitude = 6
Direction is defined by the unit vector in the direction of a
SOLUTION
The unit vector in the direction of a is
1, 2, 2
1
aˆ 
 1, 2, 2
12  22  (2) 2 3
The vector b is
Vectors
© Tshwane University of Technology: EL Voges
Page 15 of 35
b  b aˆ
1

 6  1; 2 ;  2 
3

 2 1; 2 ;  2
 2;4;4
 b  2 i  4 j  4k 
More examples
There are a wide variety of applications where unit vectors are used. Only a few basic
examples are given here.
Example 2.5
Express the velocity vector v = 12i - 5j as a product of its speed and direction.
COMMENTS
Speed = magnitude of velocity
Direction = unit vector in the direction of v
SOLUTION
v  122  (5)2  169  13
vˆ 
v
1
 (12 i  5 j )
v 13
5 
12
 v  13  i 
j 
13 
speed  13
direction
Example 2.6
A force of 6 N is applied in the direction of the vector a  2 ;  2 ;1 . Express the force as a
linear combination of unit vectors.
SOLUTION
Let the force be F. Then F  6 N. The unit vector in the direction of F is
aˆ 
2 ;  2 ;1
a
1

 2 ;  2 ;1
2
2
a
2  (2)  1 3
 F  F aˆ
1

 6  2 ;  2 ;1 
3

 2 2 ;  2 ;1
 4;4;2
 4 i  4 j  2k 
Remember the underscores indicating vectors!
Vectors
© Tshwane University of Technology: EL Voges
Page 16 of 35
Example 2.7
Vector c has the same direction as vector d = –2i + 5j + k but has the same modulus as vector
e = 10i – 2 j + 4k. Write down c.
COMMENTS
The direction of c is determined by the unit vector in the direction of d.
The modulus (length) of c is the same as the length of e.
SOLUTION
The unit vector in the direction of d is
2 ; 5 ;1
2 ; 5 ;1
d
dˆ 


d
30
(2) 2  52  12
Thus, c is a vector in the direction 2 ; 5 ;1
30
c  e  102  (2)2  42  120
 c  c dˆ
 2 ; 5 ;1 
 120 

30 

2 ; 5 ;1

2
 1i  52 j  12 k 
ACTIVITY 2
1.
Express the following vectors in terms if i, j and k.
1.1
1.2
1; 2 ; 3
0, 2 ;  1, 6 ; 3,3
3
1
i  j is a unit vector.
2
2
2.
Show that a 
3.
Construct a unit vector parallel to a  i  j  3k .
4.
4.1
Determine the unit vectors in the direction of each of the following vectors. Simplify
answers and round off to three decimal places where necessary
1;1
3; 4
4.2
4.3
0,5 ;1; 0, 2
3.
Scalar product

The scalar product, also called the dot product, is one of two methods of "multiplying"
vectors (the other method is called the cross product). The result of this type of
"multiplication" is a single number, hence the name "scalar product".
Definition
If the vectors a  a1 ; a2 ; a3 and b  b1 ; b2 ; b3 are at an angle  , where 0    180 ,
then the the scalar product is given by
Vectors
© Tshwane University of Technology: EL Voges
Page 17 of 35
a b  a1b1  a2b2  a3b3 .

Thus, to obtain the scalar product of two vectors we multiply the corresponding i, j and
k components, and add the results.
The dot product is called the scalar product because the result of the multiplication is a
real number, that is, a scalar.
Why this product is sometimes called a dot product: The dot operator, , indicates the
type of multiplication.
The product written as a b (no "dot") has no meaning.



Example 3.1
Calculate the dot product if:
a)
a  1; 3 and b  4 ;  2
a  1; 3 ; 2 and b  4 ;  2 ;1
b)
SOLUTION
a)
a b  1; 3 4 ;  2
 (1)(4)  (3)(2)
 46
 2
b)
a b  1; 3 ; 2
4 ;  2 ;1
 (1)(4)  (3)(2)  (2)(1)
 462
 0
Properties of the dot product
For any nonzero vectors a, b and c and scalar k,
 a bb a


a  kb    ka  b

a b  c   a b  a c

a a a

a0
2
Remember, 0 ≠ 0 because 0 is a scalar and 0  0 ; 0 ; 0 is a vector.
Geometrical interpretation
Consider the two nonzero vectors a and b in Figure 9. The angle between the two vectors is
 ,0     .
Vectors
© Tshwane University of Technology: EL Voges
Page 18 of 35
Figure 9 Geometrical interpretation of a scalar product
The scalar may also be calculated by using
a b  a b cos  .

The proof of this relationship between the dot product of two vectors and the angle
between the vectors is beyond the scope of this course.
Example 3.2
Calculate the scalar product of two vectors a and b with magnitudes 5 and 7 respectively. The
angle between the vectors is 50°.
SOLUTION
a b  (5)(7) cos 50
 22, 49756
 22,5 
Special cases
Remember the following scalar products of the standard unit vectors:
 i i  j j  k k since i  j  k  1 and   0  cos  1

i j  j k  k i  0 since    2  cos  0
Angle between vectors
From the geometrical interpretation of the dot product,
a b  a b cos 
we have
 ab 
  cos 1 

a b
Example 3.3
Calculate the angle, in radians, between the following vectors:
a  8 ; 5 and b  11;17
a)
Vectors
© Tshwane University of Technology: EL Voges
Page 19 of 35
a  3 i  5 j  2k and b  2 i  2 j  2k
b)
COMMEMTS
Check: Is your calculator in RAD mode?
SOLUTION
a)
a b  (8)(11)  (5)(17)  3
a  82  52  89
b  112  172  410
3


  cos 1 
  1,587
 89 410 
Note that only the answer is rounded off. Your calculator is "clever" –
enter the expression the way it is typed here and your answer will be
correct.
b)
a b  (3)(2)  (5)(2)  (2)(2)  0
 0 
  cos 1 

a b
 cos 1 0
 
There is no need to calculate the magnitude of the vectors! Thus, always
calculate the dot product first; it may save time.
Parallel and orthogonal vectors
Two nonzero vectors a and b are orthogonal (perpendicular) if and only if a b  0.
Two nonzero vectors a and b are parallel if and only if a  kb where k is a nonzero scalar.
Alternatively, calculate the angle between the two vectors using the formula.
Example 3.4
Determine whether the following pairs of vectors are orthogonal, parallel or neither:
a  2 ;18 and b  32 ;  16
a)
a  j  6k and b  i  2 j  k
b)
SOLUTION
a b  (2)( 32 )  (18)( 16 )  0
a)
Thus, a and b are orthogonal since both are nonzero vectors and their dot
product is zero.
b)
Vectors
a b  (0)(1)  (1)(2) _(6)(1)  8
Thus, the vectors are not orthogonal.
© Tshwane University of Technology: EL Voges
Page 20 of 35
There exist no scalar k such that ka  b . Thus, the two vectors are not parallel.
Alternatively, calculate the angle:
a  0  1  62  37
b  1  22  1  16
8 
  122,5
 37 6 
Thus, the two vectors are neither orthogonal nor parallel. 

  cos 1 
Unit vectors and components
Consider the vector b in Figure 10. The unit vector u in the direction of b has magnitude 1.
Figure 10 Th eunit vector in the direction of b
Thus, to calculate u,
u
b
.
b
Now consider the two nonzero vectors a and b in Figure 11. Let c represent the component of
a in the direction of b (the red arrow in the sketch).
Figure 11 The component of a in the direction of b
Using trig definitions, the magnitude of c is
c  a cos 


a b cos 
b
ab
b
Thus, the component of a vector a in the direction of another vector b is given by
ab
Component of a in the direction of b 
b
Vectors
© Tshwane University of Technology: EL Voges
Page 21 of 35
In the sketch the component is indicated as a vector. Why do we then say the component is a
scalar? We said the "component of a in the direction of b"; the "direction" is specified!

There is another quantity known as the projection of a onto b, which is beyond the
scope of this course.
Example 3.5
Calculate the component of a in the direction of b where a  3 ;  5 ; 2 and b  7 ;1;  2 .
SOLUTION
a b  (3)(7)  (5)(1)  (2)(2)  12
b  72  1  22  54
Thus, the component of a in the direction of b 
12
.
54
Work
A constant force F acts on an object and moves it from a point P to a point Q as shown in
Figure 12. Assume the force is at an angle θ with the horizontal and the displacement is s.
Figure 12 Work done by a force F
From previous courses in Physics we know that
Work done  W   component of the force in the direction of the displacement  distance 
The component of F in the direction of the displacement is F cos  and "distance" is the
magnitude s of displacement. Thus,
W   F cos   s 
 F s cos 
F s
Thus, the work, W, by a force, F, to move an object through a displacement s is given by
W F s.
Example 3.6
A force F = (2i + 7k) causes a body to move from a point A with coordinates (1 ; 1 ; 2) to
B (7 ; 3 ; 5). Calculate the work done by the force.
SOLUTION
s  7  1; 3  1; 5  2  6 ; 2 ; 3
Vectors
© Tshwane University of Technology: EL Voges
Page 22 of 35
W  F s
 2;0;7
6;2;3
 (2)(6)  0  (7)(3)
 33
Thus, the work done by the force is 33. 
No units in the question  no units in the answer
ACTIVITY 3
1.
Evaluate the following scalar products.
1.1
1.2
1;  2 3 ; 3
9;2
1; 6
2.
Simplify i ( i  j ) .
3.
If a  5 ; 3 ;  2 and b  8 ;  9 ;11 , calculate a b .
4.
If p  4 ;  3 ; 7 and q  6 ;  1; 2 , calculate p q .
5.
5.1
Determine the angle between the following pairs of vectors.
5.2
1;  2 and 5 ;1
2 ;  1 and 4 ; 9
6.
Determine the acute angle between the vectors a  1; 2 ;1 and b  0 ; 2 ; 3 .
7.
Determine the angle between a  5 ; 3 ;  2 and b  8 ;  9 ;11 .
8.
Determine the angle between p  3 i  j and q  4 i  6 j .
9.
Determine the component of the vector a  2 i  3 j in the direction of the vector
n  i 5 j .
10.
What is the component of the vector 2 i  7 j in the direction of the vector i?
11.
If a  5 i  3 j  7k and b  i  2 j  7k , determine b a .
12.
Show that the vectors 7 i  2 j  k and i  4 j  k are perpendicular.
13.
Determine the component of the vector 7 i  2 j  k in the direction of the vector
i  j  2k .
14.
Vectors
Show that 2 ;  1 and 0,5 ;  1 are orthogonal.
© Tshwane University of Technology: EL Voges
Page 23 of 35
15.
15.1
Show that the following pairs of vectors are orthogonal using vector methods.
15.2
2 ;1 and 1; 2
6 ; 3 and 1; 2
16.
16.1
Determine the component of the vector 1; 2 ; 7 in the direction of the following
vectors:
16.2

1;1;1
6;0;2
4.
Vector product
The vector product of two vectors, also called the cross product, is one of two methods of
"multiplying" vectors (the other method is called the dot product). The result of this type of
"multiplication" is a vector, hence the name "vector product".
Definition
If the vectors a  a1 ; a2 ; a3 and b  b1 ; b2 ; b3 are at an angle  , where 0    180 ,
then the the cross product is given by
a  b   a b sin   n
where n is the unit vector in the direction determined by the right-hand rule. If the angle
between the two given vectors is not known, then
i
j k
a  b  a1
b1







a2
b2
a3
b3
Note that the result is a vector, hence the name "vector product".
The vector product is also called the cross product because of the  as operator.
The cross product is perpendicular to both a and b.
Remember, the  is read as "cross" and not as "multiply".
The cross product of vectors in two dimensions doesn't exist.
Also note the application of matrix theory – the evaluation of a third-order determinant.
The i, j and k are unit vectors and must be written with underscores.
Properties
Assume a, b and c are any vectors, and p and q are two scalars. Then
 a  b  b  a

a b  c   a  b  a  c

 pa    qb   ( pq)(a  b)


aa  0
0b  0
Vectors
© Tshwane University of Technology: EL Voges
Page 24 of 35



a (b  c )  ( a  b ) c
a  b  0 if and only if a and b are parallel.
Remember, 0  0
Geometrical interpretation
Consider the two vectors a and b shown in Figure 13.
Figure 13 The cross product of two vectors
Those two vectors are in the same plane. The cross product a  b is a vector orthogonal
(perpendicular) to this plane. Thus, the cross product is perpendicular to both a and b. The
direction of the cross product is indicated by the unit vector n in Figure 13.
Mathematically,
a  b   a b sin   n
where  is the angle between a and b. Hence, the magnitude of the cross product is
a  b  a b sin 
since n  1 . The direction of the cross product is determined by the right-hand rule.
The right-hand rule
Consider the two vectors a and b in space shown in Figure 14. The angle between these two
vectors is θ.
Let the fingers of your right hand curl through the angle θ from a to b. The direction of your
thumb indicates the direction of n, the vector perpendicular to the plane (flat surface)
containing both a and b.
In Figure 1(a) the thumb points upward when the fingers curl from a to b.
In Figure 1(b) the thumb points downward when the fingers curl from a to b.
Vectors
© Tshwane University of Technology: EL Voges
Page 25 of 35
n
b
a


a
(a)
b
n
(b)
Figure 14 The right-hand rule
In applications of the vector product n is a unit vector in the direction of a  b , the vector
product of a and b .
Geometric properties of the cross product
Memorize the following properties of the unit i, j and k. You should also be able to prove
them!
 i  i  j  j  k  k  0 since   0 and sin 0  0

i  j  k,
jk  i, k  i  j

j  i  k , k  j   i , i  k   j
Example 4.1
Calculate a  b and b  a if a  2 i  j and b  2 i  j  k .
COMMENTS
The angle between the vectors is not given. We'll thus use the determinant to calculate the vector
product – remember the answer must be a vector!
SOLUTION
i
j
k
ab  2 1 0
2 1 1
Note the use of the vertical lines representing a
"determinant".
  (1)(1)  (0)(1)  i   (2)(1)  (1)(2)  j  (2)(1)  (1)(2)  k
 i  2 j  4k
i
j
k
b  a  2 1 1
2 1 0
  (1)(0)  (1)(1)  i   (2)(0)  (2)(1)  j  (2)(1)  (2)(1)  k
  i  2 j  4k 
Vectors
© Tshwane University of Technology: EL Voges
Page 26 of 35
 It is best to expand the determinant along the first row. That way like terms are
already grouped together.
 Note that a  b  b  a but a  b    b  a 
 The order of the rows in the determinant is important:
o The unit three unit vectors are always in the first row.
o The three components of the first vector are always in the second row.
o The three components of the second vector are always in the last row.
Example 4.2
If a  3 , b  20 and   70 , calculate the magnitude of a  b .
COMMENTS
Since only the magnitude is asked and the angle between the vectors is given we use
a  b   a b sin   n. Keep in mind that n is a unit vector, that is n  1 .
SOLUTION
a  b   a b sin   n
 (3)(20)(sin 70)(1)
 56, 4 
Example 4.3
Determine a vector c that is perpendicular to both a and b where a  i  2 j  k and
b  2 i  2 j  k .
COMMENTS
The cross product of two vectors is always a vector perpendicular to both the vectors. We thus set c
equal to the cross product of a and b.
SOLUTION
c  ab
i
j
k
 1 2 1
2 2 2
 6 i  6k 
Let’s look in detail at one application of vector product.
Moment
Some resources such as James (2001:225) claim that the vector product of two vectors arose
from the need to calculate the moment of a three-dimensional force.
In mechanical and civil engineering "moment" and "torque" have different meanings, while
they are synonyms in physics. To avoid confusion we'll refer to the moment of a force.
Vectors
© Tshwane University of Technology: EL Voges
Page 27 of 35
What is "the moment of a force"?
When a force acts on a solid body, the force has two effects:
 The force tends to accelerate or slow down the object; or/and
 The force tends to rotate the object.
In the context of this section the moment of a force is defined as the tendency of the force to
twist or rotate an object.
Figure 15 shows a typical example of the moment of a force: A wrench is used to loosen a
bolt.
Direction of force F
applied by hand
Position vector r

Loosen the bolt
Figure 15 An example of the moment of a force
A man applies a force F at the one end of the wrench in order to turn the bolt. Let A be the
centre of the bolt, and B the point at which the force F is applied. Let r denote the
displacement from A to B. The moment vector M of F about A is then given by the formula
M  r  F.
Remember, r is the vector drawn from the point of rotation to the point where the force is
applied. The formula is often written with a subscript to clarify the meaning of r:
M  r AB  F .
Analysis of the implications of the formula
Looking at the formula,
 the moment of a force about a point quantifies the tendency of the force to rotate an
object about that point,
 the magnitude of the moment specifies the magnitude of the rotational force, and
 the direction of the moment, determined by the right-hand rule, indicates the axis of
rotation associated with the rotational force.
Example 4.4
A bolt is loosened by applying a force of 40 N to a wrench of 0,25 m as shown in Figure 4.
Calculate the magnitude of the moment about the centre of the bolt.
Figure 16 The bolt and wrench
Vectors
© Tshwane University of Technology: EL Voges
Page 28 of 35
COMMENTS
Only the magnitude is asked! And the angle is given.
The number of decimal places where not specified; I decided to round off to two decimal places –
note the “approximately equal” sign.
Note the unit in the answer – unit in the question  unit in the answer!
If the bolt is right-threaded, then the moment vector is M  (9,66n) Nm where n is the unit vector into
the page.
SOLUTION
M  r F sin 
 (0, 25)(40)(sin 75)
 9, 6592
 9, 66 Nm
Thus, the magnitude of the moment is approximately 9,66 Nm. 
Example 4.5
A force of 4 units acts through the point B(4 ; –1 ; 2) in the direction of the vector 2i – j + 4k.
Determine the moment about the point A(3 ; –1 ; 4).
COMMENTS
The force acts through the point B causing a rotation about the point A. Thus, r is the vector from A to
B. The force has
 a magnitude of 4 and
 is in the same direction as the vector 2i – j + 4k.
Thus, to calculate the actual force F (magnitude AND direction) we must calculate a unit vector, say
uF, in the direction of 2i – j + 4k and multiply it by the magnitude 4.
SOLUTION
r  r AB  4  3 ;  1  (1) ; 2  4  1; 0 ;  2
The unit vector uF in the direction of 1; 0 ;  2 is
uF 
2 ;  1; 4

2 ;  1; 4
21
22  12  42
4 2 ;  1; 4
8 ;  4 ;16
 F  4u F 

21
21
M  rF
i
j
k

1
1 0 2
21
8 4 16

1
 8 i  32 j  4 k 
21 
Thus, the moment of the force about the point A is

1
 8 i  32 j  4k  . 
21 
No units in the question no units in the answer!
Vectors
© Tshwane University of Technology: EL Voges
Page 29 of 35
ACTIVITY 4
1.
Calculate a  b if a  1;  1; 2 and b  2 ; 0 ;1 .
2.
If a  i  k , b  2 i  j  3k and c  i  2 j  3k , evaluate
2.1
ab
3.
If a and b are two parallel vectors, show that a  b  0 .
4.
Show that the vectors 0, 2 ;  5 and 1; 25 are parallel.
5.
Simplify i  5k .
6.
Evaluate the vector product b  a if a  3 ;  2 ; 5 and b  7 ;  4 ; 8 .
7.
Determine a  b if a  9 i  j  2k and b  i  3 j  2k .
8.
Points A, B and C have coordinates (9 ; 1 ; –2), (3 ; 1 ; 3) and (1 ; 0 ; –1) respectively.
Calculate the vector product AB  AC .
9.
Determine a vector that is perpendicular to the plane containing the vectors 6 i  k
and 2 i  j .
10.
Determine a unit vector that is perpendicular to the plane containing the vectors
i  2k and 2 i  j  7k .
2.2
a  (b  c )
2.3
ac
2.4
bb

Vectors
© Tshwane University of Technology: EL Voges
Page 30 of 35
Answers
Activity 1.1
1.
Sketch
2.1
2 ; 8
2.2
3; 5
2.3
3 ;  3
2.4
1;1
3.
 9 ; 8
4.1
Sketch
4.2
Sketch
4.3
Discussion
Activity 1.2
1.
5
2.
29
3.
0; 4; 9
Activity 1.3
1.1
3 ; 3 ; 1;  2 ; 5
1.2
3 ; 9 ; 6 ;18 ;12, 6 1.3
6 ;  14 ; 8 ;14
2.1
10 ;  12 ; 18 ; 23
Activity 1.4
1.
4;5
2.1
2 ; 8
2.2
3 ;  3
3.1
3,606; 0,983
3.2
4,123; –1,816
3.3
2,417; –1,144
3.4
5.946, 1.914
4.1
1;  4 ; 5 ; 3 ; 8 ; 2
5.1
29
6.
0 ; 2 ; 2 ; 1;  2 ;1 ;
7.1
0;5
7.2
2 ;1
7.4
3; 2 ; 3
7.5
12 ; 32 ;13
8.1
x  13 , y   23 , z  13
8.2
x  0, y  2, z  1
Activity 2
 i  2 j  3k
1.1
1.2
0, 2 i  1, 6 j  3, 3k
2.
Show that …
3.
1
11
4.2
0, 6 ; 0,8
4.3
0, 440 ; 0,880 ; 0,176
4.
2.1
Vectors
9 ;  3 ;1
5.2
9 ;  12 ; 6 ;  8 ; 0
77
4.2
42 ;
7.3
1; 7
4.1
0,707 ; 0,707
59
2
( i  j  3k )
© Tshwane University of Technology: EL Voges
Page 31 of 35
Activity 3
1.1
–3
1.2
3
2.
1
3.
–9
4.
41
5.1
74,7°
5.2
92,6°
6.
37,6°
7.
95,1°
8.
142,1°
9.
3,334
10.
2
11.
–50
12.
Show …
13.
14.
Show …
15.1
Show …
15.2
Show …
8
3
3
6
16.2
4
10
Activity 4
 i  3 j  2k
1.
2.1
i  j k
2.2
i 3 j  k
2.3
2 i  2 j  2k
2.4
0
3.
Show that …
4.
Show that …
5.
–5j
6.
4 i  11 j  2k
7.
8 i  20 j  26k
8.
5 i  34 j  6k
9.
 i  2 j  6k or i  2 j  6k
16.1
10.
Vectors
1
14
(2 i  3 j  k ) or
1
14
(2 i  3 j  k )
© Tshwane University of Technology: EL Voges
Page 32 of 35
QUESTIONS FROM PREVIOUS ASSESSMENTS
The is memo available on myTUTor.
FROM EXAM PAPERS
1.
The velocity v and the acceleration u of a particle are given by v  4i  8 j  4k and
u  2i  j  4k respectively.
1.1
1.2
1.3
Calculate the scalar product v u .
Calculate the angle between v and u.
Obtain the vector product v  u .
[2]
[2]
[3]
2.
A force F   i  3 j  k is applied to an object which moves from point A with
coordinates  3; 1; 2  to B 1;1; 1 . Determine the angle  (in degrees, accurately to
2 decimal places), between F and the displacement vector r  AB .
3.
[5]
Two forces F1 and F2 are applied to an object. Determine the resultant force
F  F1  F2 if F1 has a magnitude of 27 in the direction of vector c  i  2 j  2k and
4.
F2 has a magnitude of 20 in the direction of d  3 j  4k .
[5]
Determine the magnitude of the moment about the origin, of a vector force
F   i  3 j  k passing through the point with position vector r  3 i  2 j  k .
[5]
QUESTIONS FROM AN ASSIGNMENT
1.
Given that u  i  2 j  3k and v  3 i  6 j  9k , show that u and v are parallel. [4]
2.
Given a  2 i  j  k and b  3 i  x j  2k , solve for x if a is perpendicular to b. [3]
3.
3.1
3.2
A force F  2 i  7k causes a body move from the point with coordinates (1;1;2) to
the point (7;3;5).
Determine the displacement r of the body.
[1]
Determine the work done by the force.
[2]
4.
Determine the angle between p  i  j  k and q  2 i  j  2k.
[5]
5.
Determine a unit vector in the direction of vector v  3 i +4k.
[2]
QUESTIONS FROM A SEMESTER TEST
1.
Vectors
Consider two forces F1 and F2 acting upon an object as in Figure 1.
© Tshwane University of Technology: EL Voges
Page 33 of 35
The magnitude of vector F1  AB is 12 kN and the magnitude of vector F2  AC is
1.1
15 kN, while the angle between them is 65 . Determine the magnitude of the resultant
vector R  F1  F2 accurately to one decimal place.
[4]
1.2
Use the definition of the scalar product to determine F1 F2 accurately to one decimal
place.
[3]
2.
Use the definition of the vector product to determine i  2 i
[3]
3.
Use a vector product to determine a vector that is perpendicular to both vectors AB
and BC , if A  (1;3;1), B  (0;2; 2) and C  (2;0;1) .
[5]
Determine the work done by force F  3i  j  4k (magnitude in Newton) in moving
4
an object through a displacement vector r  2i  j  3k (distance in meters).
[3]
VECTORS: Tutorial
Basics
1.
Obtain the moment M about a point A of a force F passing through a point B if A has
coordinates (1;2;3), B has coordinates (5;6;7) and F  2; 1;1 .
2.
True or false? If it is true, prove it. Otherwise explain why it is false or give an example
to show it is false.
2.1 If a b  a c and a  0 , then b  c .
2.2 If a and b are both orthogonal to c, then (a + b) is also orthogonal to c.
3.
Calculate p q if p  5 , q  8 and the angle between p and q is  .
3
4.
Determine whether a and b are orthogonal, parallel or neither. Show all calculations.
4.1 a  2;18 and b  32 ;  16
4.2
a   j  6k and b   i  2 j  k
4.3
a  6; 2; 1 and b  2;5; 2
5.
Calculate a b and b  a if a  1;3; 2 and b  4; 2;1 .
6.
Calculate the angle  between m  8 i  5 j and n  11i  7 j .
7.
Calculate the component of a  3 i  5 j  2k in the direction of b  7 i  j  2k .
8.
Given the points A (1;1;1), B (2;2;2) and C (-1;0;1) determine the angle between CA
and CB .
Determine the component of the force F  3 i  j  k in the direction of the position
9.
vector r  2 i  3 j  6k .
10.
A force F  1; 1; 2 acts at a point P (1;2;1). Calculate the magnitude of the moment
of the force about the point Q (2;1;1).
Vectors
© Tshwane University of Technology: EL Voges
Page 34 of 35
11.
12.
Determine a unit vector in the direction of the vector p  2;7 . Verify that its length
is indeed 1.
A vector a has a length of 3 and makes an angle of 600 with the horizontal. Write a as a
linear combination of the unit vectors i and j.
Applications
1.
The velocity v and acceleration a is given by v  2 i  3 j and a  2 i  2 j . Calculate
the angle between the velocity and acceleration.
2.
Determine the force F that has a magnitude of 10.5 kN in the direction of DC where
C (1.1;-2.2;-4.4) and D (2.4;5.5;7.0).
3.
Calculate the magnitude and direction of the resultant R of the two forces shown in
the sketch.
4.
Four forces act on an object such that the object is at rest. Three of the forces are
given by F1  2 i  j  3k , F2  7 i  7 j  7k and F3  3 i  k . Determine the fourth
force F4.
5.
A force F  5;3; 2 moves an object from D (1;1;1) to C (5;-1;2). Determine the
work done by F.
6.
A box is pulled 10 m across a frictionless surface using a force of 85 N. Calculate the
work done by the force if the direction of the force is 600 above the horizontal.
7.
Calculate the work done by a force F  2 i  7 j to move an object from A (1;1;2) to
B (7;3;5).
Vectors
© Tshwane University of Technology: EL Voges
Page 35 of 35