THE UNIVERSITY OF NEW SOUTH WALES

SCHOOL OF MATHEMATICS AND STATISTICS

MATH1131 Algebra

Section 1: - Introduction to Vectors.

One of the most powerful developments in Mathematics came from the simple idea of theco-ordinate plane. Indeed 2-dimensional co-ordinate geometry was crucial in the develop-ment of the Calculus.

How do we generalise this to higher dimensions?

Vectors are used in physics, engineering, weather maps etc to represent quantities whichhave both direction and mignitude.

In this chapter we will be looking in some detail as to what a vector is and how we canuse vectors to extend our knowledge of geometry from 2 to 3 (and higher) dimensions.

Definition: A vector is a directed line segment which represents a displacement fromone point P to another point Q.

The word vector comes from the Latin veho (cf. vehicle), meaning to carry.

We represent a vector either using the notation

PQ or by using v. In the algebra notes(and in these notes), vectors are represented using bold letters, v. You should representvectors by underlining the letter, viz v. This is important, because you will need to carefullydistinguish between vectors, scalars (and later matrices).

v

P

R

S

PQ

SR

Q

w

A vector has both direction and length (or magnitude). Two vectors are equal ifthey have the same direction and the same magnitude. Hence in the diagram v = w.

We will denote the length of the vector v by |v|.

1

Two vectors are parallel if they have the same direction.

Position Vectors: We choose a fixed point O, in whatever dimensional space we hap-pen to be and call this the origin. The position vector of a point in any number ofdimensions will be represented by a vector from the origin to that point.

O

P

Hence the vector

OP in the diagram is called the position vector of the point P . A positionvector gives the position of a point in space, whereas a direction vector is simply a vectorhaving direction and magnitude (length).

Addition of vectors: To geometrically add two vectors there are two different methods(each important). If we think of a force vector, then, the obvious way to add two vectorsis to put them tip to tail and join the tail of the first to the tip of the second, as in thediagram.

vw

v +w

w

To add the vectors v and w, we move w and then complete the triangle. This method ofaddition is known as the triangle law of addition.

You can see from this that one could obtain the same vector by forming a parallelogramfrom the two vectors and taking the diagonal (often called the resultant) as the sum of thetwo vectors.

2

vw

v +w

This method is known as the parallelogram law.

Subtraction of vectors is performed in a similar way:

w

v

w v

To check this makes sense, add the vectors that are tip to tail, v+(w v) = w as expected.Observe that the vector labelled w v is not a position vector.

Thus if P and Q have position vectors v and w respectively, then

PQ= w v. In gen-eral,

PQ=

OQ

OP .

w

v

w v

O

P

Q

3

Example: Suppose ABCDEF is a regular hexagon with the vector p on the side AB andvector q on the side BC. Express the vectors on the sides: CD,DE,EF, FA and the diag-onals AC,AD,AE in terms of p and q.

Example: In a paralleogram ABCD,

AB= a,

AD= b, and M is the intersection of the

diagonals. Express, in terms of a and b the vectors,

MA,

MB,

MC,

MD.

4

The Triangle Inequality:

Let us restrict ourselves, for the moment, to the plane.

Since the sum of any two sides of a triangle must exceed the third side, we can write

|u+ v| |u|+ |v|

for any vectors u and v.

u

v

u+ v

Q: When do we have equality?

Scalar Multiplication:

We can multiply a vector by a scalar (generally just a real number).

This has the geometric effect of stretching the vector if > 1, stretching and reversingits direction if < 1.

u

u

Commutative and Associative Laws:

The commutative law of vector addition states that a+ b = b+ a.

Geometrically this is obvious:

5

ab

a+ b = b+ a

b

a

The associative law of vector addition states that a + (b + c) = (a + b) + c. Again thefollowing geometric proof (?) will suffice.

a

b

b+ c

a+ b

c(a+ b) + c =

a+ (b+ c)

Ex: A body is acted on by two forces, 3 Newtons in direction NE, 4 Newtons in directionW. Find the resultant force and direction.

6

Geometric Proofs:

Ex: Prove (using vectors) that the line joining the midpoint of two sides of a triangle isparallel to the third side and half its length.

Co-ordinates:

Thus far, much of what we have done works in any number of dimensions. We are nowgoing to define n dimensional space and introduce a co-ordinate system in which to placeour vectors.

The point in R2 with coordinates say (2, 3) can be identified with the position vector

(23

)

obtained by moving 2 units to the right along the x-axis and 3 units up the y-axis. This canbe extended to any number of dimensions.

We take an n-tuple

a1a2.

.

an

of real numbers and think of each ai as lying on an axis xi. In

2 and 3 dimensions, we identify these axes as the XY and XY Z axes respectively, which

7

are mutually orthogonal.

The set of all such n-tuples will be called Rn.

We say that the vector

PQ in Rn has co-ordinates

a1a2.

.

.

an

, if we move a1 units along

the x1 axis, a2 units along the x2 axis, and so on, when moving from P to Q.

Hence an n-tuple

a1a2.

.

an

in R

n can be interpreted as the position vector of a point P

in Rn.

For example, in R3, the point P in the diagram has position vector

23

1

.

P

23

1

0 =

00

0

We can then define the addition of two vectors (algebraically) in Rn by

a1a2.

.

an

+

b1b2.

.

bn

=

a1 + b1a2 + b2

.

.

an + bn

and multiplication by a scalar to be

a1a2.

.

an

=

a1a2.

.

an

8

Multiplying a vector by a scalar merely stretches the vector (if > 1 ) or shrinks it if0 < 1. If is negative then the vector reverses direction. (Note: The algebraic definitionof addition agrees with the geometric definition.)

v

2v

0

v

b

Note that we can now prove such rules as the commutative law algebraically, viz:

a+ b =

a1a2.

.

an

+

b1b2.

.

bn

=

a1 + b1a2 + b2

.

.

an + bn

=

b1 + a1b2 + a2

.

.

bn + an

=

b1b2.

.

bn

+

a1a2.

.

an

= b+ a.

Parallel Vectors.

Two vectors are defined to be parallel if one is a non-zero multiple of the other. Thatis, v is parallel to w if v = w for some scalar 6= 0.

For example,

12

3

is parallel to

24

6

.

Ex: Find the vectors

PQ, and

QP if P =

71

3

and Q =

21

3

.

9

Ex: Suppose that A =

(00

)B =

(14

), C =

(35

), D =

(21

)are the position vectors

for four points A,B,C,D. Prove that the quadrilateral ABCD is a parallelogram.

Ex: ABCD is a parallelogram with vertices A,B,C,D which have the following position

vectors: A =

(12

), B =

(34

), C =

(26

). Find the three possible position vectors of

D.

10

Basis Vectors:

The standard basis vectors in R2 are the vectors

(10

)and

(01

)which are often

denoted by i and j. Observe that every vector in R2 can be written in terms of these basis

vectors. For example

(12)can be written as i 2j.

In 3-dimensions, the basis vectors, i, j,k are

10

0

, 01

0

, 00

1

.

Note that every vector in R3 can be expressed in terms of these basis vectors, viz:

a1a2

a3

can be expressed as a1i + a2j + a3k.

In higher dimensions, we label the basis vectors as e1, e2,... and so on.Thus, in R4, we have

e1 =

1000

, e2 =

0100

, e3 =

0010

, e4 =

0001

.

Once again, we can represent any vector in Rn in terms of the standard basis vectors in Rn.

Distances and Lengths:

Given a vector x =

(a

b

)in R2, we can use Pythagoras Theorem to compute the length

of this vector asa2 + b2. We use the notation |x| = a2 + b2. In R3, given a vector

x =

ab

c

, we can see from the diagram that |OP | = a2 + b2 and then in OAP we

have |OA| = |x| = a2 + b2 + c2.

11

XO

b

a

c

A

P

In higher dimensions, we can define the length of a vector by generalising this formula, i.e.

Definition: A vector x =

a1a2...an

in Rn has length |x| given by |x| =

a21+ a2

2+ + a2

n.

Ex: Find the lengths of a =

13

2

and b =

1234

.

12

The distance between two points A andB in Rn will be defined as the length of the vector

AB,

in other words, if A has position vector a =

a1a2...an

and B has position vector b =

b1b2...bn

,

then the length of

AB is

|

AB | = |b a| =(b1 a1)2 + + (bn an)2.

Ex: Find the distance between

(12)and

(34)and between

12

5

and

36

1

.

The length function | |, (sometimes called a norm) has the following properties:

1. |a| 0.

2. |a| = 0 if and only if a = 0.

3. |a| = |||a|, for R.

A vector which has unit length is called a unit vector. Any vector can be made into aunit vector by dividing by its length.

Ex: Find a unit vector parallel to the vector

23

1

.

13

Ex: Suppose A and B are points with position vectors a and b.Find a vector (in terms of a and b) which bisects the angle AOB, where O is the origin.

Equations of Lines:

We seek to find the equation of a line in vector form. The vector equation of a line is aformula which gives the position vector x of every point on that line. This equation issometimes referred to at the parametric vector form of the line. I will generally just sayvector equation of the line.

Suppose we have a line passing through the origin which contains a vector u in R2. Ev-ery point on that line will have a position vector which is a multiple of u. Conversely,every multiple of u will correspond to the position vector of a point on that line. Hence theequation of the line can be written as x = u where is any real number.

14

ux = u

0

For example, if u were the vector

(23

)then the equation of the line through u passing

through the origin would be x =

(23

), R.

Another way of denoting the set of all real multiples of a given vector is to call it thespan of the vector. Thus we could write {u : R} as span(u). This idea of span isextremely important.

15

Ex: In R2 what is the span of

(10

)? What is span

(11

)?

The advantage of this definition of the equation of a line is that it easily generalises to anynumber of dimensions. For example, the equation of the line in R3 which passes through the

origin and is parallel to the vector

13

6

is simply x =

13

6

.

If the line does not pass through the origin, then we proceed as follows: To find the vectorequation of a line we need to know two things:

1. The position vector a of a point A on the line

2. The direction of the line, i.e. a vector

AB= b parallel to the line.

a

x = a+ b

0

b

b

A

B

X

x

Thus, the span of b will give a line through the origin parallel to b and adding a will shiftthe line to its proper position. Thus we can find the position vector x of any point X on theline by going from the origin along the vector a and then moving along the line, by adding

some multiple of b until we reach X . Thus, the vector

OX is given by

OX=

OA +

AX and

AX is some multiple of b. Hence the equation of the line is x = a+ b, R.

16

Ex: Find the vector equation of the line passing through the point P with position vec-

tor

23

5

and parallel to the vector

16

4

.

Ex: Find the vector equation of the line passing through the two points P,Q with position

vectors P =

12

6

and Q =

42

3

.

Ex: Find the vector equation of the line in 2-dimensions with cartesian equation y = 2x+1.

17

Ex: Does the point

12

3

lie on the line x =

34

11

+

21

4

?

Configurations:

In R3 two lines can

meet at a point

be parallel

neither meet nor be be parallel.

Line Segments:

The set

S =

x R3 : x =

12

3

+

23

1

,1 3

represents that line segment from

31

4

to

511

0

.

Cartesian Equations of the Line: Given the vector form of a line in 3-dimensions, we canwrite down the Cartesian equations (note the plural) as follows. For example, suppose the

vector equation is x =

12

3

+

32

1

. Recall that x is simply an abbreviation for

x1x2

x3

. Hence, equating co-ordinates, we can write x1 = 1+ 3, x2 = 2 2, x3 = 3+ .

18

Eliminating from these equations we have

x1 + 1

3=x2 22 =

x3 31

.

These are called the cartesian equations of the line. Clearly this can be done for any suchline and so the general form is

x1 a

=x2 b

=x3 c

,

where (a, b, c) is a point on the line and (, , ) is a direction vector of the line, providedthat none of the numbers , , is zero.

Hence in vector form this would be x =

ab

c

+

.

The following example tells us what to do when one of the components of the directionvector is zero.

Ex: Convert x =

21

5

+

20

1

into cartesian form.

Observe that two lines will be parallel if their direction vectors are parallel. Two directionvectors are parallel if and only if one is a nonzero multiple of the other. For example thelines

x =

15

6

+

32

1

and x =

24

7

+

64

2

are parallel since their direction vec-

19

tors are

32

1

and

64

2

respectively and the second vector is simply a multiple of the

first. Observe also that the equations x =

15

6

+

32

1

and x =

15

6

+

64

2

represent the same lines, since they are parallel and pass through the same point.

Ex: Find the equation of the line passing through

12

5

and parallel to

x+ 1

3=y 12 =

z + 6

4 .

Ex: Find the intersection (if possible) of the lines x =

10

3

+

12

1

and

x =

35

2

+

21

3

.

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Equations of Planes:

In 3-dimensions and higher, we can construct planes. Suppose we seek the vector equa-tion of the plane passing through the origin parallel to two given non-parallel vectors a andb.

a a0

b

b

X

P

Q

x

To reach any point X with position vector x on the plane we need to stretch the vector a

to P and stretch the vector b to Q in such a way that

OX=

OP +

OQ. Thus, x = a+b.Conversely, if we take the vector which results from adding a multiple of a and a multipleof b then this will be the position vector of a point on the plane.

Ex: Find the vector equation of the plane passing through the origin parallel to the vectors 15

3

and

24

7

.

The plane generated by two such vectors is called the span of the two vectors. So for example,

the span of

15

3

and

24

7

is simply the set

15

3

+

24

7

: , R

.

21

This set is also referred to as the set of all linear combinations of the two vectors. Thus,given any two vectors a,b, span{a,b} = {a+b : , R} and we say that x is a linearcombination of a and b if x = a+ b for some particular and .

Ex: Describe the span of

12

3

, 13

5

. Repeat for span

12

3

, 24

6

.

As with the equation of a line, to get the vector equation of a plane not through the origin,we simply shift the plane by adding any position vector of a point which lies on the plane.Thus to obtain the vector equation of a plane we need:

1. The position vector of a point on the plane.

2. Two non-parallel vectors which are parallel to (or lie on) the plane.

Ex: Find the vector equation of the plane passing through the point P with position vector 23

5

and parallel to the vectors

16

4

and

25

1

.

22

Ex: Find the vector equation of the plane passing through the three points P,Q,R with

position vectors P =

1262

, Q =

4231

and R =

1725

.

Configurations:

In R3, two (distinct) planes can be

parallel

meet in a line

In R3, three (distinct) planes can be arranged so that

the three planes are parallel

two planes are parallel and the third plane is parallel to neither the first two.

they meet at a single point

they meet in a line

none are parallel, but no point lies on all three planes.

In the next chapter we will learn how to analyse these scenarios algebraically.

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Cartesian Equation of a Plane:

As with lines, we can find the cartesian equation of a plane by eliminating the two pa-rameters and . This is generally quite fiddly to do algebraically. Later in this course, youwill see a much better method, but for the moment, we will do it by algebra.

Ex: Find the cartesian equation of the plane x =

12

3

+

24

0

+

10

3

.

The procedure can be reversed to find the vector equation of a plane from the cartesianequation.

Ex: Find the vector equation of the plane 3x 6y + 2z = 12.

24

Further Examples:

Ex: Find the intersection of the planes 2x+ y z = 10 and 3x+ 4y + 2z = 29.

Ex: Show that the line x =

115

3

is parallel to the plane

x =

25

6

+

42

6

+

13

5

.

25

Ex: Find the intersection of x =

12

3

+

14

2

and 2x+ 3y z = 29.

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