1

CHAPTER ONE 1. Vectors. Examples and Uses.

For example, a wind may be described as being of 10 m.p.h. in a northerly

direction. We say its size (or magnitude) is 10 m.p.h. and its direction is north. Other

examples of vectors are a force, tension in a string, velocity of a car, current in a river,

acceleration, momentum and impulse.

It is understood in vector study that the magnitude and direction of a vector will

completely determine the vector. For example the location, temperature etc. of a wind

will have no relevance with regard to its being considered as a vector.

Many other objects can be thought of as vectors with a little imagination e.g.

i) – 3 has magnitude 3, direction “to the left”. ii) an ordered pair (a,b), its magnitude is

said to be how far from the origin it is,

i.e. 22ba + and its direction is

! where

tan ! =b

a. This means for example, that (3,4)

has magnitude 5 and direction 53.13°.

A Vector is an entity which has both size and direction.

!

(a,b) 22ba +

2

iii) a complex number a + bi can be thought of as in example ii) as having magnitude

22ba + and direction Arctan

a

b (angles being measured as usual counterclockwise

from positive horizontal axis)

Not all objects are vectors. Age has no notion of direction for example (except

maybe up!) and is consequently not a vector. Kinetic energy, work quantity of heat are

other examples which are not vectors.

1.1 Notation

A vector will be written a!

Its magnitude will be written a! . Note that this looks like absolute value. In fact

for real numbers it is. –3 has magnitude 3 (its distance from the origin) because 3! = 3.

Saying (3,4) has magnitude 5 is written 5 |)4,3(| =

!. The arrow over the )4,3(

!tells us we

are thinking of it as a vector. Note that a! is a positive real number and must be distinguished from the vector

a! itself. It makes little sense to talk about a! + a! . After all, what does 5|)4,3(| +

! mean? It

does make sense to talk of 3 + |-3| of course, but this is the exceptional circumstance.

This means, for example that a northerly wind is never equal to an easterly wind,

regardless of their magnitudes.

Two vectors are to be said to be equal if and only if their magnitudes and directions are equal separately.

3

Position of a Vector is irrelevant.

Diagrammatically a vector will be represented by

The length of the arrow represents the magnitude of the vector and

the direction of the arrow is the direction of the vector.

In the diagram the vector has magnitude

|OA|!

and direction north-east.

It is easy to confuse the notation with that of early geometry class. There, OA =

AO but here!!AOOA ! because

!AO and !OA have different directions. In geometry OA

usually means the length of the line segment OA and since the distance from O to A is

the same distance from A to O, then OA = AO. In our vector notation this is the

equivalent of saying |AO| |OA|!!

= which is true.

A 10 m.p.h. northerly wind in Timbuctoo

is identically equal to a 10 m.p.h. northerly wind in Toronto. This is a very important

concept and cannot be overemphasised.

Exercise 1.1

1. Find i) |AC|!

ii) |CA|!

45°

A

O W

N

S

E

A

B C

3

4

4

2. i) Does !!DFDE = ?

ii) Does !!|DF||DE| = ?

3. If !!

ba = does !!

|b||a| = ? Is the converse true?

4. Find all pairs of equal vectors represented in the diagram. ABCD is a parallelogram.

5. Find x so that i)

(x,4)!

= (3,4)!

ii)

| (x,3) |!

= | (3,4) |!

6. Draw a diagram where !!!

|BC| |AC||AB| == but none of !!!BC AC,AB, are equal as

vectors. What is the angle between !!ACand AB ?

Exercise 1.1 Answers

1. i) 5 ii) 5

2. i) No ii) Yes

3. Yes No

4. !!ECAE = ,

!!EBDE = ,

!!BCAD = ,

!!DCAB =

5. i) x = 3 ii) x = 4 or – 4

6. 60° (in an equilateral triangle)

D

E F

2 2

D

A B

C

E

!!EBDE e.g. =

5

150

50

1.2 Addition of Vectors

To add vectors it is clear that we don’t simply add directions and magnitudes

separately. For example, an aeroplane capable of 150 m.p.h. in still air, flying north in an

easterly wind of 50 m.p.h. is certainly not travelling at 200 m.p.h. (or 100 m.p.h. for that

matter) relative to the ground. Addition is best represented diagrammatically.

Draw a diagram so that the plane vector and the wind vector are joined together, as

follows:

The dotted vector represents the addition of the wind

and aeroplane vectors. In fact the plane is flying at

158.1 m.p.h. (Pythagoras) in a direction 18.43° west of

north relative to the ground.

Plane

150 m.p.h. 50 m.p.h.

Wind +

Plane 150

50 Wind

6

Example

To find the addition of the two vectors in the diagram it is best to redraw the diagram as

Note that the flow of the arrows is continuous.

The addition of these two vectors produces a vector (called a resultant) whose magnitude

is 10 (Pythagoras). The direction of the resultant depends upon the directions of the

original vectors.

Example

To add redraw as

6 8

6

8 10

The dotted line represents the vector a +b +c

a

b

c

a

b

c

a

b

c

a + b +c

7

Note that it is important that arrows of a , b and c on the diagram be continuous, i.e. if a

person were to ‘walk’ from the beginning of a to the end of c he would not be ‘stabbed’

i.e.

Example

An object is being pulled by 3 forces as in the diagram.

To find the resultant of these forces draw a new diagram as follows:

The dotted line represents the addition of the three vectors. It has magnitude 8.02 lbs in

fact.

20

25

10

45°

a

b

c

Is not a representation of a +b +c

because the person would be

‘stabbed’ at the first ‘corner’.

10 45°

20

25

8

Example

This is an important idea and will be used often, later on, where O is considered as the

origin.

Exercise 1.2

1. i) Does BCAB+ always equal toAC?

ii) Does CDBCAB ++ always equal toAD ?

iii) Does BCAB + always equal AC ?

2. Show by diagram that ba + = ab + for all a , b .

3. Show by diagram that ba + rarely equals ba + .

4. Find a condition on a and b so that ba + does equal ba + . Is it good

enough that a and b be parallel?

A

B

O

Note that OBABOA =+ regardless of the

positions of O,A,B. An important way of

thinking of this is OAOBAB !=

9

5. In parallelogram ABCD

Represent each of the following as a single vector.

i) BEAB+

ii)

AE + EC

iii) ECBE +

iv) DACDEC ++

v) EABC +

vi) ADAB+

vii) EABADE ++

viii) BCDEAE ++

6. True or False?

i) aba !+

ii) DCAB = means ABCD is a parallelogram

iii) ba + = ca + means cb =

iv) ba + > ba +

v) . )cb(ac)ba( ++=++

D

A B

C

E

10

7.

Express each of the following as a single vector:

i) GCAGOA ++

ii) DEOB +

iii) ADOC +

iv) GFEA +

v) EGGFGA ++

Exercise 1.2 Answers

1. i) Yes ii) Yes iii) No

4. a and b must be in the same direction. No.

5. i) AE ii) AC iii) BC iv)EA v) BE vi) AC

vii) CA viii)AC

6. i) False ii) True iii) True iv) True v) True

7. i) OC ii) OF iii)OF iv)EB v) EB

O

A

B

C

D E

F

G

OAGCFEDB is a cube.

11

Example

Parallelogram ABCD

Example

When subtracting vectors it is best to think of adding (-1) times the vector, e.g.

CBAB! means CB)1(AB !+ which is BCAB+ i.e. AC .

Example

Here BA2

1AB2

1AM !==

BC2

1MN =

A

B C

M N

D

A B

C

E

Here CDAB !=

DB2

1DE =

AC2

1AE =

CA2

1EC !=

A

B

C

CBAB! = BCAB+ =AC

12

Example

ABCD is a parallelogram.

i) AE.BEABEBABDB2

1AB =+=!=!

BC !DC+ ED = BC + CD + ED

= BD + ED = BD+1

2BD =

3

2BD

Exercise 1.3

1.

.

2.

D

A B

C

E

A

B C

M N

M,N are mid-points of AB, AC respectively.

Express MN in terms of scalar multiples of

AC and AB only.

O

A

B

C

D E

F

G

OAGCFEDB is a cube.

i) Express GO in terms of OC,OB,OA

ii) Express CD in terms of OC,OB,OA

ii)

13

3.

4.

C D

E

A

B

M

p

q

r

s

M is the mid-point of AE

Let s,r,q,p be DE,CD,BC,AB respectively

i) Express BM in terms of s,r,q,p

ii) Express MD in terms of s,r,q,p

A

B

D

C a

bO

Let aOA = , let bOB = . C,D are mid-points

of AB, OC respectively.

i) Express AB in terms of a and b

ii) Express OD in terms of a and b

14

5.

6.

7. Draw a geometric figure illustrating the idea:

2( ba + ) = b2a2 +

8. True or False?

i) a2a2 =

ii) a3a3 =!

iii) baba !"+

iv) baba !=+ implies b is the zero vector

O

P Q R S

R bisects PQ. PS = 3

1 PQ

Let pOP = , qOQ =

i) Express OR in terms of p and q

ii) ExpressOS in terms of p and q

iii) Express RS in terms of p and q

p q

AD

BA

C

B

D

E

F

p

q

ABCD is a parallelogram

Let pAD = , qDC =

F is the mid-point of BC

Express the following in terms of p and q

i) AC ii) BD iii) DB

vi) ED v) FD vi) FE

15

v) am = bn implies bm

na = (assume m ≠ 0)

vi) am + bn = cp implies cm

pb

m

na =+ (assume m ≠ 0)

vii) ba + = ca + implies cb =

viii) ra + s a = (r + s) a for any scalars r and s

9. i) What is the magnitude of a

a

1 ?

ii) What is the angle betweena and b if 2b

b

1a

a

1=+

iii) What is the angle betweena and b if 2b

b

1a

a

1=+

10. Forces of magnitude 5, 10 and 10 ‘act’ along different edges of a rectangular

3-dimensional “brick”. Find the magnitude of their resultant.

11. D,E,F are the mid-points of the sides BC, AC and AB respectively in triangle

ABC. Show that

AF+ BD + EAequals the zero vector, i.e. has magnitude

zero.

12. ABCDEF is a regular hexagon. Show that EFFADE =+ .

13. In quadrilateral ABCD DCsAB = where s is some positive scalar. What can

you say about the figure ABCD? What happens if s = 1?

14. Let

! be the angle between a and b (see diagram)

θ

a!

b

!

i) Express ba + in terms of a , b and

cos

! .

ii) Express ba ! in terms of a , b and

cos

! .

16

Exercise 1.3 Answers

1. AB2

1AC2

1!

2. ! OCOBOA,OCOA ++!

3. i) s2

1r2

1q2

1p2

1+++! ii) s

2

1r2

1q2

1p2

1!++

4. i) ab ! ii) b4

1a4

1+

5. i) q2

1p2

1+ ii) q

3

1p3

2+ iii) q

6

1p6

1!

6. i) qp + ii) qp ! iii) pq !

iv) q2

1p2

1! v) qp

2

1! vi) q

2

1!

7.

8. i)T ii) T iii)F iv) F v) T vi) T vii)T viii)T

9. i) 1 ii) 0° iii) 90°

10. 15

13. Trapezoid, parallelogram

14.

a + b =! a 2

+

!

b

2

+ 2a b cos!

a !b =! a 2

+

!

b

2

! 2a b cos"

a!

b

!

2

!b

!a

17

1.4 Geometry Proofs Using Vectors

To prove ABCD is a parallelogram it suffices to show DCAB = because this

guarantees that AB is both parallel and equal to DC in length. (see P. 2)

Note that it also guarantees the coplanarity of A, B, C, D.

Example

“To prove that if diagonals of a quadrilateral bisect each other then the quadrilateral is a

parallelogram”.

Proof

We know that AC and DC bisect each other at E.

Hence: ECAE = and DEEB =

ram.parallelog a is ABCD

DCECDEAB

DEECEBAE

!

=+=!

+=+!

Example

“To prove that the line segment joining the mid-points of two sides of a triangle is

parallel to the third side and equal in length to one half of it”.

D

A B

C

E

18

!DE is parallel to BC and equal to one half of BC.

Exercise 1.4

1. ABCD is a quadrilateral in which AC = BD. M, N, O, P are mid-points of AB,

BC, CD, DA respectively. Show that MNOP is a rhombus.

2. In triangle ABC, D, E, F are mid-points of AB, BC, AC respectively. Medians

AE, BF, CD are concurrent at point O.

Prove OCOBOAOFOEOD ++=++

3. ABCD is a parallelogram. E, F lie on diagonal BD so that BF = ED. Show, using

vectors, that AFCE is a parallelogram.

4. Let A, B, C, D be four non-coplanar points. Let M, N, O, P be mid-points of AB,

BC, CD, DA respectively. Show MNOP is a parallelogram.

5. ABCD is a quadrilateral. M, N are mid-points of diagonals BD and AC

respectively.

Show NM4CDCBADAB =+++

(Hint: - expressNM in four different ways and add)

A

B C

D E

Given ABC! with D,E mid-points of

AB and AC respectively.

BC2

1

)ACBA(2

1

AC2

1BA2

1

AEDADE

=

+=

+=

+=

19

1.5 Geometric Applications of Vectors

Example

Let P, Q, R be three collinear points and O be any point not on the line.

Let Q divide PR in the ratio

! :1- !

For example if Q divides PR into the ratio 5

2:5

3 then PQ = 5

3 PR (i.e.

! =3

5)and hence

.OP5

2OR5

3OQ +=

Note that the scalar 5

3multiplies OR (notOP as one might at first think) i.e. there

is a cross-over effect of the scalars. A good way to think of this is to imagine

! as being

nearly equal to 1 e.g. let

! =7

8. Q is then close to R.

P Q R

O

i.e.

PQ

PR= ! . This means that

PQ = !PR

Note : OQ = OP+ PQ

= OP+ !PR

= OP+ !(OR "OP)

= !OR + (1" !) OP

P Q R

O

8

1

8

7

i.e. OQ is “nearly equal” to OR hence

OQ = OR8

7OP8

1+

20

Note that the converse is also true; namely that if OQ = ORnOPm + where m + n = 1

then P, Q, R are collinear. Furthermore if m or n is negative the converse statement is still

true. For example, if OQ = OR2OP3 ! then this can be easily rearranged to

OR3

2OQ3

1OP += . It is best to do this because this rearrangement avoids all difficulties

related to external division and immediately tells you that P is the point which is between

the other two.

Exercise 1.5

1. If AB5

3DA5

2AC =+ find out which three points are collinear and which point is

between the other two.

2.

3. 3OP! "!!

= mOR! "!!

! 2OQ! "!!. Find m, a scalar, so that P, Q, R are collinear. Determine

which point is between the other two.

In particular if Q is the mid-point of PR then OQ = OR2

1OP2

1+

X Z

O

Y 7

4

7

3

Write OY in terms of OX and OZ

21

4.

Express in terms of a and b .

i) BD ii) DE Make a geometric deduction concerning DE.

5. ORnOQmOA and OR2

1OA2OP +=+=

P, Q, R are collinear. Find m + n.

6. If OR6OQ5OP +!= explain why P, Q, R are collinear and find the ratio into

which R divides PQ.

7. Show that if G is the centroid of triangle ABC and O is any point that

OC3

1OB3

1OA3

1OG ++=

(Hint: medians of a triangle intersect at a point dividing medians in the ratio

)3

1:3

2

8. Convince yourself that if O,P,Q, R are four points such that P, Q, R are collinear,

then OORpOQnOPm =++ where m + n + p = 0.

A

F

B

D

E C

F,D are mid-points of AB, CF

respectively. CE: EB = .4

3:4

1

Let aAC = and bAB =

22

9.

Hint: - G is intersection of four line segments joining vertices to centroids of opposite

faces but it turns out to be better to use the fact that G is also the intersection of the three

line segments joining mid-points of opposite edges.

You might like to show the truth of the hint using vector methods.

10. Let ABCD be a square whose centroid is G. Let O be any external point. Show

that

OG! "!!

=1

4(OA! "!!

+OB! "!!

+OC! "!!

+OD! "!!)

11. i) Show that in any triangle ABC

OGCGBGA =++ where G is the centroid.

ii) Show that in any tetrahedron ABCD

OGDGCGBGA =+++ where G is the centroid.

12. Investigate whether )OA....OAOA(n

1OG n21 ++= where A1, A2, A3… An is a

regular n-gon whose centroid is G and where O is any point. Does it make any

difference if the n-gon is not regular?

D

C

B A

G

Let ABCD be a tetrahedron whose centroid

is G. If O is any point show that

OD4

1OC4

1OB4

1OA4

1OG +++=

23

Exercise 1.5 Answers

1. B – C – D

2. OZ7

4OX7

3OY +=

3. m = 5. P – R – Q

4. i) b4

3a2

1! ii) a

4

1 . DE is parallel to AC.

5. m + n = 4

1

6. Because 6

1:

6

5 .OQ

6

5OP

6

1OR +=

12. i) True ii) Yes, conjecture is now false. (You might like to think about the

possibility of an irregular n-gon in n – 1 space having the required property.)

1.6 More Geometric Applications with Vectors

Example

In the diagram below BD and CE are drawn so that

A

B C

D E

F

6

5:6

1DA:CD =

and

BE: EA = 5

3:5

2

Find BF: FD and CF : FE 6

1

6

5

5

2

5

3

24

Method

Let b.AC and aAB == Join AF.

Let BF: FD = m:1 – m and CF: FE = n: 1 – n

Since B – F – D are collinear and using A as an external point

am)1()b6

5m(AF !+=

Similarly since E- F- C are collinear

)a5

3(nbn)-(1AF +=

In equations and we equate the scalars of b and a separately then we would have a

solution for m and n. Later on we shall see that this is the only solution.

i.e. m6

5 = 1 – n (equating scalars of b in and )

and 1 – m = n5

3 (equating scalars of a in and )

Solving these equations by elementary algebraic methods yields n = 3

1 and m = 5

4 .

i.e. F divides BD into ratio 5

1:5

4

F divides CD into ratio 3

2:3

1

25

Exercise 1.6

1.

Find i) BF:FD ii) CF: FE

2.

Find i) BF:FD and ii) CF:FE

3. 2 ORmOQOP =+ and P, Q, R are collinear.

Find m. Which point is between the other two and into what ratio does it divide

the line segment?

A

B C

D E

F

In triangle ABC, D and E are on AC

and AB respectively so that D is the

mid-point of AC and BE:EA = 3

2:3

1

A

B C

D E

F

In triangle ABC, D and E are on AC

and AB respectively so that AD: DC =

5

2:5

3 and AE:EB = 7

3:7

4

7

3

7

4

5

3

5

2

26

4. If Ocpbnam =++ and m + n + p = 0 show that ac ! and cb ! are parallel

vectors.

5.

i) Prove that CF: FE = 3

1:3

2 and BF: FD =3

1:3

2

ii) Join AF and extend to point G on BC. Show that AG is a median and hence

deduce medians of a triangle are concurrent.

6. True or False? ( means “implies”)

i) ba = ba =

ii) ba + = ab +

iii) ba = ba =

iv) ba ! = ab !

v) ba + = ba !

vi) bma = a and b are parallel

vii) ( )BCAD and BCAB == ABCD is a rhombus

viii) m( ba ! ) = bmam !

ix) c)ba()cb(a +!=!!

A

B C

D E

F

In triangle ABC, BD and CE are

medians intersecting at F.

27

x) If a = 3 and b = 4 and ba + = 5 then a is perpendicular tob .

xi) If a = 3 and b = 4 then ba + < 7

xii) a > 0 for all a

xiii) bmbm = for all scalars m.

xiv) CDAB = BDAC =

7.

8. Three forces act on a body so that it is in equilibrium (i.e. resultant is zero vector).

The magnitude of the forces are 7, 3, and 8 units. Find the angle between lines of

directions of the smallest and largest force.

9.

A B

C D

E

F

P

ABCD is a square. E, F are mid-points

of AB and AD respectively.

Find DP: PE.

A

B C

D E

F

In the diagram, F is the mid-point of

EC and BE: EA = 3

2:3

1

Find BF: FD and AD: DC 3

2

3

1

2

1

2

1

28

10.

Exercise 1.6 Answers

1. i) 23

8:

23

15 ii) 23

9:

23

14

2. i) 2

1:2

1 ii) 4

1:4

3

3. m = 3, P – R – Q. R divides PQ into ratio of 3

2:3

1

6. i)F ii) T iii) T iv) T v) F vi) T vii) T viii) T ix) T x) T xi) T

xii) T xiii) False (m < 0)

7. 5

3:5

2

8. 120°

9. 3

1:3

2 and 4

1:4

3

10. y2x2 !

A B

C F

E D

x

y

ABCDEF is a regular hexagon.

xCD = and yDE =

Find AD in terms of y and x