1. VECTORS IN 2 DIMENSIONS
§1.1. Geometry Through the Ages
For the Greeks, geometry was an art form in which a small number of “self-evident”
propositions were cleverly used to build up elegant proofs about space. René Descartes, in the
seventeenth century, introduced coordinates and this enabled geometric problems to be solved
algebraically in a systematic, though not always elegant, way. The
clumsiness that often arises in Cartesian geometry is due partly to
the fact that each point is represented by two (or three, for 3dimensional geometry) coordinates while in Euclidean geometry a
point is a single entity. What is needed is a way of representing a
point by a single algebraic object.
An ordered n-tuple of real numbers (x1, x2, …, xn) is called a
row vector and the numbers are called the components of the
vector. The number of components is called the dimension of the
vector and we call a vector of dimension n an n-dimensional vector.
Example 1: The so-called “vital statistics” of a woman can be regarded as a 3-dimensional vector
(b, w, h) where b is the bust measurement, w is the circumference of
the waist h is the hip measurement (all in centimetres). It’s a 3-tuple
(or triple) and, moreover, it is an ordered triple. The order of the
components is important. There is a considerable difference between
two women whose vital statistics are (90, 50, 85) and (50, 85, 90)!
A point in the plane can be represented by a row vector (x, y)
where x and y are its coordinates. But, in Cartesian geometry,
isn’t this the way we normally represent points? That’s true. The
important difference is that in vector geometry we think of the vector as a single object and
combine vectors as single entities. Instead of referring to a point (x, y) we can refer to it as the point
corresponding to the vector v. If necessary we can split v into its coordinates, by writing v = (x,
y) but if possible we avoid doing so.
We distinguish between vectors and scalars (real numbers) by underlining a vector with a
v
tilda “~” if we are writing it by hand, or using bold type in print. We would write a vector as ~
while in print it would appear as v.
x
A vector, written as (x, y), can be called a row vector. Often we write it vertically y, and
then it is called a column vector. There’s more than mere typographic significance in our being
able to write vectors in rows or columns. There’s a mathematical reason for having both types, as
will become apparent later.
Frequently we will need to convert a row vector into a column vector or vice-versa. To do
this we have an operation called transposition. We denote it by the symbol “T”. Thus
x
x T
x
(x, y)T = y and y = (x, y). We say that y and (x, y) are transposes of one another. Note
that (vT)T = v for all vectors v.
Usually we write column vectors in the form v and row vectors in the form vT. (The
reason for regarding column vectors as the “normal” kind will become apparent later.) In
accordance with this we’ll usually represent points by column vectors, even though this takes up
more space.
5
§1.2. Addition of Vectors and Scalar Multiplication
We define addition of vectors component by component.
 xx12   yy12  xx12 ++ yy12
… + … =  …..  .
 xn   yn  xn + yn
Similar definitions apply for row vectors and for subtraction of row or column vectors. Note
that we can only add or subtract vectors if they are of the same type and have the same dimension.
3 6
9
Example 2: 2 + 1 = 3 ;
(2, 4) + (5, 0) = (7, 4);
3
5
T
4 + (2, 3) = 7 .
We define the product of a vector by a scalar in a similar way.
1
 xx12  kx
kx2
k … =  …  .
 xn  kxn
100
Example 3: Suppose Sue’s vital statistics vector is v =  70  and the amount she expects to lose
 95 
2
per week, during a diet, is represented by the vector d = 1 . Then her vital statistics vector after 6
2
88
weeks would be v − 6d = 64 .
83
 00 
The n-dimensional zero vector is the vector 0 = … .
0
Theorem 1: The following hold for all vectors u, v, w (where defined):
(1) (Commutative Law) u + v = v + u;
(2) (Associative Law) u + (v + w) = (u + v) + w;
(3) (Additive Identity) u + 0 = 0;
(4) k(u + v) = ku + kv;
(5) (k + h)u = ku + hu;
(6) (kh)u = k(hu);
(7) 0v = 0;
(8) k0 = 0.
Proof: All eight properties are quite straight-forward and should be obvious. To prove them simply
 xx12 
put u = … etc and evaluate both sides of each equation.
 xn 
6
§1.3. Vectors and Geometry
A 2-dimensional real vector can represent a point in the plane.
It can also represent the directed line segment (piece of a line with a
given direction) from the origin to that point. Moreover we often
identify the vector with the point or the line segment. Thus we might
speak of the distance between two vectors (here we are thinking of
them as points) or the length and direction of a vector (here we are
thinking of the vector as a directed line segment). The context will
always make clear which of the alternatives is meant.
The following demonstrate the geometric significance of vector addition and multiplication of
vectors by scalars.
(1) If a parallelogram is formed from the three points 0, u, v with the points u, v on opposite ends
of a diagonal, the fourth point is u + v.
u+v
u
y1
v
y1
y2
0
x1
x1
x2
x1
x2
x1 + x2
[If u = y  and v = y  then the above diagram shows that u + v = y + y  is the fourth point of
1
2
1
2
the parallelogram.]
(2) The point dividing the directed line segment joining u to v in the ratio k :1 is
 1 
 k 
w = 1+k u + 1+k v.
 


1
k
u
[w − u has the same direction, and k times the length, of v − w so w − u = k(v − w).
Thus w(1 + k) = u + kv.]
In particular, the midpoint of the line segment joining u to v is ½ (u + v).
7
w
v
(3) The set of points of the form
w = (1 − λ)u + λv
k
is the straight line joining u and v. (Just put λ = 1+k ).
u
For 0 < λ < 1, w lies between u and v.
For λ < 0, w lie beyond u.
For λ > 1, w lies beyond v.
λ<0
λ=0
v
λ>1
λ=1
λ=½
(4) The vector v − u has the same length and direction as the directed line segment from u to v.
v
v−u
u
0
Example 4: A median of a triangle is a line joining the vertex to the midpoint of the opposite side.
Prove that the three medians of a triangle intersect at a point 2/3 of the way down each median.
Solution:
u
1/3(u+v+w)
w
v
½(v+w)
Let the vertices be u, v and w. The midpoint of the line segment from v to w is ½ (v + w). The
point 2/3 of the way down the median from u to this midpoint divides the median in the ratio 2 : 1
1
2 1
 1
and so is u +  (v + w) = (u + v + w) . By symmetry this is the same point that is 2/3 of the
3
3 2
 3
way down each of the other two medians. Hence all three medians intersect at this point.
§1.4. Lengths and Angles
a1
b1
The inner product of the two real vectors a = a  and b = b  is defined to be:
2
2
a . b = a1b1 + a2b2.
Notice that the inner product of two vectors is a scalar.
x
If a = y then a . a = x2 + y2.
The length of the vector a is x2 + y2 = a . a . It is
denoted by |a|. So a . a = |a|2.
The distance between two vectors a and b is the
length of a − b, that is, |a − b|.
Example 5: Find the distance between the points a = (1, 2) and b = (−1, 6).
Solution: The distance = |a − b| = |(2, −4)| = 4 + 16 = 20 = 2 5 .
8
Theorem 2: For all vectors a, b and c (of the same dimension):
(1) a . b = b . a;
(2) a . (b + c) = a . b + a . c.
a1
Proof: Let a = a  etc and evaluate both sides of each equation.
2
We measure the angle, θ, between two vectors in the range 0 ≤ θ ≤ 180 (in degrees).
θ
Theorem 3: If θ is the angle between the vectors a and b, then:
a.b
.
cos θ =
|a|.|b|
Proof: From the cosine rule, applied to the triangle:
b
|b|
|b − a|
θ
0
|a| + |b| − 2|a|.|b| cos θ = |b − a|
= (b − a) . (b − a)
=b.b−a.b−b.a+a.a
= |a|2 + |b|2 − 2 a . b and so the result follows.
2
2
2
a
|a|
Two vectors a and b are orthogonal if a . b = 0. It follows that two non-zero vectors are
perpendicular if and only if they are orthogonal.
Example 6: Find the angle between the vectors a = (4, −3) and b = (1, 1).
Solution: a . b = 4.1 + (−3).1 = 1; |a| = 22 + 32 = 5; |b| = 2 .
1
2
Thus the required angle is θ where cos θ =
= 10 ≈ 0.1414, so θ ≈ 81°52′.
5 2
Example 7: Prove that the diagonals of a rhombus (parallelogram with four equal sides) intersect at
right angles.
Proof: Let the rhombus have vertices 0, a, b, a + b where |a| = |b|.
Now (b − a) . (b + a) = b . b + b . a − a . b − a . a
a+b
= b . b − a . a since b . a = a . b
b
=|b|2− |a|2
= 0.
a
0
9
We have shown that b − a is orthogonal to a + b. But these vectors represent the lengths and
directions of the two diagonals. Hence the diagonals are perpendicular.
The projection of a vector a onto a vector b is the signed distance from the origin to the
foot of the perpendicular that goes from a onto the line joining 0 to b. The sign is positive if the
foot of the perpendicular is on the same side of the origin as b and negative if it is on the opposite
side.
a
|a|
b
θ
0
The projection of a onto b is |a| cos θ =
a.b
by theorem 3.
|b|
§1.5. An Application of Vectors to Statistics
A good deal of what was done for 2dimensional vectors extends to n-dimensional ones.
We may not be able to visualise the point (1, 2, 4, 0, 3)
in 5-dimensional space, but we can still carry out
vector addition, scalar multiplication and inner
products. In other words we can talk about ndimensional space algebraically, even if we can’t
imagine more than 3 dimensions.
Vector algebra can be used in statistics.
Suppose that a set of n measurements of some attribute is collected, for example, the weights of
fish in a catch, or the test marks of students in some class. We can represent that sample by the row
vector (x1, x2, …, xn).
If the measurements are x1, x2, … , xn the mean (or average) is defined to be
_ x1 + x2 + … + xn
x =
.
n
While capturing some of the important information from the data the mean takes no account
of the spread of the data. For example the mean of a set of test marks out of 100 might be 42. This
might also be the mean neck measurement (in centimetres) of a sample of men. Yet the spread
would be very different. A test mark of 10 out of 100 would not be uncommon, while a man with a
neck circumference of 10 cm would be quite a freak! Test marks have a greater spread than neck
measurements.
A measure of the amount of spread in a set of data is the standard deviation. The standard
deviation of the sample of measurements x1, x2, …, xn is defined to be
(x − x ) + (x
2
2
1
)
(
− x 2 +  + xn − x
n
)
2
.
A set of measurements x1, x2, …, xn can be thought of as the n-dimensional row vector
x = (x1, x2, …, xn). If 1 denotes the vector (1, 1, …, 1) then
(x − x ) + (x
2
1
2
)
(
− x 2 +  + xn − x
10
)
2
is the distance, in n-dimensional space, between the vectors x and x 1 or alternately, the length of
_
_
|x − x 1|
the vector x − x 1. Thus the standard deviation is given by σx =
.
n
If two sets of measurements are taken on a sample of size n, there is often a correlation
between them. For example if we measure the heights and weights of a sample of people we would
tend to find that taller people were heavier and shorter people were lighter. The connection
between height and weight is not one of perfect correlation however, since there are tall thin people
who are light in comparison to their height.
A measure of the correlation between two sets of measurements is the correlation
coefficient. For two sets of readings, x1, x2, …, xn and y1, y2, …, yn the correlation coefficient is
defined to be
_
_
_
_
_
_
(x1 − x )(y1 − y ) + (x2 − x )(y2 − y ) + … + (xn − x )(yn − y )
ρ=
.
nσxσy
_
_
The numerator is simply the inner product of the vectors x − x 1 and y − y 1. Moreover since
_
_
(x − −
x 1) . (y − −y 1)
nσx = |x − x 1 | and n σy = |y − y 1| we may write ρ =
= cos θ
|x − −
x 1| . |y − −y 1|
A
A
EA
A
A
A
A
A
A
A
A
A
EA
E
A
E
A
A
E
A
A
E
EA
EA
A
E
A
A
E
A
A
E
A
A
E
A
_
_
where θ is the angle between the vectors x − x 1 and y − y 1 in n-dimensional space. (We are
taking a lot for granted about n-dimensional space. You should not concern yourself with trying to
justify the extensions of geometric concepts from 2-dimensional space to n-dimensional space.)
Because the correlation coefficient is the cosine of an angle it ranges between −1 and +1. A
correlation of +1 represents perfect correlation and a correlation of −1 represents perfect anticorrelation.
A
E
A
A
E
A
x12 + x22 + … + xn2 _ 2
−x .
n
_
_
_ _
_
_
_
_
Solution: (x 1).(x − x 1) = ( x , x , …, x ).(x1 − x , x2 − x , … , xn − x )
_
_
_ _
_
= x (x1 + x2 + … + xn − n x ) = x (n x − nx ) = 0.
_
_
Thus the vectors x 1 and x − x 1 are orthogonal.
_
_
By Pythagoras’ Theorem |x|2 = |x − x 1|2 + |x 1|2 and so
_
x12 + x22 + … + xn2 = nσx2 + nx 2.
Rearranging, we get the desired formula for σx2.
Example 8: Prove that σx2 =
A
E
A
A
E
A
A
A
A
E
A
E
E
E
A
A
E
A
A
E
A
A
A
A
E
A
A
A
E
A
A
A
A
E
E
A
A
E
A
E
A
A
A
E
A
A
E
E
_
x−x1
A
A
x
A
A
E
A
A
E
A
A
E
_
x1
A
Example 9: A report of a survey of school students performance in various subjects claimed that on
a sample of 200 students, the correlation coefficients between marks in mathematics, music and
science were given by the following table.
Maths Music Science
1.0
0.7
0.8
Maths
0.7
1.0
Music
−0.1
0.8
1.0
Science
−0.1
Why is this report erroneous?
11
Solution: Let u, v, w be the 200-dimensional vectors whose components are the maths, music
and science marks respectively. Let α be the angle between u and v, let β be the angle between u
and w and let γ be the angle between v and w.
If we extend our geometric intuition from two dimensions to two hundred we would conclude that γ
cannot be greater than α + β. In fact γ can only equal α + β if all three vectors u, v and w lie
in the same plane.
v
u
w
0
Now the correlation coefficient between maths and music is cos α = 0.7, and so α = 45°34′
and the correlation coefficient between maths and science is cos β = 0.8, and so β = 36°52′
and the correlation coefficient between music and science is cos γ = −0.1, and so γ = 95°44′.
Since γ > α + β we have a contradiction.
EXERCISES FOR CHAPTER 1
−3
12
Exercise 1: If u =   and v =  5  find
4
(i) u + v; (ii) |u + v|; (iii) |u| + |v|; (iv) u.v; (v) |u|.|v|.
A
E
A
A
E
A
7
1
Exercise 2: If a = 1 and b =   find
−7
(i) |b − a|; (ii) |b| − |a|; (iii) a.b; (iv) a.(a + kb).
A
E
A
A
E
A
2
1
Exercise 3: (a) Find the angle between the vectors u = 1 and v = 3 ;
A
E
A
A
E
A
1
1
2
(b) Find the angle between the lines AB and AC where A = 2 , B = 1 and C = 3 .
A
E
A
A
E
A
A
E
A
Exercise 4: Find the four angles of the quadrilateral with vertices A(1, −3), B(2, 4), C(3, −1) and
D(0, 5).
Exercise 5: Prove, using vectors, that the angle subtended by a diameter of a circle at the
circumference is a right angle.
Exercise 6: In the following diagram, OC is perpendicular to AB. OA is parallel to, and has twice
the length of, BC. If O, A, B and C are represented by the vectors 0, a, b, c respectively prove that
(i) c = ½ a + b;
(ii) a.c = b.c;
|a|2 + 2(a.b)b
(iii) cos θ =
where θ is the angle COB.
A
2|b|.|c|
C
A
E
A
E
B
O
12
Exercise 7: Prove that if a is equidistant from b and c if and only if |b|2 − 2a.b = |c|2 – 2a.c
and hence prove that the triangle with vertices a, b, c is equilateral if and only if
|a|2 + 2b.c = |b|2 + 2a.c = |c|2 + 2a.b.
Exercise 8 (Harder): Prove that the vector (a.b − |b|2)a + (a.b − |a|2)b is orthogonal to the line
joining a to b. Prove that its length is |a|.|b|.|a − b| sin θ where θ is the angle between a and b.
SOLUTIONS FOR CHAPTER 1
9
Exercise 1: (i) 9 ; (ii) 9 2 ; (iii) 18; (iv) −16; (v) 65.
A
E
A
A
E
A
Exercise 2: (i) 10; (ii) 6 2 ; (iii) 0; (iv) 2.
A
E
A
5
≈ 0.98058, so θ ≈ 11° 21′.
26
−1
0
1
(b) This is the angle between the vectors   and 1 , which is cos−1
= 135°.
−1
2
Exercise 3: (a) u.v = 5, |u| = 2 , |v| = 13 so cos θ =
A
E
A
A
A
E
A
E
A
A
E
A
A
E
A
A
E
A
Exercise 4:
D
B
C
A
∠ DAC = 52°8′, ∠ ACB = 123°41′, ∠CBD = 127°52′, ∠ BDA = 56°19′.
As a check we see that the sum of these angles is 360°, as it should be for any quadrilateral.
Exercise 5: Let the centre be 0 and let the ends of the diameter be −u and u.
Let v be a typical point on the semicircle.
Then (v + u).(v − u) = v.v − u.u = |u|2 − |v|2 = 0 since u, v lie on the circle.
Hence v + u and v − u, and so the lines joining v to the ends of the diameter, are orthogonal.
Exercise 6: (i) OA is represented by a and BC is represented by c − b.
Since OA is parallel to BC, and twice its length, a = 2(c − b).
Hence c = ½ a + b.
(ii) Since AB is perpendicular to OC, a − b is orthogonal to c.
Hence (a − b).c = 0.
∴ a.c = b.c.
b.c
(iii) cos ∠COB =
|b|.|c|
a.c
=
|b|.|c|
A
A
E
E
13
a.( ½a + b)
|b|.|c|
a.(a + 2b)
=
2|b|.|c|
|a|2 + 2(a.b)b
=
2|b|.|c|
=
A
E
E
A
E
E
A
E
E
Exercise 7: Since |a − b| = |a − c|, |a − b|2 = |a − c|2.
∴ (a − b).(a − b) = (a − c).(a − c).
∴ |a|2 + |b|2 − 2(a.b) = |a|2 + |c|2 − 2(a.c).
∴ |b|2 − 2(a.b) = |c|2 − 2(a.c).
For an equilateral triangle we have |b|2 + 2(a.c) = |c|2 + 2(a.b) = |a|2 + 2(b.c).
This argument is reversible, so the condition is both necessary and sufficient (provided we allow a
single point, where a = b = c, to be considered to be an equilateral triangle).
Exercise 8: Let x = a.b − |b|2 and y = a.b − |a|2 so that the vector is xa + yb.
Hence (xa + yb).(b − a) = x(a.b) − x|a|2 + y|b|2 − y(a.b)
= (x − y)(a.b) + y|b|2 − x|a|2.
Now x − y = |a|2 − |b|2 so
(xa + yb).(b − a) = (|a|2 − |b|2)(a.b) + (a.b − |a|2) |b|2 − (a.b − |b|2)|a|2
= |a|2(a.b) − |b|2(a.b) + (a.b) |b|2 − |a|2|b|2 − (a.b)|a|2 + |a|2|b|2
= 0.
Hence xa + yb is orthogonal to b − a.
Now |xa + yb|2 = (xa + yb).( xa + yb)
= x2|a|2 + y2|b|2 + 2xy(a.b).
x2 = (a.b)2 + |b|4 − 2(a.b)|b|2,
y2 = (a.b)2 + |a|4 − 2(a.b)|a|2 and
xy = [(a.b)2 − (|a|2 + |b|2)(a.b) + |a|2|b|2
∴ |xa + yb|2 = [(a.b)2 + |b|4 − 2(a.b)|b|2]|a|2 + [(a.b)2 + |a|4 − 2(a.b)|a|2]|b|2
+ 2[(a.b)2 − (|a|2 + |b|2)(a.b) + |a|2|b|2](a.b)
= (a.b)2|a|2 + |a|2|b|4 − 2(a.b)|a|2|b|2 + (a.b)2|b|2 + |a|4|b|2 − 2(a.b)|a|2|b|2
+ 2(a.b)3 − 2|a|2(a.b)2 + 2|b|2(a.b)2 + 2|a|2|b|2(a.b)
= − (a.b)2|a|2 + |a|2|b|4 − 2(a.b)|a|2|b|2 − (a.b)2|b|2 + |a|4|b|2 + 2(a.b)3
Now [|a|.|b|.|a − b|.sin θ]2 = |a|2.|b|2.|a − b|2.sin2 θ
= |a|2.|b|2.|a − b|2.(1 − cos2 θ)
(a.b)2
= |a|2.|b|2.|a − b|2 − |a|2.|b|2.|a − b|2. 2 2
|a| .|b|
= |a|2.|b|2.[(a − b).(a − b)] − [(a − b).(a − b)].(a.b)2
= |a|2.|b|2.[|a|2 + |b|2 − 2(a.b)] − [|a|2 + |b|2 − 2(a.b)](a.b)2
= |a|4.|b|2 + |a|2|b|4 − 2(a.b)|a|2|b|2 − |a|2(a.b)2 − |b|2(a.b)2 + 2(a.b)3.
Hence |xa + yb|2 = [|a|.|b|.|a − b|.sin θ]2.
Taking positive square roots, |xa + yb| = |a|.|b|.|a − b|.sin θ.
A
E
E
14
A