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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 3-dimensional geometry) coordinates while in Euclidean Geometry a point is a single entity. What’s 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’s 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 tilda “~” if we are writing it by hand, or using bold type in print. We write a vector as v~ while in print it appears as v.

A vector, written as (x, y), can be called a row vector. Often we write it vertically x

y , and then it’s 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’ll 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, y)T = x

y and x

yT = (x, y).

We say that x

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.

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§1.2. Addition of Vectors and Scalar Multiplication We define addition of vectors component by component.

x1x2

…xn

+

y1y2

…yn

=

x1 + y1x2 + y2

…..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.

Example 2: 3

2 + 6

1 = 9

3 ; (2, 4) + (5, 0) = (7, 4); 3

4 + (2, 3)T =

5

7 . We define the product of a vector by a scalar in a similar way.

k

x1x2

…xn

=

kx1kx2

…kxn

.

Example 3: Suppose Sue’s vital statistics vector is v =

100

7095

and the amount she expects to lose

per week, during a diet, is represented by the vector d =

2

12

. Then her vital statistics vector after 6

weeks would be v − 6d =

88

6483

.

The n-dimensional zero vector is the vector 0 =

00

…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

put u =

x1x2

…xn

etc and evaluate both sides of each equation.

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§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. hus we might speak of the distance between two vectors (here we’re thinking of them as points) or the length and direction of a vector (here we’re 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.

[If u =

x1

y1 and v =

x2

y2 then the above diagram shows that u + v =

x1 + x2

y1 + y2 is the fourth point of the parallelogram.] (2) The point dividing the directed line segment joining u to v in the ratio k :1 is

w =

1

1+k u +

k

1+k v.

[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).

u + v

u

v

0

x2

y1

x1

y1

y2

x1

u

v

w k

1

8

(3) The set of points of the form w = (1 − λ)u + λv

is the straight line joining u and v. (Just put λ = k

1+k ).

For 0 < λ < 1, w lies between u and v. For λ < 0, w lie beyond u. For λ > 1, w lies beyond v.

(4) The vector v − u has the same length and direction as the directed line segment from u to v. 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: 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

and so is 13 u +

23

1

2( )v + w = 13( )u + v + w . By symmetry this is the same point that is 2/3 of the

way down each of the other two medians. Hence all three medians intersect at this point. §1.4. Lengths and Angles The inner product of the two real vectors a =

a1

a2 and b =

b1

b2 is defined to be: a . b = a1b1 + a2b2.

Notice that the inner product of two vectors is a scalar.

If a = x

y then a . a = x2 + y2.

The length of the vector a is x2 + y2 = a . a . (We can prove this by Pythagoras’ Theorem.) 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|.

λ > 1

λ = 1

λ = 0 λ < 0 λ = ½

u

v

u

v

0

v − u

u

v

w

½(v+w)

1/3(u+v+w)

9

0

|b| θ

b |b − a|

|a| a

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 . 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.

Proof: Let a =

a1

a2 etc and evaluate both sides of each equation.

We measure the angle, θ, between two vectors in the range 0 ≤ θ ≤ 180 (in degrees).

Theorem 3: If θ is the angle between the vectors a and b, then:

cos θ = a . b|a|.|b| .

Proof: From the cosine rule, applied to the triangle: |a|2 + |b|2 − 2|a|.|b| cos θ = |b − a|2 = (b − a) . (b − a) = b . b − a . b − b . a + a . a = |a|2 + |b|2 − 2 a . b and so the result follows. 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 .

Thus the required angle is θ where cos θ = 1

5 2 = 10

2 ≈ 0.1414, so θ ≈ 81°52′.

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 = b . b − a . a since b . a = a . b =|b|2− |a|2 = 0.

a

0

b a + b

θ

10

We’ve 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. Example 8: A parallelogram has diagonal of 30cm and 40cm. One side is twice as long as the other. What are the lengths of the sides? Solution: Let the vertices of the parallelogram be 0, a, b and a + b, with a being the longer side, and let the sides have lengths x and 2x. Now 1600 = |a + b|2 = (a + b)(a + b) = a.a + b.b + 2a.b = |a|2 + |b|2 + 2a.b = 5x2 + 2a.b and 900 = |a − b|2 = (a − b)(a − b) = a.a + b.b − 2a.b = |a|2 + |b|2 − 2a.b = 5x2 − 2a.b. Adding we get 2500 = 10x2 Hence x = 5 10 ≈ 15.81 so the sides are approximately 15.81cm and 31,62cm. 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.

The projection of a onto b is |a| cos θ = a . b|b| by theorem 3.

§1.5. An Application of Vectors to Statistics

A good deal of what was done for 2-dimensional 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 n-dimensional space algebraically, even if we can’t imagine more than 3 dimensions. Vector algebra is 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

x_ =

x1 + x2 + … + xnn .

|a| b

a

0

θ

11

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:

(x1 −x_)2 + (x2 −x

_)2 + … + (xn −x

_)2

n

. 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

(x1 −x_)2 + (x2 −x

_)2 + … + (xn −x

_)2

is the distance, in n-dimensional space, between the vectors x and x 1 or alternately, the length of

the vector x − x_ 1. Thus the standard deviation is given by σx =

|x − x_ 1|

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

ρ = (x1 − x

_ )(y1 − y

_ ) + (x2 − x

_ )(y2 − y

_ ) + … + (xn − x

_ )(yn − y

_ )

nσxσy .

The numerator is simply the inner product of the vectors x − x_ 1 and y − y

_1.

Moreover since nσx = |x − _ x 1 | and n σy = |y − y

_ 1| we may write:

ρ = (x − x− 1) . (y − y− 1)

|x − x− 1| . |y − y−1| = cos θ

where θ is the angle between the vectors x − x_ 1 and y − y

_1 in n-dimensional space. (We’re taking

a lot for granted about n-dimensional space. You shouldn’t 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 anti-correlation.

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The square of the standard deviation is called the variance. A well-known formula in statistics, that makes it easier to compute variance, and hence standard deviation, is the following:

σx2 = x12 + x22 + … + xn2

n − x_ 2.

Theorem 4: The variance of a sample (x1, x2, …, xn) is σx2 = x12 + x22 + … + xn2

n − x_ 2.

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 10: A report of a survey on the 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 Maths 1.0 0.7 0.8 Music 0.7 1.0 −0.1 Science 0.8 −0.1 1.0

Why is this report erroneous? Solution: Let u, v, w be the 200-dimensional vectors whose components are the maths, music and science marks respectively. Extending the 2-dimensional formula for the angle between two vectors to 200 dimensions, we can calculate the angles between the vectors u, v and w. 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 and three dimensions to two hundred, we’d conclude that γ can’t 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. 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. The report is wrong! Who would have thought that geometry could be useful in statistics!

v u

0

w

x

x_ 1

x − x_ 1

13

EXERCISES FOR CHAPTER 1

Exercise 1: If u =

−3

4 and v =

12

5 find

(i) u + v; (ii) |u + v|; (iii) |u| + |v|; (iv) u.v; (v) |u|.|v|.

Exercise 2: If a = 1

1 and b =

7

−7 find

(i) |b − a|; (ii) |b| − |a|; (iii) a.b; (iv) a.(a + kb).

Exercise 3: (a) Find the angle between the vectors u = 1

1 and v = 2

3 ;

(b) Find the angle between the lines AB and AC where A = 1

2 , B = 1

1 and C = 2

3 .

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;

(iii) cos θ = |a|2 + 2(a.b)b

2|b|.|c| where θ is the angle COB.

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

Exercise 1: (i) 9

9 ; (ii) 9 2 ; (iii) 18; (iv) −16; (v) 65. Exercise 2: (i) 10; (ii) 6 2 ; (iii) 0; (iv) 2.

A

B

C

O

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Exercise 3: (a) u.v = 5, |u| = 2 , |v| = 13 so cos θ = 526

≈ 0.98058, so θ ≈ 11° 21′.

(b) This is the angle between the vectors

0

−1 and 1

1 , which is cos−1−1

2 = 135°.

Exercise 4: ∠ 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.

(iii) cos ∠COB = b.c

|b|.|c|

= a.c

|b|.|c|

= a.( ½a + b)

|b|.|c|

= a.(a + 2b)

2|b|.|c|

= |a|2 + 2(a.b)b

2|b|.|c|

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).

A

B

C

D

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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|2.|b|2.|a − b|2 − |a|2.|b|2.|a − b|2.(a.b)2

|a|2.|b|2

= |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.

16