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composed by J.M.E.Hyland
*
From its beginning mathematics has been concerned with abstract ideas: points, lines, circles and other geometrical constructions; numbers, their addition, multiplication and factorisation into prime factors. As mathematics has developed, less obvious abstract notions have been discovered, whose value is that they can be applied to solve problems in a wide variety of areas. Here we shall discuss elementary aspects of one such very important notion, that of vectors.
Vector in Latin means carrier and it also has that meaning in English: the mosquito is the vector of malaria. The original notion of vector in mathematics was of an operation which carried one point in space to another. We shall make this precise in terms of elementary
Euclidian geometry, so we assume that you know a bit about geometry in the plane. One can do a lot of interesting things with vectors in the plane, but we shall also discuss three dimensional vectors just a little. (As you may suppose, we can consider vectors in all dimensions, but we shall not worry about that here.)
We start by describing vectors abstractly and initially you may find that hard to understand. Please be patient. After some practice using vectors in geometry, the ideas should become clearer.
The idea of a vector can be abstracted from that of a translation . A translation (of the plane in the 3D space) is an operation moving each point a fixed distance in a fixed direction; and the vector captures the notion of a fixed distance in a fixed direction. (One often says
‘vectors have magnitude and direction’.) In the picture:
A`
B`
A
B
The translation taking A to A` also takes B to B` . Geometrically this means that the lines AA` and BB` are parallel and of equal length. In other words, ABB`A` is a parallelogram.
A`
B`
A
B
* Please, send comments, corrections and queries at kf262@dpmms.cam.ac.uk
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We write
→
AA` for the vector corresponding to the translation taking A to A` . Any line segment with the same length and direction as AA` represents the same vector. In particular, in the diagram above
→
AA`
→
= BB` .
Test Question 1 And also
→
AB =
→
A`B` . Why?
We use boldface letters which we write a , b , c , … (with underline) to denote vectors.
Write T a for the translation by the vector a . (Note that a translation is completely determined by what it does to a single point.) To say a =
→
AA` is to say that a is the vector determined by the translation taking A to A` ; and that is equivalent to saying T a
(A)=A`.
The zero vector.
Do not overlook the translation which leaves things as they are! We write 0 =
→
AA and observe that T
0
(B)=B for all B ; so T
0
is the identity translation. (Note that it is a bit odd to say that the vector 0 has magnitude and direction : but that is a loose way of talking.)
The negative of a vector.
Suppose a is a vector. Take points O , A so that a =
→
OA .
Then we define a by -a=
→
AO : we have the opposite direction.
Test Question 2 It does not matter which points we took with a =
→
OA . Why?
Addition of vectors Suppose a and b are vectors. Take points O , B so that b =
→
OB ; and let C = T a
( B ) so that a=
→
BC . Then we define a+b a+b by a+b=
→
OC .
C b a
O B
Test Question 3 Again it does not matter which points O , B we started with. Why?
One way to understand the last two exercises is to observe
(i) that a is defined so that T
-a
( T a
( P ))= P
(ii) that a + b is defined so that T a + b
( P )= T a
( T b
( P )).
For that to make sense you should prove the following.
Test Question 4 If you take one translation followed by another than the result is still a translation.
We now have some laws of vector additions:
( a + b ) + c = a + ( b + c ) a + b = b + a a + 0 = a a + -a = 0
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Test Question 5 Interpret and prove these laws. (If you find these confusing, wait until we have given a more concrete view of vectors.)
Remark.
An abstract idea of great importance in mathematics is that of a group. The laws above say that the collection of vectors forms an abelian (that is, commutative: a + b = b + a ) group.
Scaling vectors Suppose we have a vector a and a real number λ . For λ ≥ 0 it is intuitively clear that we can consider a vector λ a whose direction is the same as that of a but whose magnitude is λ times that of a : for λ =1 we get a of course and for λ =0 we get 0 . We can extend this to the case when ≤ 0 the direction, i.e. ⋅
λ if we take the view that multiplying by -1 should reverse
− 1 a = - a . This gives us a scalar multiple λ a of a for all λ . We write some laws for this scalar multiplication:
λ ( a + b ) = λ a + λ b
( λ + µ ) a = λ a + µ a
0 ⋅ a = 0
λ ⋅ a = 0
Test Question 6 Interpret and prove these laws. (Again you may wish to wait until we have a more concrete view of vectors.)
Remark . We have now given the basic laws of vector algebra. They say that vectors form a vector space , another important abstract notion. Do not worry about that now. The main thing is to get used to vector algebra. It is no harder than ordinary algebra.
Our first description of vectors was abstract and it is often better to think of them more concretely.
Pick a point O (in the plane or 3D space) to be our origin of reference. Then a vector a can be identified with the point A such that
→
OA = a , or equivalently, such that T a
( O )= A ; that is, a corresponds to the point A which is a away from the origin O . This enables us to think of vectors as points; and from that point of view the structure of the vectors makes clear geometric sense.
• The zero vector corresponds to the origin O .
• The negative of a vector is obtained by “reflecting” points in the origin:
b a
a
O b
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• Addition of vectors is given by a parallelogram from the origin:
O b a a+b
• Scalar multiplication is given by contraction or expansion along the line of the vector:
λ
− a
∞
for
< λ < − 1
-a
λ
− 1 a for
< λ < 0
O
λ
0 a
<
for
λ < 1 a
λ
1 < a for
λ < ∞
The geometric view of vectors as points is particularly useful: it should make the laws of vector algebra seem more or less geometrically obvious. We shall see also that we can use vectors to do geometry. So often we assume an origin O is given and use the convention that points A , B , C correspond to vectors a , b, c . The basic observation then is that vector b a as in the diagram b
→
AB b - a
→
AB is the
O
This makes sense as ( b – a )+ a = b . a
You should be asking what happens if we chose a different point P say as origin of reference. Then of course the vectors corresponding to points will be different. Suppose that
→
OP = p . Then the picture
A
O p a
P makes it clear that if A has vector a with O as origin, then it has vector a – p with P as origin.
That is because
→
PA = a – p .
It is important to realize that while vectors change with a change of origin, the geometrical relations between the corresponding points will remain the same. One can often exploit this.
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Example 2.1
Show that ABCD forms a parallelogram if a + c = b + d .
Solution.
Take A as new origin of reference. With A as origin, B , C , D correspond to vectors b` = b – a c` = c – a d` = d – a respectively.
D
C d a
A b - a
B
But the rule for vector addition with origin A gives us that ABCD is a parallelogram if and only if c` = b` + d` , that is, if and only if c – a = b – a + d – a .
Rearranging we get ABCD a parallelogram if and only if a + c = b + d .
Suppose A ≠ B are distinct points so that they determine a unique line ( in the plane or in 3D-space, it makes no difference). Any point in the extended line AB can be got by going from to A and then moving along AB by some scalar multiple of
→
AB . So identifying points with their vectors, it is a vector of form a + λ ( b – a ) = (1λ ) a + λ b for some λ . We can say that v = (1λ ) a + λ b is the equation for the line through the points a and b . (You should think of v as a variable vector – varying with λ - but we shall not use this formalisation here.) Note the following.
• When λ =0 we get the point a .
• When λ =1 we get the point b .
• When 0< λ <1 we get a point between a and b .
Test Question 7 What happens if λ >1? What if λ <0?
As an alternative formulation we could say that the points in the extended line AB are of the form x a + y b where x+y=1. That is more symmetric, but for the moment we stick with the formulation in terms of the single parameter.
Now suppose C given by c = (1λ ) a + λ b lies on AB . Then
→
AC = c – a = λ ( b – a )
→
CB = b – c = (1λ )( b – a ).
Thus the ratio
→
AC :
→
CB is λ :(1λ ). Note that in the ratio λ is associated with A and (1λ ) with B while in the expression c = (1λ ) a + λ b it is the other way round. (We will come back to this.)
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Example 2.2
What if the point C is such that the ratio
→
AC :
→
CB is 2:3?
Solution.
We must scale the ratio to get fractions adding up to 1.
2 : 3 =
2
:
3
.
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Then the answer is
3 a +
2 b .
5 5
To illustrate the matter we mark some points in the line determined by a and b .
2 a - b a
1
2
( a + b )
Note in particular that the midpoint of AB is
1
2 a +
1
2 b =
1
4
1
2 a +
( a
3
4
+ b b b ).
2 b - a
Example 2.3
Show that the diagonals of a parallelogram bisect each other.
Solution.
Suppose ABCD is a parallelogram with corresponding vectors a , b , c , d .
D
C
A
B
We know that a + c = b + d . Now the midpoint of AC is
1
2
( a + c ) and the midpoint of BD is
1
( b + d ). We have
2
1
( a + c )=
1
2 2
( b + d ), and so these midpoints coincide.
Take a triangle ABC and let A` , B` , C` be the midpoints of the sides BC , CA and AB respectively.
A
B
C` G
A`
B`
C
(We may just say A` , B` , C` are the midpoints of the opposite sides.)
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As usual A , B , C are given by vectors a , b , c . Then A` is
1
( b + c ) and so on. Now we
2 notice something. The line AA` has points of the form
(1λ ) a + λ ⋅
1
2
( b + c )
Set λ =
2
3
and we see that
1
( a + b + c )
3 lies on AA` . But the situation is completely symmetric so the point G given by
1
( a + b + c ) also lies on BB` and CC` . (If you do not like the appeal to symmetry then you
3 can just argue similarly.) We deduce the following famous result of elementary geometry.
Theorem 3.1
The three lines AA` , BB` and CC` are coincident (i.e. pass through the same point).
Another way to state the theorem is this. Let G be the point of intersection of AA` and
BB` . Then G also lies on CC` (i.e. C , G , C` are collinear). It is instructive to note how even simple geometric facts have nice vector proofs.
Example 3.1
Show in the above diagram that
→
C`B` .
Solution.
→
C`B` =
1
2
( a + c ) -
1
2
( a + b ) =
1
2
( c – b ) =
1 →
BC .
2
The point G thorough which the lines AA` , BB` and CC` pass is called the centroid .
We can write the vector
1
2
→
= BC
1
( a + b + c ) =
1 a +
2
(
1
( b + c )).
3 3 3 2
This shows that the ration AG : GA`= 2:1, that is, G is two thirds of the way along AA` .
Similarly for BB` and CC` .
Test Question 8 Show that the centroid of the triangle AC`B` is one third of the way along
AA` .
We have seen that if a , b are vectors for points A , B with respect to the origin O , then
1
( a + b ) is the midpoint of AB . Similarly if a , b , c are vectors for the vertices A , B , C of a
2 triangle, then
1
( a + b + c ) is the centroid. In each case we have a kind of average of the
3 vectors in question; and we can see geometrically that the point defined is independent of the origin of reference.
Clearly there is a simple generalization. If a
1 , …
, a n are vectors representing A
1
, …, A n
.
Then
1 n
( a
1
+ … + a n
)
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http://extramaths4fun.com is a vector representing some point. If we change the origin to P , the average of the new vectors is
1 n
(( a
1
+ p ) + … + ( a n
– p )) =
1 n
( a
1
+ … + a n
) – p which is the new vector of the average. So again the average point is independent of the origin of reference.
Now we generalize a bit more and use some intuitive mechanics. Suppose we place masses m
1
, …, m n
at the points a
1 , …
, a n .
Then the centre of mass of this collection of point masses is defined to be i n
∑
= 1 i n ∑
= 1 m i a i m i
.
If we let M = n
∑ i = 1 m i be the sum of the masses, then we can write the centre of mass as i n
∑
= 1 m i
M a i
; i.e. it is the average of the vectors a i weighted by the proportion of the total mass at each a i .
Test Question 9 Show that the centre of mass is independent of the origin of reference of the vectors.
The centre of mass has a number of remarkable properties of which we mention a couple.
1.
Suppose external forces given by vectors F i
are applied to the points a i
. If the system behaves in accordance with Newton’s Laws, then the centre of mass moves as if the vector sum F = n ∑ i = 1
F i were applied at it to a mass M = i n ∑
= 1 m i
. (This is true in the presence of internal forces so long as they satisfy Newton’s Third Law.) It is a consequence of this kind of analysis that a system of forces on a rigid body in 3D is equivalent to a force at the centre of mass together with a couple.
2.
Suppose that gravity acts uniformly in some direction in space which we call vertically downwards. (In case we are dealing with the plane we may take gravity to act at right angles to the plane.) Now take a line l in a horizontal plane, so that the meaning of the moment of the gravitational forces about l should be clear. Then the sum of the moments of the gravitational forces of magnitude m i g acting downwards through a i
is equal to the moment of a force Mg ( M = i n ∑
= 1 m i
) acting downwards through the centre of mass.
In virtue of the second property the centre of mass is often called the centre of gravity .
Properties of vectors make the existence of the centre of mass having properties as above not hard to establish. (We shall shortly say a bit more about moments.) But it should not be regarded as trivial. The idea goes back at least to Archimedes, who certainly appreciated its importance, as it plays a role in many of his most remarkable discoveries in elementary mathematics.
The main issue as regards the centre of gravity is already clear in one dimension. Take a line in which we mark a point O to serve as origin: points A , B , C in the line correspond to
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http://extramaths4fun.com/ vectors
→
OA ,
→
OB ,
→
OC . The only choice of direction in a line is which is the positive direction: so
→
OA and so on are magnitudes, positive when we draw A to the right of O , negative otherwise.
If we place a mass m at A (and set g = 1 ), then the moment about a point P is m
→
PA .
The moment of a collection of masses m
1
, …, m n at A
1
, …, A n
is i n
∑
= 1 m i
→
PA i
, that is the sum of the moments about P . The total mass is
And if we set λ i
= m i
/ M
M = i n ∑
= 1 m i
, then the centre of mass is i n ∑
= 1
λ i
→
OA i
.
Now observe that the equation from placing m i
≥ 0 at a i
: set λ i i n
∑
= 1 m i
→
OA i
= M n
∑ i = 1
λ i
→
OA i says that the moment about the origin of the masses equals the moment of M placed at the centre of mass. But the centre of mass is independent of the origin of reference, so we can choose it to be anywhere we like. Hence the moment of the masses about any point equals the moment of M placed at the centre of mass.
Test Question 10 Think about the problem of the moment of masses about a line in the plane: show that it reduces to the one dimensional problem.
Given vector points a
1 , …
, a n we can characterize the possible centres of mass arising
= m i
/ M where M = n ∑ i = 1 m i
, and we see that we have all points of the form
∑ λ i a i where
λ i
= 1 λ i
≥ .
This is the convex hull of the points. Here is a picture in the plane.
The convex hull of six market points
Test Question 11 Explain why the points inside do not affect the convex hull.
There are good reasons to consider
∑ λ i a i with
λ i
= 1 but without insisting on
λ i
≥ 0 . Even the idea of negative mass makes sense in some mechanics.
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Example 3.2
Suppose in a light beam we place a weight and apply an upward force in different places in the diagram.
1 b a
2
We can regard this as a matter of positive and negative mass, thus a b
-1
2
Thus there is an effective net mass of 1 which acts at the centre of mass
2
1
− 1
( − 1 ⋅ a + 2 ⋅ b ) = 2 b − a , that is as indicated: a b 2 b -
1 a
There is also a couple tending to turn the beam, but we do not go into that.
Test Question 12 What happens when the sum of positive and negative masses or weights is
0?
Recall that any point in the extended line AB can be written (uniquely) in the form
(1λ ) a + λ b , that is in the form x a + y b where x+y=1.
We interpret x a + y b as the centre of mass of a system with positive or negative masses x , y at a , b , the system being normalised to that the total mass is 1. We are about to extend this idea from the line to the plane.
triangle ABC as lying in the plane, but if they were in
3D space, they would line in the plane through them and the discussion would go through in the same way. We investigate the centre of mass of masses placed at A , B , C . We normalize so that the total mass is 1. We place masses x at A , y at B and z at C with x + y + z = 1: then the centre of mass is x a + y b + z c .
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We can describe a geometric construction for the centre of mass x a + y b + z c .
A
S
B
P
First take a point P in BC so that the ratio
→
BP :
→
PC = z : y
C
. We can do this so long as y and z are not both 0; but in that case x =1 and the centre of mass is A . Otherwise the point P is y y
+ z b + y z
+ z c .
Now (and we can certainly do this) take a point S on AP so that the ration
→
AS :
→
SP =
( y + z
)
: x . We see that S is the point x a + ( y+z ) ( y
+ b + z
+ c ) = x a + y b + z c y z y z as required.
Test Question 13 Show that any point S in the triangle ABC is of the form x a + y b + z c for some x + y + z = 1 with x ≥ 0 , y ≥ 0 , z ≥ 0 . What is the condition that S lie on the side
AB ? On BC ? On CA ?
The construction which we have just given tells us something interesting about the coordinates x , y , z for S .
B
A
S
P
C triangles and SBC have the same base BC and the ratio of their heights is as AP : SP = 1 : x . It follows that area ( SBC ) = x area ( ABC ).
But there was nothing special about our choice of P in BC for the analysis: similarly we get
Thus the masses x , y , z for S are proportional to the areas of the opposing triangles SBC ,
SCA , SAB ; and the normalization x + y + z = 1 corresponds to x = area ( area ( SCA ) = y area ( ABC ), area ( SAB ) = x area( ABC ).
SBC ) area ( ABC )
, y = area area (
( SCA
BCA
)
)
, z = area ( SAB ) area ( CAB )
.
These considerations also answer the last exercise, though not perhaps in the most obvious way!
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Recall that we are happy to allow negative masses for our centre of gravity, so long as the total mass is not zero. We normalize the total mass as always and so consider points of the form x a + y b + z c with x + y + z = 1, but no other assumptions on x , y , z .
Test Question 14 Show that any point in the plane can be written uniquely in the form x a + y b + z c with x + y + z = 1. coordinates areal coordinates . (Of course they depend on a choice of reference triangle ABC .) If more flexibly you determine a point in the plane by saying it is the centre of mass of m
3
at C , with m
1
+ m
2
+ m
3
≠ 0 , then you are using m
1
at A , barycentric coordinates m
2 m
1
,
at B and m
2
, m
3
(with respect to ABC ). (Barycentric coordinates are not unique, but are determined up to scalar multiple.)
Test Question 15 Draw a diagram to show where are the points x a + y b + z c with x + y + z = 1 in the cases when ( x , y , z) has signs
(+, +, -) (-, +, +) (+, -, +) (-, +, -) (+, -, -) (-, -, +).
Let us return to the centroid G of a triangle ABC : it is the centre of mass of those equal masses placed at the vertices. Now suppose we consider instead the triangle to be a flat surface made of a uniformly dense material. Then there is an extraordinary fact: G is still the centre of mass.
We can make this plausible as follows. Suppose there is a centre of mass G of the triangular surface satisfying properties as indicated earlier for the centre of mass of point masses. Then it must be in any line about which the moment of the triangle, when placed in a horizontal plane with gravity acting vertically downwards, is zero. So it suffices to show that the medians, the lines joining vertices to the midpoints of opposite sides, are such lines.
A h h
B
A`
C
Regard the triangle as made up of lines parallel to AA` : then it should be enough to show that the moments of the individual lines (or very thin slices if you prefer) are the same, in the sense of the picture.
Test Question 16 Show that the lengths of pairs of lines parallel to AA’ in the triangle and equidistant from AA` are the same. Deduce that the moment of ABA` about AA` equals the moment of AA`C about AA` and so the moment of ABC about AA` is zero. Finally, deduce that if a centre of mass exists it must be G .
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In this plausibility argument there are two assumptions: that we can regard the triangle as made up of lines (or thus slices); and that the centre of gravity of uniform plane bodies exists. The first is certainly not trivial and is usually explained in terms of calculus.
The second is also an issue: for example, how do we know that if we draw an arbitrary line through the centre of gravity it divides the triangle into two pieces whose moment about the line is equal and opposite?
Test Question 17 Prove the moment property of the centre of mass of point masses in the plane. (This may be easier after the next section on coordinate systems.)
Remark.
If you make the side of a triangle out of uniform one-dimensional medium (like the triangle as a musical instrument) then the resulting figure generally has a quite different centre of mass.
We do not need to consider general coordinate systems, but we say a little for completeness. We concentrate on the plane.
Suppose b are two vectors in the plane not scalar multiples of each other.
(In particular, they are not the origin O.
) We extend the lines OA and OB as skew axes in the plane.
C
B
O
A
Take any point C . The line through C parallel to OB will intersect the OA axis at some scalar multiple multiple
λ a of a , while that parallel to
µ b of b . We see at once that
OA will intersect the OB axis at some scalar c = λ a + µ b and that the coordinates for c viz λ , µ in this coordinate system are uniquely determined.
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We can arrive at the same conclusion another way. Recall that the points in the extended line AB are of the form x a + y b where x + y =1. Now as by hypothesis O is not in
AB , the extended lines OC and AB will meet at a point P ≠ O (assuming here that C ≠ 0).
C
B P
O
A
Let P be x a + y b with x + y = 1, and let OC = k OP . Then c = k ( x a + y b ) = λ a + µ b with λ = kx and µ = ky again uniquely determined. (Of course if C = O , then c = 0 = 0 a + 0 b .)
Either way we have seen that every c in the plane can be written c = λ a + µ b for unique λ , µ . One says that a , b form a basis and that λ , µ are the coordinates of c for the basis (or coordinate system).
Test Question 18 Take points A ≠ B and C ≠ D in the plane. Show that AB is parallel to CD
(in the sense that a – b and c – d are scalar multiples of each other) if and only if the extended lines AB and CD either coincide or do not meet.
Test Question 19 The condition that a and b are not multiples of each other says that they are linearly independent. What is the analogous condition for those vectors in 3D space?
Remark Up to this point we have made few assumptions about vectors; specifically
(i) we have not assumed that we can compare lengths of vectors in different directions, and
(ii) we have not assumed any information about angles between vectors.
These related points mean that we have been doing affine geometry. We now introduce a notion which allows us to do Euclidean geometry and also enables many applications of vectors.
The scalar product or inner product a , b of two vectors a and b is defined to be the scalar ab cos θ ,
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http://extramaths4fun.com/ where a is the length of a , b is the length of b , and θ is the angle between a and b . b
O θ a
θ θ , it does not matter in which direction we measure the angle; and we see at once that a ⋅ b = b ⋅ a , so the scalar product is commutative. (One later meets an operation, the vector product which is not commutative.)
From the definition of scalar product it follows at once that a ⋅ b is positive if the angle between a and b is acute. b
O a
O a and negative if the angle between a and b is obtuse b
O a and
O b are orthogonal a b
(or at right angles) written a ⊥ b just when a ⋅ b b a
=0. Finally note that the length a = a of vectors is determined by the scalar product. As cos 0 = 1 , a 2 = a
2 = a ⋅ a .
(Note that the length of a sum a + b is generally not equal to the sum of the lengths of a and b . So the convention that a has length a can get misleading and we shall not use it.) product
It is immediate from the definition that if we scale the vectors we scale the scalar
( λ a ) ⋅ b = λ ( a ⋅ b ) a ⋅ µ b = µ ( a ⋅ b ) and generally
This is clear for λ , µ ≥ 0 ; and since
λ a
− a ⋅
⋅ µ b b
=
= λµ ( a ⋅ b )
− ( a ⋅ b )
.
, it holds for all λ µ product
We can explain the scalar product in terms of orthogonal projections. The scalar a ⋅ b is the product of the length a of a with the length b cos θ of the orthogonal projection of b in the direction a .
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O b
θ cos θ b a
Equally it is the product of the length b of b with the length a cos θ of the orthogonal projection of a on b . These readings of the scalar product justify the following natural laws for a product:
( a + b ) ⋅ c = a ⋅ c + b ⋅ c a ⋅ ( b + c ) = a ⋅ b + a ⋅ c .
Test Question 20 Justify these laws by showing that the orthogonal projection of a + b in the direction c is the sum of the orthogonal projections of a and of b . [It is instructive to draw the picture, not just when a ⋅ c and b ⋅ c are positive, but also when one is positive and the other negative.]
The main force of these properties is that we can multiply out scalar products just like ordinary ones:
( a + b ) ⋅ ( c + d ) = a ⋅ c + a ⋅ d + b ⋅ c + b ⋅ d . c 2
2
Test Question 21 Recall that a = a ⋅ a
(
and prove that a ⋅ b =
1 a + b
2 − a
2 − b
2
)
.
2
Explain the connection of this with the theorem in geometry that in a triangle ABC with lengths of sides as indicated
B c
A a b
C
= a 2 + b 2 − 2 ab cos C .
We say a bit more about orthogonality. a ⊥ b if and only if a ⋅ b =0 generally has the meaning expected, though you should note the degenerate case a ⋅ 0 = 0 so a ⊥ 0 for all a .
Now a ⋅ a = a
2
, the square of the length of a , which is 0 only if a = 0 . Thus the only vector a with a ⊥ a is 0 . From this we can derive an important principle. Suppose we are in the plane with independent vectors a , b as in the section in coordinates. Now suppose that we have c in the plane with c ⋅ a = 0 and c ⋅ b = 0 . Then c ⋅ ( λ a + µ b ) = λ c and since any point in the plane can be written as plane. In particular c ⊥ c and so c = 0 .
⋅ a + µ c ⋅ b = 0
λ a + µ b , c is orthogonal to all points in the
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Test Question 22 Show in like circumstances that c ⋅ d = d ⋅ a and c ⋅ b = d ⋅ b implies c = d .
We give a traditional application of the scalar product to geometry. Let ABC be a triangle.
A
E
F H
B
D
C
Drop perpendiculars from A to BC , from B to CA and from C to AB meeting at D , E , F as shown. Suppose AD and BF meet in H . Then AH ⊥ BC and BH ⊥ AC ; so the vector h for H satisfies
( h − a ) ⋅ ( b − c ) = 0 ( h − b ) ⋅ ( c − a ) = 0
Test Question 23 Deduce from this that
( h − c ) ⋅ ( a − b ) = 0 , so that the perpendicular from C to AB also passes through H .
You have proved another famous result of elementary geometry.
Theorem 4.1
The altitudes of a triangle are coincident. The point H of coincidence is called the orthocentre of ABC .
There are a couple more points associated with a triangle of which we shall need to be aware. The circumcentre.
The perpendicular bisectors of the sides of a triangle are coincident.
A
B
O
C
The point of coincidence O is the centre of the unique circle (the circumcircle) pointing through A , B , C . It is often convenient in geometry to take O as origin.
Test Question 24 Prove that the perpendicular bisectors are coincident.
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The Incentre.
The bisectors of the interior angles of a triangle are coincident.
A
I
B C
The point in question is the centre of a circle touching AB , BC , and CA .
Test Question 25 (Geometry not vectors!) Show that the angle bisectors are coincident.
Now that we have notions of length and angle we can choose special coordinate systems based on an orthonormal system of vectors.
In the plane . Take a vector i of unit length another j of unit length and orthogonal to i , where by convention we take the direct angle of
π
90 o
2
= from i to j in the counterclockwise direction.
We have i ⋅ i = j ⋅ j = 1 and i ⋅ j = j
O
0 i so by definition i , j form an orthogonal system.
Any c in the plane can be written in the form c = λ i + µ j
(as i , j do not lie in a line through the origin); but now we can find the coordinates λ µ directly. Taking the scalar product of the equation with i gives c ⋅ i = λ i ⋅ i + µ j ⋅ i = λ and with j gives c ⋅ j = λ i ⋅ j + µ j ⋅ j = µ .
Thus the coordinates of c with respect to i and j are c ⋅ i , c ⋅ j , which are the lengths of the projections of c on the i and j directions as expected.
In 3D space Take a vector i of unit length, another j of unit length and orthogonal to i , and finally a further k of unit length and orthogonal to both i and j . Here we choose directions so that turning from i to j induces a right handed screw in the direction k . The usual picture is
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http://extramaths4fun.com/ k j where j is thought of as going into the page. We have i ⋅ i = j ⋅ j = k ⋅ k = 1 and i ⋅ j = i j ⋅ k = k ⋅ i = 0 , so by definition again i , j , k form an orthogonal system.
In view of an earlier exercise it should be plausible that as i , j , k do not lie in a plane through the origin, any vector c in 3D space can be written uniquely as c = λ i + µ j + v k .
Taking scalar products as before we calculate the coordinates:
λ = c ⋅ i , µ = c ⋅ j , ν = c ⋅ k .
Again they are the lengths of the orthogonal projections of c onto the various axes.
The basic facts about vector introduced above are quite useful for the solution of various geometrical problems. We suggest that you polish up your skills by trying problems below. Good luck!
Problem 1 Answer the test questions in the text.
Problem 2 Let a , b , c be successive vectors if a regular hexagon. Give expressions in terms of a , b , c for (i) the centre of symmetry and (ii) the other vertices.
Problem 3 Let ABC be a triangle with A' , B' , C' the mid points of the opposite sides.
Establish the vector equation
→
AA' + B
→
B' +
→
CC' = 0 .
Problem 4 Suppose ABCD is such that AC and BD bisect each other. Show that ABCD is a parallelogram.
Problem 5 Let ABCD be a parallelogram and let E be the midpoint of AD. Show that the lines EB and AC bisect each other.
Problem 6 Let ABCD be a quadrilateral and P, Q, R, S the mid points of the sides AB, BC,
CD, DA. Show that PQRS is a parallelogram.
Problem 7 Let ABC be a triangle with A' , B' , C' the mid points of the opposite sides. Show that the centroid of A' B' C' is the centroid of ABC.
Problem 8 Let ABC be a triangle; P on BC divides BC so that divides AB so that
→
AR :
→
RB = 2 : 1
→
BP :
→
PC = 1 : 2 and R on AB
. Let S be the point of intersection AP and CR; and Q the intersection of the (external) BS with CA. Find the ratios
→
CQ :
→
QA ,
→
BS :
→
SQ and
→
AS :
→
SP .
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Problem 9 Let ABCD be a quadrilateral. Show that the centre of mass of equal masses at the vertices is
(i) the point of intersection of the bimedians (the lines joining midpoints of opposite sides) and
(ii) the mid point of the line joining the mid points of the two diagonals.
Problem 10 Prove Ceva’s Theorem : Let ABC be a triangle, P, Q, R points on BC, BQ and
CR are coincident if and only if the product of ratios
→
PC
→
BP
⋅
→
QA
→
CQ
⋅
→
RB
→
AR
= 1 .
Problem 11 Let ABC be a triangle.
(i) Let P, Q, R be chosen on BC, CA, AB so that the distances round the perimeter from A to P, from B to Q and from C to R in ether directions are equal. Show that AP, BQ and CR are coincident.
(ii) Let P, Q, R be the tangent points of the incircle to BC, CA, AB respectively. Show that AP, BQ and CR are coincident.
Problem 12 In the diagram
F
A
S
E show that
→
AS :
→
SD
B
=
→
ΑΕ :
→
EC +
→
AF :
→
FB
D
.
C
Problem 13 Two points are chosen at random on a line of length 1. What is the probability that the three resulting segments form a triangle?
Problem 14 Prove Menelaus’s Theorem : Let ABC be a triangle, P, Q, R points on the
(possibly extended) lines BC, CA, and AB respectively. Show that P, Q, R are collinear if and only if the product of ratios
→
PC
→
BP
⋅
→
QA
→
CQ
⋅
→
RB
→
AR
= − 1 .
Problem 15 Let ABCD be a quadrilateral with
→
AD
1
2
→
= BC . Suppose the surface ABCD is made of a uniform material. Find the centre of mass.
Problem 16 (i) Let ABC be a triangle and let the sides BC, CA, and AB have lengths a , b , c , respectively. What is the centre of mass of masses a placed at A, b at B and c at C?
(ii) Suppose that the sides of a triangle are made of uniform material. Identify the centre of mass.
Problem 17 Suppose that ABCD is a parallelogram. Show that it is a rhombus (i.e. the lengths of all sides are equal) if and only if AC ⊥ BD.
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Problem 18 Prove that the sum of the squares on the six edges of a tetrahedron is equal to four times the sum of the squares on the three lines joining mid points of opposite sides.
Problem 19 Let O be the circumcentre, G the centroid end H the orthocentre of a triangle
ABC. Show that O, G, H are collinear with
→
OG :
→
GH = 1 : 2 .
Problem 20 Let ABC be a triangle with A' , B' , C' the mid points of the opposite sides. Let
O be the circumcentre and H the orthocentre of ABC. Show that the circumcentre of bisects OH.
A' B' C'
Problem 21 (A Steinhaus problem) It is required to build a common school for villages A, B,
C with 50, 70 and 90 children respectively. Where should the school be situated so that the total sum of time to go to school be minimal? And why?
Problem 22 (Mathematical Tripos 1882) Suppose that ABCD is a quadrilateral with sides
AD and BC meeting in X and AB and CD meeting in Y. Show that with the right choice of angles at the points, the intersections of the angle bisectors of X, Y, of A,C and of B, D are collinear.
Problem 23 Show that if the diagonals of a quadrilateral are orthogonal then any other quadrilateral with the same length of sides has orthogonal diagonals.
Problem 24 Consider four points A, B, C and D on a plane. Prove that
→
AB ,
→
CD +
→
BC ,
→
AD +
→
CA ,
→
BD = 0 .
Problem 25 Let O be the centre of the circumcircle of the triangle ABC, and a point H satisfies
→
OH =
→
OA +
→
OB +
→
OC . Prove that H is the orthocentre of ABC.
Problem 26 Consider a quadrilateral ABCD. Let u =
U = BD 2 + CD 2 − BC 2 , V = AD 2 + CD 2 − AC 2 , W = AD 2 +
AD 2
BD 2 −
, v
AB 2
= BD 2 , w
. Prove that
= CD 2 , uU 2 + vV 2 + wW 2 = UVW +
Problem 27 Prove that for any four points A, B, C and D
4 uvw .
AB 2 + BC 2 + CD 2 + DA 2 ≥ AC 2 + BD 2 with an exact equality achieved if and only if ABCD is a parallelogram.
Problem 28 Prove that among any five vectors there are two with the length of their sum not greater than the length of the sum of the other three vectors.
Problem 29 Let O be a point inside a triangle ABC. Prove that area (BOC) ⋅
→
OA + area (AOC) ⋅
→
OB + area (AOB) ⋅
→
OC = 0 .
Problem 30 Three tourists walk along three straight roads with constant speeds. At the initial moment of time the tourist are not in a straight line. Prove that it can happened no more than twice during their journeys that the tourist are in a straight line.
The materials are prepared and can be used for non commercial purposes only.
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