VECTORS AND CARTESIAN PLANE

Part I

Fundamentals of vectors

Foreword It sure becomes hard to grasp geometry when we first encounter it

in high school and it becomes harder when we start using

equations to describe the geometric bodies and their behavior in a

referential system.

Analytic geometry and Euclidean geometry are practically the

same, with the difference being the referential system and their

way to represent the geometric bodies and places in question.

The theorems and formulas from Analytic geometry are derived

and proven by solving a generic problem which gives as a result, a

formula which describe the behavior of variables, points and even

geometric bodies under a certain conditioned variability or

pattern.

I write this little tutorial, to show some of the groundings of analytic

geometry which then can expand to the more complicated

algebra we see in university.

Actually, I would dare to say this is also applicable and teachable

for any student in High school that is taking an analytic geometry

class and wishes to expand upon some techniques that give result

with less given information.

And these techniques involve vectors, which I will discuss later on

this tutorial but for now, all you need to know is that vectors can

make your life easier when trying to represent a geometric body

with a referential system.

I will also tackle the idea of using vectors to define a line and how

a line also, can define a vector.

I will also briefly describe circles and some of their characteristics

but I won’t go in depth, I’ll leave it to someone else so they can

complete this tutorial and describe the rest of the conical sections.

In order to keep things flowing, I will show a simple and then a

complex application of the concepts discussed in this tutorial, so

that may simplify and make more evident they application in any

Analytical Geometry problem/exercise you’ll be facing in your

academic life.

Parallelograms We call “parallelograms” to those quadrilaterals whose opposite pair of sides are

parallel.

A consequence of this definition is:

“The sufficient and necessary condition for a quadrilateral to be a

parallelogram is that its diagonals intersect in their mid-points”. This is easily provable with Euclidean Geometry.

Theorem I (H) ADCB parallelogram

(T) 𝑂 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶

𝑂 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐷𝐵

Proof:

By definition of parallelogram we know that 𝐷𝐶 ∥ 𝐴𝐵 𝑎𝑛𝑑 𝐴𝐷 ∥ 𝐵𝐶 and since 𝐷𝐵 is a diagonal intersecting two parallel lines,

we conclude that: ∠𝑶𝑫𝑪 ≃ ∠𝑶𝑩𝑨∠𝑶𝑩𝑪 ≃ ∠𝑨𝑫𝑶

And, by vertical angles we can conclude that: ∠𝑨𝑶𝑫 ≃ ∠𝑩𝑶𝑪∠𝑨𝑶𝑩 ≃ ∠𝑫𝑶𝑪

𝐴𝐶 is a diagonal that intersects two parallel lines 𝐴𝐷 𝑎𝑛𝑑 𝐵𝐶 which analogically

as before, we can conclude that: ∠𝑶𝑨𝑫 ≃ ∠𝑶𝑪𝑩∠𝑶𝑨𝑩 ≃ ∠𝑶𝑪𝑫

Giving a theorem which is perfectly provable and I’ll be leaving that to the

lector, is that the opposite sides that compose a parallelogram are indeed

congruent, meaning: 𝑨𝑫 ≅ 𝑩𝑪

𝑫𝑪 ≅ 𝑨𝑩

With all that information and the property of congruence ASA we can draw the

conclusion: △ 𝑨𝑶𝑫 ≃△ 𝑩𝑶𝑪△ 𝑫𝑶𝑪 ≃△ 𝑨𝑶𝑩

And from this congruence we can draw the data: 𝑂 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐶

𝑂 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐷𝐵 Q.E.D.

This may seem like a normal Geometry exercise, indeed it does, but it holds quite

the mathematical truth to it, which allows us to create one of the most important

groundings for this tutorial.

D C

A B

O

Definition

The previous theorem forms the grounding for this definition, since

we can also do the reciprocal proof, which I’ll leave for the lector.

This definition is:

ABCD is parallelogram only if 𝑨𝑪 𝒂𝒏𝒅 𝑫𝑩 have the same mid-point.

Observe that this definition is applicable to any parallelogram in

the Euclidean Geometry, but also includes the parallelograms

where all the vertices are aligned, and that happen as long

as the segment AC and DB share their mid-point.

These parallelograms whose vertices are aligned are

called “Degenerated parallelograms”.

One must be very careful in naming these

Parallelograms, because if ABCD is a parallelogram, then ACDB is

not a parallelogram and neither is ABDC. Application “Let ABCD and AECF be two parallelograms. What can be

affirmed about these two parallelograms?”

This is a clear application of the definition I have just taught you:

ABCD parallelogram 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 AC and BD have the same mid-point.

AECF parallelogram 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 AC and EF have the same mid-point.

Therefore, BD and EF must have the same mid-point 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 BEDF is a parallelogram.

As appreciable, the previous proof is valid let the parallelograms

be degenerated or not.

A consequence of the mid-point being shared by the diagonals

imply that the mid-point is the center of symmetry, from which we

can draw the conclusion:

“The pair of opposite sides on a parallelogram are congruent”.

Vectors I will consider all the points in a plane π (A, B, C, D…) and effectuate the

Cartesian product 𝝅 × 𝝅 to all the elements 𝑨, 𝑩 ∈ 𝝅 × 𝝅 . We can associate the oriented segment AB which the notable elements “Origin

A” and “Extreme B” pop up.

A

D

O

B

C

This gives us as a result the oriented segment AB and BA which are completely

different geometrical entities, this is a consequence of establishing a distinction

between Origin and Extreme.

The line AB is named “support” of the oriented segment AB.

Equipollence between oriented segments

Definition: The segment AB is equipollent to segment CD only if ABCD is a parallelogram.

Notation:

𝐴𝐵~𝐶𝐷 ⬚ 𝐴𝐵𝐶𝐷 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚

There are two important consequences in this definition and they are:

(1) 𝑖𝑓 𝐴𝐵~𝐶𝐷 ⬚ 𝐴𝐷 𝑎𝑛𝑑 𝐵𝐶 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑚𝑖𝑑 − 𝑝𝑜𝑖𝑛𝑡

(2)𝐴𝐵~𝐶𝐷⬚

𝑎 𝑇ℎ𝑒𝑖𝑟 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑏 𝑇ℎ𝑒𝑖𝑟 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙 𝑐 𝑇ℎ𝑒𝑖𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒

(3) Equipollence is a relation of equivalence.

Indeed, all this shows that the relationship follows the rules: Symmetry,

Reflexive and transitive.

The first two are immediate, while the third, “transitive” is a follow up of (2)

having the conditions (a), (b) and (c) satisfying the transitive property.

Of course, I am storming through this, but all of this actually took years and

years of studying and meticulously analyzing the very result of operating two

Cartesian planes. This will all give us the grounding we need to define, finally,

what a vector actually is:

Definition of Vector:

The Equipollence of the oriented segments establish, in the set of all the oriented

segments included in the plane π, is a partition in the classes of equivalence.

This means, any oriented segment inside this equipollent space can represent any

of the oriented segments inside π.

All these classes of equivalences were given the name of “Vector”. Vectors are notated with an arrow on the top of the oriented segment in question,

let it be “vector AB”: 𝑨𝑩 A more precise way to notate a vector definition without committing a fault on

the definition of Equipollence is: 𝑨𝑩 = 𝑪𝑫 𝑻𝒉𝒆𝒐.𝑰 𝑨𝑩 ~𝑪𝑫

I will represent with 𝕍 the set of all vectors defined by the oriented segments

inside the plane π.

We can associate any element 𝒖 ∈ 𝕍 to any other AB that belongs to the same

class.

This also allows us to define the oriented segment AA as the null element 𝒐 .

Structure of the Vectorial Space

It is evident that there will be countless amounts of vectors on a plane, and to

make this things simple, a Vectorial space is constructed, and in order to do that:

An operation in 𝕍 must be defined , in other words, a law of internal binary

composition (addition):

+:𝕍 × 𝕍 → 𝕍

A law of external composition is defined between

a body of scalars ℝ and𝕍.

For this to happen, the following was proposed:

“Let 𝑶 ∈ 𝝅. For all vector 𝒖 ⊂ 𝝅 there exists and

it’s

unique, a point M such that 𝑶𝑴 = 𝒖 .”

Indeed, if 𝑢 = 𝑜 𝑀 ≡ 𝑂.

If 𝑢 = 𝐴𝐵 , point M will be that point in π such that

ABMO is a parallelogram.

The unicity of M is provided by the fact that if 𝐎𝐌′~𝐀𝐁

𝐀𝐁𝐌′𝐎 𝐢𝐬 𝐚 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥𝐨𝐠𝐫𝐚𝐦 and if two parallelograms have three vertices in

common, then the fourth will also be common, allowing to deduce: 𝑴′ ≡ 𝑴.

From this construction, some interesting operations can be defined, these being:

(a) Vector addition

In the set of 𝕍 we can define the operation of “sum”, we can denominate

“addition of vectors”:

+:𝕍 × 𝕍 → 𝕍

This makes a correspondence each pair of vectors in 𝕍 to a new vector in

the same 𝕍 called the sum of the first two vectors:

𝑢 , 𝑣 → 𝑤 = 𝑢 + 𝑣 We can define 𝑤 as follows: We

choose a point 𝑂 ∈ 𝜋 and two

representative vectors for 𝑶𝑨

and 𝑨𝑩 , these being 𝑢 and 𝑣 respectively.

𝑶𝑩 Is represented by 𝒘 .

Unicity of the sum.

The sum vector 𝒘 does not depend on the representative vectors taken for

𝒖 and 𝒗 .

Let, as effect, a point 𝑂′ ∈ 𝜋 𝑂′⁄ ≠ 𝑂 ∧ 𝑂′𝐴′ ∧ 𝐴′𝐵′ 𝑟𝑒𝑝.→ 𝑢 ∧ 𝑣 𝑟𝑒𝑠𝑝.

because 𝑂𝐴~𝑂′𝐴′ and 𝐴𝐵~𝐴′𝐵′, the quadrilaterals OAA’O’ and ABB’A’

are parallelograms, therefore the segments OO’, AA’ and BB’ are

equipollent: 𝑑𝑒𝑓. 𝑂𝑂′~𝐵𝐵′, in other words, both of these segments define

one and only one 𝒘 . From this, we can draw the conclusion of the relationship:

𝑨𝑩 + 𝑩𝑪 = 𝑨𝑪

(Michael Chasles relation)

We can further generalize this result to an N-number of points inside a

plane, giving as a result, the sum of all the vectors defined by each pair of

points:

𝑨𝟏𝑨𝟐 + 𝑨𝟐𝑨𝟑 + ⋯+ 𝑨𝒏−𝟏𝑨𝒏 = 𝑨𝟏𝑨𝒏

(b) Algebraic properties of the addition Let 𝒖 , 𝒗 , 𝒘 be any three vectors on the plane π

Commutative property:

𝑢 + 𝑣 = 𝑣 + 𝑢 It seems intuitive indeed, but as the mathematician Bolzano said; even

trivial claims like these requires a proof, so we’ll begin by considering

two vectors 𝐴𝐵 and 𝐵𝐶 which we’ll represent by 𝑢 and 𝑣 respectively,

we’ll be able to construct a parallelogram ABCD:

𝑢 + 𝑣 = 𝐴𝐵 + 𝐵𝐶 = 𝐴𝐶

𝑣 + 𝑢 = 𝐴𝐷 + 𝐷𝐶 = 𝐴𝐶 𝑢 + 𝑣 = 𝑣 + 𝑢

This also proves the “parallelogram rule” we use in physics when adding

force vectors.

Associative property:

𝑢 + 𝑣 + 𝑤 = 𝑢 + 𝑣 + 𝑤

Let 𝐴𝐵 , 𝐵𝐶 and 𝐶𝐷 be represented by 𝑢 , 𝑣 , 𝑤 respectively.

𝒖 + 𝒗 + 𝒘 = (𝑨𝑩 + 𝑩𝑪 ) + 𝑪𝑫 = 𝑨𝑪 + 𝑪𝑫 = 𝑨𝑫

𝒖 + 𝒗 + 𝒘 = 𝑨𝑩 + (𝑩𝑪 + 𝑪𝑫 ) = 𝑨𝑩 + 𝑩𝑫 = 𝑨𝑫 Proving the

prop.

Neutral additive: ∃𝑜 ∈ 𝕍 ∀𝑢 ⁄ ∈ 𝕍: 𝑢 + 𝑜 = 𝑜 + 𝑢 = 𝑢

Let 𝐴𝐵 be represented by 𝑢 .

𝑢 + 𝑜 = 𝐴𝐵 + 𝐵𝐵 = 𝐴𝐵 𝑢 + 𝑜 = 𝑢

Existence of an opposite vector:

∃𝒖 ∈ 𝕍, ∃ −𝒖 ∈ 𝕍 ∀⁄ +:𝒖 × −𝒖 𝒖 + −𝒖 = 𝒐

If 𝑢 = 𝐴𝐵 then, 𝐴𝐵 + 𝐵𝐴 = 𝐴𝐴 = 𝑜 −𝒖 = 𝑩𝑨 = −𝑨𝑩 (c) Multiplication of a Real and a vector

ℝ × 𝕍→ 𝕍

Definition

Let A, B be given points and a real number “k”, a given value.

There exists an “M” point such that: 𝐴𝑀 = 𝑘. 𝐴𝐵 This is only true if:

(1) A, B and M are aligned.

(2) 𝐴𝑀 = |𝐾|. 𝐴𝐵

(3) If K>0 then 𝐴𝑀 , 𝐴𝐵 have the same direction.

If K<0 then 𝐴𝑀 , 𝐴𝐵 have opposite direction.

If k=0 then 𝐴𝑀 = 𝑜 𝑀 ≡ 𝐴

With 𝐴𝐵 representing the longitude of 𝐴𝐵 , in other words: 𝑑𝑖𝑠𝑡 𝐴, 𝐵

The result does not depend on the chosen representatives.

Indeed, if we let 𝐴𝐵 ∼ 𝐴′𝐵′. Then from 𝐾. 𝐴𝐵 = 𝐴𝑀 and 𝑘. 𝐴′𝐵′ = 𝐴′𝑀′

we can deduce that 𝐴𝑀 ∥ 𝐴′𝑀′ and as an immediate result, their direction

and longitudes are the same.

Then 𝐴𝑀 = 𝐴′𝑀′ .

Observe that the relationship 𝑘. 𝐴𝐵 = 𝐴𝑀 characterizes the alignment of

three points and parallelism of two lines.

(d) Algebraic properties of the multiplication by a real and a

vector. I’ll let the lector of this handbook, prove that being 𝑥 ∈ ℝ, 𝑦 ∈ ℝ and 𝑢 ∈𝕍, 𝑣 ∈ 𝕍 that the following properties are true:

(1) 1. 𝑢 = 𝑢 (2) 𝑥 + 𝑦 𝑢 = 𝑥. 𝑢 + 𝑦. 𝑢 (3) 𝑥 𝑦. 𝑢 = 𝑥. 𝑦 𝑢 (4) 𝑥 𝑢 + 𝑣 = 𝑥. 𝑢 + 𝑥. 𝑣

Example exercises

(1) Let two points A, B and 𝑢 = 𝐴𝐵 . Also given a point O and a real

k=2, construct the point M such that 𝑂𝑀 = 𝑘. 𝑢 . We can begin the exercise by representing

the point “O” anywhere in the plane and

after draw a line parallel to the segment

AB on the point O.

Given the fact that the number given is

“2”, and we seek the point “M” such that

the vector “u” fits two times in the vector

“OM”, so we’ll transport the vector with a

compass, to the line drawn on the point

“O” two times.

The head of the second transported vector

becomes then “M” and the sum is evident:

𝑢 + 𝑢 = 𝑂𝑀 𝑝𝑟𝑜𝑝. 1 + 1 𝑢 = 𝑂𝑀

𝑜𝑝. 2𝑢 = 𝑂𝑀 therefore giving:𝑂𝑀 = 𝑘. 𝑢

(2) Let A, B and C such that 𝐴𝐵 = 𝑘. 𝐴𝐶 . Calculate the real numbers m, n and p such that:

𝒊 𝐵𝐴 = 𝑚. 𝐴𝐶 𝒊𝒊 𝐵𝐶 = 𝑛. 𝐴𝐵 𝒊𝒊𝒊 𝐵𝐶 = 𝑝. 𝐴𝐶

The strategy here is to get the value of these vectors, by using the given

relationship, since that is the only “true” information we have.

(i) 𝐵𝐴 = 𝑚. 𝐴𝐶 We can begin by rewriting the “BA” vector, after all, it is the

opposite of AB, so therefore:

−𝐴𝐵 = 𝑚. 𝐴𝐶

And we can simplify further by dividing both sides by (-1):

𝐴𝐵 = −𝑚.𝐴𝐶

And look at that, if we compare it to the given relationship it is

very similar, except for that “-m”, so, we can only conclude it is

the “k” value:

𝐴𝐵 = 𝑘. 𝐴𝐶

𝐴𝐵 = −𝑚 . 𝐴𝐶 𝑘 = −𝑚 𝑚 = −𝑘

(ii) 𝐵𝐶 = 𝑛. 𝐴𝐵 We are going to use the same strategy in order to solve this, just

algebraically play with the relation so we can find that “n”

value, something important to note is:

𝐴𝐶 = 𝐴𝐵 + 𝐵𝐶 𝐵𝐶 = 𝐴𝐶 + −𝐴𝐵

𝐴𝐵 = 𝐴𝐶 + −𝐵𝐶

These is just taken from the given points and relationships,

remember we discussed it earlier.

Let’s plug those values in:

𝐴𝐶 + (−𝐴𝐵 ) = 𝑛. 𝐴𝐵

And rewriting the “-AB” vector using the original given

relationship:

𝐴𝐶 + (−𝑘. 𝐴𝐶 ) = 𝑛. 𝐴𝐵

𝐴𝐶 −𝑘 + 1 = 𝑛. 𝐴𝐵

−𝑘 + 1 𝐴𝐶 = 𝑛. 𝑘 . 𝐴𝐶

Leading to the conclusion: 𝑛 =−𝑘+1

𝑘

(iii) 𝐵𝐶 = 𝑝. 𝐴𝐶 Yet again, the vector that is giving us problem is that vector BC,

so we are going to change that:

𝐴𝐶 + (−𝐴𝐵 ) = 𝑝. 𝐴𝐶

And transform again the vector AB:

𝐴𝐶 + −𝑘 . 𝐴𝐶 = 𝑝. 𝐴𝐶

And since the vector AC is the only one present, we do the remaining transformations and conclude the exercise:

−𝑘 + 1 𝐴𝐶 = 𝑝. 𝐴𝐶

𝑝 = −𝑘 + 1

As appreciable, vectors can be algebraically related, saving us the time to look at

a diagram to draw conclusions.

Independence and collinearity

Observe that if 𝑣 = 𝑘. 𝑢 , there must exist two representatives for 𝑢 and 𝑣 on a

same line.

Definition We’ll say that vectors 𝒖 and 𝒗 are collinear only if there exists a real

number “k” such that: 𝒖 = 𝒌. 𝒗

It can also be said that 𝑢 and 𝑣 have the same direction. The real number “k” is

also called “coefficient of linearity”.

If the vectors 𝑢 , 𝑣 are not collinear, they are called “independent”.

- For example, vectors 𝑢 and 2𝑢 are not independent because they are

collinear.

- Vector 𝑜 is collinear with any vector inside 𝕍.

- From the geometrical interpretation of 𝐴𝐶 = 𝑘. 𝐴𝐵 , three points A, B and

C surge. Aligned only if 𝐴𝐵 , 𝐴𝐶 are collinear.

From this last observation, we can also say:

“The sufficient and necessary condition for 𝒖 and 𝒗 to be independent

is that 𝒂. 𝒖 + 𝒃. 𝒗 = 𝒐 𝒂 = 𝒃 = 𝑶′′

Example exercise “Let ABCD be a generic quadrilateral.

Let P, Q / 𝐵𝑃 =1

2𝐴𝐵 , 𝐴𝑄 = 3. 𝐴𝐷 .

(a) Prove that: (I) 𝐶𝑃 =1

2𝐴𝐵 − 𝐵𝐶

(II) 𝐶𝑄 = 2. 𝐴𝐷 − 𝐷𝐶 ” Just as in classic geometry, we will

have an auxiliary diagram to help us

visualize the exercise, but the proof will

be mostly algebraic.

The strategy here is to express the

desired vector as a sum of other two,

and do the necessary transformations to

obtain the desired relationship.

(I) 𝐶𝑃 =1

2𝐴𝐵 − 𝐵𝐶

The desired vector to express is the

vector CP, but the vector CP is the sum of other two:

𝐶𝑃 = 𝐶𝐵 + 𝐵𝑃

Remember the commutative property of the sum of vectors:

𝐶𝑃 = 𝐵𝑃 + 𝐶𝐵

Given the relationship 𝐵𝑃 =1

2𝐴𝐵 and the equivalent to CB being −𝐵𝐶 = 𝐶𝐵

𝐶𝑃 =1

2𝐴𝐵 − 𝐶𝐵

(II) 𝐶𝑄 = 2. 𝐴𝐷 − 𝐷𝐶 This can be solved using the same strategy, representing the desired vector

and transforming the components algebraically to the desired expression.

𝐶𝑄 = 𝐶𝐷 + 𝐷𝑄 (1).

This one requires a little mid-step because DQ is not an evident vector, but we

can calculate it:

𝐴𝑄 = 𝐴𝐷 + 𝐷𝑄 3𝐴𝐷 = 𝐴𝐷 + 𝐷𝑄 𝐴𝐷 3 − 1 = 𝐷𝑄 𝐷𝑄 = 2𝐴𝐷

Now that we know for certain the value of the vector DQ, using the commutative

property and the opposite of vector CD:

𝐷𝑄 = 2. 𝐴𝐷

𝐶𝐷 = −𝐷𝐶

𝑖𝑛 1 𝐶𝑄 = 2. 𝐴𝐷 − 𝐷𝐶

To finish off this first part, I’ll leave to the lector the following exercise:

“Deduce that if ABCD is a parallelogram 𝑃, 𝑄, 𝐶 are aligned.”

Hint: It is the same scenario with a parallelogram instead of a generic

quadrilateral and the vector 𝐴𝑄 does not have the same direction of AD but

everything else is the same.

Try to represent the vector: 𝐶𝑄 = 𝑘. 𝐶𝑃