www.mathcentre.ac.uk/.../Cartesian%20components%20of%20vectors.mp4

The natural way to describe the
position of any point is to use

Cartesian coordinates. In two
dimensions it's quite easy.

We just have. Picture like
this and so we have an X

axis and Y Axis.

Origin oh where they cross and
if we want to have vectors in

that arrangement, what we would
have is a vector I associated

with the X axis and a vector Jay
associated with the Y axis.

What all these vectors I&J?

Well, they have to be unit

vectors. A unit vector I under

unit vector. J.

In order to make sure that
we do know that they are

unit vectors, we can put
little hat on the top.

So if we have a point P.

And let's say the coordinates of
that point are three, 4, then

the position vector of P which
remember is that line segment

joining oh to pee is 3 I.

Plus four jazz.

Notice the crucial difference.
That's a set of coordinates

which refers to the point
that's the vector which refers

to the position vector. So
point and position vector are

not the same thing.

We can write this as
a column vector 34.

And sometimes. This is
used sometimes that one

is used, just depends.

What about moving then into 3

dimensions? We've got XY
and of course the tradition

is to use zed.

So let's have a look. Let's draw
in our three axes.

So then we've got XY.

And zed.

I'll always write zed with a bar
through it that so it doesn't

get mixed up with two. I don't
want the letters Ed being

confused with the number 2.

So I've got these three axes or
at right angles to each other

and meeting at this origin. Oh,
and of course I'm going to

describe any point P by three
coordinates XY and Z.

Now when I drew up this set of
axes, I indicated them.

Quite easily.

I could of course Interchange

X&Y. I might choose to

interchange Y&Z. But this is the
standard way. Why is it the

standard way? What is it about
this that makes it the standard

way? It's standard because it's
what we call a right?

Hand.

System.

Now. How can we describe
workout? What is a right hand

system? Take your right hand and
hold it like this.

Middle finger. Full finger and
thumb at right angles.

This is the X axis.

The middle finger. This is the Y
axis, the thumb.

Now rotate as though we were
turning in a right handed screw.

And we rotate like that.

And so the direction in which
we're moving this direction

becomes zed axis. So we rotate

from X. Why?

And we move in the direction of
the Z axis. So right hander

rotation as those screwing in a
screw right? Handedly notice

that it works whatever access we
choose. So if we take this to be

Y again, the thumb and we take
this to be zed then if we make a

right handed rotation from why
route to zed, we will move along

the X axis. So let's do that.

You can see that as we rotate
it, we are moving right handedly

along the X axis and you can try
the same for yourself in terms

of rotating from X to zed and
moving along the Y axis. So

that's our right handed system.
So let's have a look at that in

terms of having a point P that's
got its three coordinates XY.

And said

X.

And why?

And said now origin, oh.

Will take a point P anywhere
there in space. What we're

interested in is this point P.
It's got coordinates, XY and

zed. And its position
vector is that line segment

OP. And so we can write down,

Oh, P. Bar is
equal to XI.

Plus YJ.
Plus, Zed and the unit vector

that is in the direction of the
Zed Axis is taken to be K.

So again, notice the
difference. These are the

coordinates XY, zed. This is
the position vector

coordinates and position
vector are different.

Coordinates signify
appoint, position vector

signifies a line segment.
We sometimes write again

as we did with two
dimensions. We sometimes

write this as a column
vector XY zed.

Now there are various things we
would like to know and certain

notation that we want to
introduce for start. What's the

magnitude of Opie bar? What's
the length of OP? Well, let's

drop a perpendicular down into
the XY plane there and then.

Let's join this up.

The axes there and
across there.

Now let's just think what this
means this length here.

Is the distance of the point
above the XY plane, so it must

be of length zed.

This length, here and here is
the same length. It's the

distance along the X coordinate,
so that must be X.

And that's also X. Similarly,
This is why and so that must be

why as well.

So if we join up from here out

to here. What we have here is a
right angle triangle, and of

course we've got a right angle

triangle here as well. So this
length here. There's are right

angle this length using
Pythagoras must be the square

root of X squared plus Y
squared, and so because we've

got a right angle here, if we
use Pythagoras in this triangle

then we end up with the fact
that opie, the modulus of Opie

Bar is the square root of. We've
got to square that and add it to

the square of that. So that's
just X squared plus.

Y squared plus Zed Square.

Now I'm going to draw this
diagram again, but I'm going to

try and miss out some of the
extra lines that we've added.

So XY.

Zedd.

We'll take our point P with
position vector OP bar.

Again.

Drop that perpendicular down on
to the XY plane.

Draw this in across here.

And that in there.

Now. This line OP
makes an angle with this

axis here.

It makes an angle Alpha.

And if I draw it out so that we

can see it. Let me call this a.

If we draw out the triangle
so that we can actually see

what we've got, then we've
got the line.

From O to a.

There. Oh, to A and we've got
this line going out here from A

to pee and that's going to be at
right angles there like that.

And so if we now join P2O, we
can see the angle here, Alpha.

Now we know the length of this
line. We know that it is the

square root of X squared plus Y
squared plus zed squared and we

also know the length of this
line, it's X.

And that is a right angle, and
so therefore we can write down

cause of Alpha is equal to X
over square root of X squared

plus Y squared plus zed squared.

Why have we chosen this? Well,
cause Alpha is what is known

as a direction.

Cosine

be cause. It is the cosine of
an angle that in some way helps

to specify the direction of P.

An Alpha is the angle that Opie
makes with the X axis. So of

course what we can do for the X
axis we can do for the Y axis

and for the Z Axis.

So we have calls Alpha which
will be X over the square

root of X squared plus Y
squared plus zed squared.

Kohl's beta which will be
the angle that Opie makes with

the Y axis, and so it will be
why over the square root of X

squared plus Y squared plus zed
squared and cause gamma.

Gamma is the angle that Opie
makes with the Z Axis, and so it

will be zed over the square root
of X squared plus Y squared

close zed square.

So these are our direction
cosines. These are expressions

for being able to calculate
them, but there is something

that we can notice about them.
What happens if we square them

and add them? So what do we
get if we take 'cause squared

Alpha plus cause squared beta
plus cause squared gamma?

So let's just calculate
this expression.

Kohl's squared Alpha is going
to be X squared over

X squared plus Y squared
plus said squared.

Call squared beta is going to
be Y squared over X squared

plus Y squared plus zed squared.

And cost squared gamma is
going to be zed squared

over X squared plus Y
squared plus zed squared.

Now we're looking at adding all
of these three expressions

together. Cost squared Alpha
plus cost squared beta plus cost

squared gamma. Well, they've all
got exactly the same denominator

X squared plus Y squared plus
said squared, so we can just add

together X squared plus Y
squared plus 10 squared in the

numerator. So that's X squared
plus Y squared zed squared all

over X squared plus Y squared
plus said squared. Of course,

that's just one.

So the squares of the direction
cosines added together give us

one. What possible use could
that be to us? Well, one of

the things it does mean is
that we have the vector,

let's say cause.

Alpha I plus
cause beta J

plus cause Gamma
K.

That vector is a unit vector.
It's a unit vector because if

we calculate its magnitude
that's cost squared Alpha plus

cost squared beta plus cost
squared gamma is equal to 1.

Take the square root. That's
one. So this is a unit vector.

Further, this is X over X
squared plus Y squared plus Z

squared Y over X squared plus
Y squared plus said squared.

And zed over X squared plus Y
squared plus said Square and so

it's in the same direction as
our original OP. Our original

position vector opi bar.

And that means that this is a
unit vector in the direction of

OP bar and that may prove to be
quite useful later on when we

want to look at unit vectors in
particular directions. For now,

let's just have a look at doing
a little bit of calculation.

Let's say we've got a point.

That has
coordinates 102.

Under point that has
coordinates 2 - 1.

4.

The question that we might ask
is if we form the vector AB.

What's the magnitude of a bee?

And what are its direction

cosines? We just have a
look at this. Let's

remember that, oh, a bar.

Is.

I.

No JS.

And two K's.

That OB bar.

Will be. Two I.

Minus one
J plus 4K.

We want to know what's the
magnitude of the vector AB bar.

Just draw quick picture just to
remind ourselves of how to get

there. There's A and its
position vector with respect to.

Oh there's B with its position
vector with respect to. If we're

wanting a baby that's from there
to there and so we can see that

by going from A to B, we can go
round AO plus OB.

And so therefore, that is OB bar
minus Oh, a bar. So that's what

we need to do here. A bar must
be OB bar minus oh, a bar.

And all we do to do the
subtraction is what you would

do naturally, which is to
subtract the respective bits

so it's two I take away I.
That's just an eye bar.

Minus J takeaway no
JS, so that's minus J

Bar and 4K takeaway
2K. That's plus 2K.

So now we have our vector AB
bar. We can calculate its

magnitude AB modulus of a bar
that's just a be the length from

A to B, and that's the square
root of 1 squared plus minus one

squared +2 squared altogether.
That's 1 + 1 + 4 square

root of 6, and the direction

cosines. Our cause Alpha.

That's The X coordinate
over the modulus, so that's

one over Route 6.

Kohl's beta that's minus
one over Route 6. the Y

coordinate over the
modulus and cause gamma.

The zed coordinate
over the modulus.

Now this is a fairly standard
calculation. The sort of

calculation that it will be

expected. You'll be able to do
and simply be able to work your

way through it very quickly.
Very, very easily, so you have

to be able to practice some of
these. You have to be able to

work with it very rapidly, very,
very easily, but always keep

this diagram in mind.

That to get from A to B to form
the vector AB bar, you go a

obarr plus Obiba and so.

It's the result, so to form a B
it's Obi bar, take away OA bar.