www.mathcentre.ac.uk/.../Introduction%20to%20vectors.mp4

Vector quantities are extremely
useful quantities in physics.

Most of the quantities that we
meet in physics and engineering

are in fact vector quantities.

First of all, we need to
explore just what that means.

What are vector quantities?

The quantities.

That have two numbers associated
with them in order to specify

them completely. A magnitude and
a direction. So any vector

quantity must have these two
things associated with it in

some way. A magnitude, and in
some way a direction.

So for instance, let's take
something quite simple.

Let's take a quantity known
as Displacement.

If we think about it,
displacement what we have to

specify is how far away we are
from a fixed point.

And in what direction we are. So
if we fix a .0 and we say we've

got a point P over here
somewhere, then in order to

specify the displacement of this
point P from the .0 exactly, we

have to say how long Opie is.

The magnitude and we have to
say in which direction we

have traveled to get from O
to pee its direction.

So we've got 2 numbers that
would specify the position of P,

the Displacement precisely, the
magnitude and the direction.

Now there are all manner
of quantities that are

vector quantities. So for
instance another one that

we will meet is velocity.

Velocity is speed in a
particular direction. So if we

say we're traveling at 60 miles
an hour, then we're saying

that's on speed, but we haven't
specified in which direction

we're going. So if we say we're
traveling 60 miles an hour due

North, then we specified a
vector because we specified a

magnitude and direction.

A quantity which doesn't have
magnitude and direction but

just has magnitude alone is
called a scalar.

So let's just have a look at
those scale us.

Scalar quantities are
things like distance.

How far away?

We are from a fixed point. How
far we've traveled, but no

mention of a direction Mass.

Is another scalar? How much
of a quantity of we got 10

kilograms, 20 kilograms?
There isn't a direction

associated with that, so
that's just a scalar

quantity.

How then can we actually
represent vector quantities?

Because they've got a magnitude
and direction, we can represent

a vector quantity by a line

segment. Because this line
segment has a magnitude, its

length and it has a direction
as well. So that line segment

is clearly different to that
line segment. They may have the

same length if I measured them,
they might have, but they've

clearly got very different
directions and so they

represent very different
vectors.

Right, let's label these AB.
Usually we put an arrow on that

to show the direction that we're
going in from A to B, because

going from B to a is going in
the opposite direction a

different direction, so that's
how you might see these things

represented in textbooks. You
might also see them.

With a little letter against
him, usually that letter is in

very heavy black type known as
Clarendon time. It's very

difficult to show that when
you're writing, and So what we

usually do is we put a bar
either underneath or on the top

of the letter.

Doesn't matter whether it's
underneath or whether it's on

top. That's up to you and the
conventions that's being used at

the time. And normally we would
say that's the vector a bar.

Quite often. These vectors
are attached or referd to

some fixed .0 an origin.

So often we might say the
position vector of a point P

with respect to an origin. Oh so
then we have our position vector

R Bar for the point P. Notice
the position vector refers to

this line segment. Opi doesn't
just refer to pee itself. If we

wanted to refer to pee itself as

a. Point, we'd have to give it

some coordinates. But the
position vector of P is this

line segment OP. And if we want
to write that down in text, what

we might say is Opie with a bar
over it to show we're going from

oh to pee and to show it to
vector is equal to R Bar. This

looks a bit odd. Bars on top and
bars underneath, so perhaps one

of the things that we might do
to keep it tidy is in fact to

agree. To put all the bars on
top of said, it doesn't really

matter which way you choose to
do it. That's entirely up to

you or up to the textbook. All
the lecture that you've been

following.

OK. We've got an idea that
vectors can be represented by

line segments, and we know how
to write them down, and we know

how to recognize them in books.
Now we need to look at a little

bit of notation and some of the
properties that that notation

allows us to write down.

What if we say that two vectors,
a bar and B bar are equal? What

does a statement like that

actually mean? Well, one of the
things that it means is we know

that the length of A.

Length of a bar is actually
equal to the length of rebar.

We also know that they are in
the same direction.

So a statement about the
equality of two vectors tells us

two things. First, that they are
equal in magnitude, the length

of the line segments which
represent them are equal.

Secondly, that they are in the
same direction. Now we need a

way of writing this down. This
is very, very cumbersome. Let's

deal with the in the same

direction. First of all,
in the same direction is

the same as saying that a
bar is parallel to be bar.

In the same direction means
they are parallel. What about

length? We need some sort of
notation for us to be able to

talk about the length.

So let's have a look at that.
How might we express it if we've

got a vector like this?

Then writing that line segment
as a vector, we write it like

that. But if we want to say the
length, maybe then we can write

it as a bee without the bar on
top. Or we can say that it's

equal to the modulus of a bar.
These two vertical lines mean

modulus or size of. If we
represent our vector in this

way. A bar.

Then the modulus of a bar we can
write as that the size of or we

can leave the bar off and This
is why it's so important when

writing vectors to keep the
notation that shows when they

are vectors with the bar on the
top and when it's just a length

in a textbook, you can tell the
difference quite easily. This

will be in heavy Clarendon type
heavy black type, and this will

usually be in light type.

Usually italic type, but when
you're doing work for

yourself, it's very, very
important to write a vector as

a vector and to write a scalar
or the modulus of vector as a

scalar or as a modulus.

OK. We've got these quantities
called vectors. We know that we

can represent them by line
segments. We know how to express

the magnitude or the length of
that line segment, and therefore

the magnitude of the quantity.
We also know something about

what equality means. So what we
have to look at now is how can

we add two vectors together?

If we can define the addition
of two vectors, then we can

define the addition of any
number of vectors simply by

repeating the process. Take
the first 2, add them

together, then add on the
third one to that and so on.

So all we really need to look
at is how do we add two of

them together. So let's take
a vector a bar.

Let's take a
vector be bar.

How can we add these
two vectors together?

One way is to think about.

Vectors as being displacements
after all displacement is a

vector quantity, and to think
about what would happen if we

were to travel along this
displacement, we'd get to there.

And then if we were to travel
the same displacement be bar.

Would come to their.

So we would have taken a bar and
in some way added on B Bar.

And so the result ought to be
this vector across here. And

that's exactly the way that we
add these two together. We put

them there for a bar.

And there for be bar we put them
end to end and the result.

Is what we get by joining up
the triangle, so that is a

bar plus B bar.

I think that was called the

triangle law. There is another
way of doing this.

Game, let's take those two
vectors a bar.

And be bar. What's the 2nd way
of adding them together the 2nd

way of adding them together is
to make use of the parallelogram

law. So we attempt to
form a parallelogram,

so there's a bar.

There's people.

And what we do is we complete.

The parallelogram now
parallelogram opposite sides are

equal and parallel, so there's
be bar and there it is repeated

there. Same magnitude, same
direction. So these two are

equal as vectors.

And then we join that up there
and again. This is now a bar

once again because opposite
sides of a parallelogram are

equal in magnitude and are
parallel. I in the same

direction, so these two opposite
sides are equivalent as vectors.

And then of course the resultant
as we sometimes call it or the

sum is there across that
diagonal. A bar plus B Bar.

So this is exactly the same as
we have previously with a

triangle law, because here we've
got that same triangle

replicated. Now this is called

the sum. And it's also
called the resultant.

And we use those two words
interchangeably, the sum or

the resultant.

Having got that, how to add
them together? How do we

subtract 2 vectors?

Well again, let's have a
look at a bar.

And.

Be bar.

Ask ourselves how would we
subtract that from that. So what

is a bar minus B bar?

Well. We need a convention here.

Let's think of this as a
bar plus minus B Bar.

And then ask ourselves what's
minus B bar? Well, if this is B

bar from there to there.

Then that which is equal in
magnitude to be bar.

But in the reverse direction is
minus B Bar.

So the question, what is a bar
minus B bar becomes? What is a

bar added to minus B bar? So
let's just have a look at that

using the triangle law there
we've got a bar.

There we've got be
bar so minus B bar

must be this one.

And so therefore we've
added minus B bar on the

end across. There would
be a bar minus B Bar.

So we've got a geometric
representation.

Having added two different
vectors together, we need to

have a look. Now what happens
when you add two of the same

vectors together. So that's a
bar and we add on another a

bar. Add it on the end.

And then again we add on another
a bar, so we've a bar, a bar. So

what do we got? Well, clearly
we've got three of them, and So

what we must have is that a bar
plus a bar, a bar is 3 a bar.

And so if we have N times by a
bar, what this must mean is a

bar plus a bar plus plus.

A bar and so we'll have here N
of these a bars added together.

1 final little bit of notation
that we just need to have a look

at and that's to do with the
modulus. The length of a vector

we know. The modulus the
size of this vector.

We would write as a
with no bar on it.

What we want is a notation for
a vector that has unit modulus

unit length. Its length
is just one.

And what we do for that is we
put a little hat on it instead

of a bar we put a hat on
it and that stands for a unit

vector, its length, its size,
its modulus is one. So if we do

that, that is equal to 1.

But actually gives
us a way of writing.

The vector a bar, because
what it means is that the

vector a bar, because it's
got magnitude little a. We

can write it as little a
times by the unit vector.

So a bar can be written as
little a. The magnitude of A the

length, the size of it.

Multiplied by this vector,
which has a unit vector which

has unit length length one, and
that's going to be very very

helpful to us indeed. OK, let's
now have a look at using these

vectors in some geometry.
Vectors are very, very

powerful. They can help us to
prove some theorems that will

take a great deal of space and
time if we working in Euclidean

geometry.

Let's begin by looking at the
theorem, which is called the

midpoint theorem. I'm going to
take two points.

A.

And be these two points are
going to be defined with respect

to an origin. Oh.

And so I've got the position
vector of a, let's call it a bar

and the position vector of B,
let's call it.

Be bar. If I join A to B
with a straight line, then I've

got a triangle.

I'm going to take the midpoints.

Of these two sides, an.

And N. I'm going
to join them up.

The question is, is there any
relationship between this line

MN and this line AB?

So let's have a look and see
if there is we can do this

very easily with vectors.

First of all, let's think how
can we write a B as a vector?

May be.

Well, one thing about vectors is
they are line segments, so they

are displacements and so we can
think of a journey from A to B

as being a journey via the
origin and so we can write that

as. AO plus
OB.

Now in going from a 20.

We're going against the
direction of a bar.

So OK is in fact this
vector minus a bar.

Going from oh to be is in the
direction of B bar and so that

is B bar and so we've got B bar
minus a bar.

What about the vector MN?

Let's write that down MN.

Or following the same
reasoning, this is Mo

plus ONMO plus ON.

From M2, Oh well, we know that M
is the midpoint of OK, so it's

half way along it and it's in
the same direction, so om must

be 1/2 of a bar where going in.
Actual fact from M2. Oh, so

we're going against that
direction, and so it must be

minus 1/2 of a bars. Plus here
we're going from O2 N.

So we're going half way along
Obi and we're going in the

direction of B bar so that
vector ON must be 1/2 of B Bar.

How else can we write this
down? Well, for a start,

there's a common factor of
1/2 that we can take out.

And then we can write that as B
bar minus a bar.

Now we can compare these two.

What we can see is that the

vector AB. And the vector MN
have the same vector part. They

both got this be bar minus a

bar. Further, MN is actually
a half of a bar minus B

bar. So what we've managed to
prove is that MN as a vector

is equal to 1/2 of a B
as a vector.

And what does that mean?
Remember any statement about

vectors tells us two things
tells us something about the

magnitude and something about
the directions. So this must

mean that these two vectors are
equal in magnitude, and so MN

must be equal to 1/2 of a bee.
In other words, the length of MN

is 1/2 the length of a B, and
they must be in the same

direction. And so M Ann is.

Carol till 1/2 of a bee and
therefore Parral to AB. Now this

result is known as the midpoint
theorem for a triangle, which

states if you join the midpoint.

Of two sides of a triangle. Then
the resulting line is equal to

1/2, the third side of the
triangle and these parallel to

it, and that's what we've got
here. Eh? Man is 1/2 of a be an

is 1/2, the third side an MN is
parallel to a be in the same

direction. So let's now
apply this result. 2

quadrilaterals indeed, to
any four points in space.

So let me take my
four points ABC.

And a.

Complete these ABC and D and

let's label. The midpoints of
each of these line segments will

call them P.

QRS.

Now let's join them up.

The question we can ask is what
sort of shape is PQ R&S?

I'm going to join a C.

Now the midpoint theorem tells
us that the vector peak you

must be equal to 1/2 of
the vector AC.

Because these two points P&Q,
they are the midpoints of two

sides of a triangle, and so
this line must be equal to 1/2

of this line in length. The
half the third side an

parallel to it. Similarly,
because of the midpoint

theorem, this vector Sr.

Most also be equal to 1/2 of a
C because again as and are the

midpoints of two sides of a
triangle, and so the line joins

them as our must be half the
length of the third side and

parallel to it, so Sr equals 1/2
of AC because of our midpoint

theorem. So therefore PQ mostly
equal to Sr as vectors.

Any statement about vectors
immediately tells us two things.

It tells us a statement about
the magnitudes, so this tells us

that the length of PQ is equal
to the length of Sr and about

direction. So it tells us that
PQ is power off to Sr.

So here we have a 4 sided

figure. And we know that one
pair of opposite sides are equal

in length and are parallel.

And that's exactly the property
of a parallelogram. So therefore

PQRS is a.

Parallel oh gram
parallelogram.

Let's look at just
one more theorem.

Again, will take two

points A&B. And their
position vectors will be a bar

and B bar with respect to an

origin. Oh. Will join them.

And what we're going to be

looking at? Is a point P.

On a bee that divides AB
in the ratio MNM parts to

end parts, and the question that
we want to ask is what's

the position vector of opie?
What is R bar the position

vector of OP?

Well, let's begin by writing
down how we might travel from O

to pee. Or in order to
go from oh to P, we might

go from Oh 2A and then from
A to pee. So that would be

OA plus AP.

OK, is fine. That's a bar. What
about AP? One of the things that

we can see is that AP is in the
direction of a B.

Not only is it in the direction
of a B, but it's M parts of

a B out of N plus N parts,
and so AP.

Is M parts out of
M plus N parts of

a bee?

Treating these as vectors, and
so we can now ask ourselves what

is AB bar?

And to go from A to B, we
can go a oh plus oh be, that

would be a O plus OB.

And we can write that down. Oh,
to be is be bar.

And a 20. He's going against the
direction of the arrow, so that

is minus a bar.

And now we can put these three
statements together this one.

Replacing AP bar by this and
replacing AB bar in here by

this so that we end up with
everything in terms of a bar

and be bar.

So if we write these three
results out again.

OPOA plus
AP.

And we saw a pee was
M parts out of M plus

N parts of a bar and

we all. So soul that AB
bar was be bar minus a

bar.

And now what we want to do is to
put these three things

altogether. So oh P will be
equal to OA is just a bar.

Plus and instead of a pea,
we're going to take this

and instead of the eh bee,
we're going to take this

so Opie Bar is a bar plus.

M over N plus N.

Times B bar
minus a bar.

We can put all this together
over a common denominator so we

have a bar times N plus N plus M
Times B bar minus a bar.

All over N plus N.

And now we need to look at these
brackets. I've got a bar times

by M. And I've got M times by
minus a bar, so those two are

going to take each other out.
They're going to cancel each

other out, and so I'm going to
be left with a bar times by N&M

times by B Bar.

And so the result I have
is NA bar plus MB bar

all over N plus N.

Now that doesn't look
particularly startling, but it

does give us a way of
calculating what the position

vector of P is when we know the
ratio in which it divides a

baby. Let's just remember what
the diagram was.

We had our two points A&B refer
to an origin. Oh and we had the

point P on the line AB.

We were calculating the position
vector when P divided AB in

the ratio M to N.

And so imagine that M was one.

And N was one as well. Then this
would say that P was the

midpoint, 'cause we've divided
in the ratio one to one equal

parts. This would say that the
position vector of the midpoint

was a plus B all over 2 sounds
are perfectly reasonable answer

if M was two and N was one. In
other words, P came 2/3 of the

way along this line. Then this
would say that this was a bar

plus. To be bar all over
three. So we've got that. A

way of calculating what the
position vector is of this

point P when it divides AB
in the ratio M tool N.