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.