www.mathcentre.ac.uk/.../The%20scalar%20product.mp4

In this unit, we're going to
have a look at a method for

combining two vectors. This
method is called the scalar

product. And it's called the
scalar product because when we

calculate it, the result that we
get or the answer is a scalar as

opposed to a vector.

So let's look at two vectors.
Here's the first vector.

That's cool that vector A.

And here's the second vector.
Let's call this vector B and

you'll notice that I've drawn
these two vectors so that the

tales of the two vectors
coincide.

Now that they coincide, we can

measure this angle. Between the
two vectors and I'm going to

label the angle theater.

So there are two vectors.

Now when we calculate the scalar
product, we do it like this. We

find the length. Or the
magnitude, or the modulus of

the first vector.

Which is the modulus of a.

We multiply it by the length.

Of the second vector, which is
the modulus of beat.

And we multiplied by the cosine
of this angle between the two

vectors. We multiplied by the
cosine of Theta.

And that's how we calculate the
scalar product. The length of

the first vector times the
length of the second vector

multiplied by the cosine of the
angle in between the two

vectors. And we have a
notation for the scalar

product and we write it like
this vector A dot vector be.

And this is the definition.
Sometimes this scalar product is

also referred to as a dot
product because of that dot in

there. So sometimes you'll hear
it referred to as a dot product.

OK, so let's have an example and
see if we can calculate the dot

product of two vectors.

Let's suppose our vector a here.

Has got length for units.

And let's suppose we have a
vector B up here and let's

suppose vector B has length
5 units.

And let's suppose that the angle
for the sake of argument between

the two vectors is 60 degrees.

And let's use this formula to
calculate the scalar product A

dot B. It's a vector there.

A dot B will be.

Well, it's the modulus or the
length of the first vector.

Which is 4.

Multiplied by the modulus
or the length of the

second vector and the
length of the second

vector is 5.

Multiplied by the cosine of the
angle between the two, and so we

want the cosine of 60 degrees.

Now the cosine of 60 degrees. If
you don't know, you can work it

out on your Calculator, but it's
a common angle an the cosine of

60 degrees is 1/2, so we've got
4 * 5 * 1/2, so if we work all

that out, we get 4 files, or 20
and a half of 20 is 10.

So that's our first example of
calculating a scalar product,

and let me remind you of
something that I said Very early

on that when we calculate the
scalar product, the answer is a

scalar and you'll see the answer
here 10 is just a single number

Whilst we started with two
vectors A&B, we finish up with

just a single number, a scalar,
and that's the scalar product.

Let me illustrate some other
features of this scalar product

by doing the calculation the
opposite way round, supposing

I'd try to calculate the dotted
with a instead. Let's work that

through and see what happens.

So I've done the operation in a
different order instead of a

baby. I've now got dot A.

Well, again, using the
definition we want to say that

the dot product of DNA.

Is the modulus or length of the
first vector which is the

modulus or the length of be this
time and the modulus of B is 5?

Multiply by the modulus or the
length of the second vector.

Which is now 4.

Multiplied by the cosine of the
angle between the two and the

angle still 60 degrees.

The cosine of 60 degrees is 1/2,
so if we work this out this time

will get five 420 1/2 of 20 is
10, so you'll see that whichever

way we did the calculation
whether we worked out a dot B or

whether we worked out B dotted
with a, we get the same answer.

10 and that's true in general
for any two vectors that we

choose. If we workout a dot B.

That will be the same as working
out B Dot A.

And that's another important
property of the scalar product,

and we call this property

commutativity. We say that the
dot product or the scalar

product is commutative.

Now, as well as the
commutativity property, I

want to tell you about
another property which is

known as distributivity.

Suppose we have a dotted with
the sum of two vectors B Plus C.

The distributivity property
tells us to expand these

brackets in the normal sort of
algebraic way that you would

expect. We work this out by
saying it's a.

Dotted with B.

Add it to.

A dotted with C.

This also works the other
way round, so suppose we

have B Plus C.

And we want to dot it with the
vector a As you might predict

the result that will get is B
dotted with a.

Attitude.

See dotted with a.

And these properties are said to
be the distributivity.

Rules.

I will need those rules along
with the commutativity rules

later on in this unit.

Now there's another property
very important property of the

scalar product that I like to
tell you about as well, and

it's the scalar product of two
perpendicular vectors.

Let's see what happens when the
two vectors that we're dealing

with A&B are perpendicular. That
means there at right angles. So

we have a situation like this.
There's a vector B. There's a

vector A and they're separated
by an angle of 90 degrees, so

the vectors are perpendicular.

Let's use our formula for the
dot product. The scalar

product, the modulus of the
first one with the modulus of

the second one times the
cosine of the angle between

the two. Let's use that
formula in this specific case,

when the vectors are
perpendicular.

Well, we want the length of
the first one. Still the

modulus of A.

The length of the second one
will be the modulus of B and

this time we want the cosine of
Theta, but theater is 90 degrees

from the cosine of 90 degrees.

Now you either should know
or you can easily check on

your Calculator that the
cosine of 90 degrees is 0.

So these two vectors, these two

moduli? Whatever they are
multiplied by zero. So the

answer that will get will be 0.

This is a very important result
that if you have two vectors

which are perpendicular, their
scalar product is 0.

So for two perpendicular
vectors a.

The.

Important result which
you should learn that

their dot product their
scalar product is 0.

Now the converse of this is also
true, and by that I mean that if

we choose any two vectors at
all, and we find their dot

product and we find that the
answer is 0, that will mean that

provided that neither a norby
was zero, those two vectors must

be perpendicular and will use
this as a test later on in this

unit to test whether or not to

given vector. Are
perpendicular or not?

I want to move on now to look at
how we calculate the scalar

product of two vectors when
they're given in Cartesian form.

Let me remind you about
cartesian form when we're

dealing with vectors in
Cartesian form, where we express

the vector in terms of unit
vectors IJ&K. So we might be

dealing with vectors of this

form 3I. Minus two J.

Plus 7K.

Expect to be might be something
like minus 5I.

Plus 4J. Minus 3K. So when
we see the vectors written down

like this with eyes and Jays and

kays in. These vectors are now
given in Cartesian form.

And we're going to learn how
to do now is figure out how

to calculate the dot
product. The scalar product

vectors like these.

Before we do that, I'd like to
introduce some other important

results. Let me remind you a
little bit about IJ&K.

In three dimensions, where we
have an X axis.

AY axis. And as Ed Access.

We did note unit vectors along
the X axis.

I I along the Y axis by J and
along the Z Axis by K and you'll

notice that the vectors IJ&K.

US are all it's 90 degrees to
each other, so the angle between

I&J is 90 degrees. The angle
between Jane K is 90 degrees and

the angle between I&K is 90
degrees and all of these vectors

I Jane KA unit vectors. That
means that their length is one.

Now let's just calculate the dot
product of I with J.

We use the result we had before
the definition of the dot

product, which says that we
write down the length of the

first vector and the length of I
being a unit vector is just one.

The length of J being a unit
vector is also one.

And we want the cosine of the
angle between the two and the

angle between the two is 90
degrees, so we want the cosine

of 90 degrees. And again, the
cosine of 90 degrees is 0, so

we find that I got Jay is
simply zero. We could have

deduced that from the
previous result that I

obtained for you, which was
that for any two vectors

which are perpendicular,
their dot product is 0.

So I got Jays 0. Now the same
argument tells us also that Jay

dotted with K must be 0.

Be cause J&K are
perpendicular vectors

there at right angles and
the angle between them is

90 degrees and the cosine
of 90 is 0.

What about INK? Well, I dotted
with K must also be 0.

What about dotting the vector
with itself? Let's just have a

look at that. Suppose we wanted
I dotted with I.

Well, the length of the
first vector is one.

The length of the second vector
is still one, and if we're

dotting a vector with itself,
the angle between the two

vectors must be 0. So this time
we want the cosine of 0.

And the cosine of 0. The cosine
of note is one.

So we have 1 * 1 * 1, which
is just one.

So if we're dotting the vector I
with itself, we get one.

A similar argument will tell us
that if we got the vector Jay

with itself, we also get one,
and if we drop the vector K with

itself, we get one.

Let me summarize those
important results here and

I've results that you should
remember that if you took the

vector I and dotted it with
itself, that would be the same

as dotting Jay with itself.

And it's the same as Dot Inc. A
with itself, and in any case we

get the answer 1.

If we don't, I with J we get
zero if we got I with K if we

get zero and if we got J with K
we also get 0.

And all those results are
particularly important, and

once that you should become
familiar with as you work

through the exercises
accompanying this unit.

Now I want to use those results
about the dot products of those

unit vectors to find a way to
calculate the dot product of two

arbitrary vectors that are given
in Cartesian form. So let's

choose two arbitrary vectors.
Let's suppose the first one.

Is a one.

I.

A2, J and
a 3K.

So this is now an arbitrary

vector. A where a one A2 and a
three are the three cartesian

components of the vector A.

And let's suppose vector B
similarly is B1I.

Plus B2J Plus V3K and
again B1B2B3 arbitrary numbers,

so this is an arbitrary
vector in three dimensions.

What I want to do is workout the
dot product of A&B.

So let's work it through a
dot B equals.

It's the first vector
which is all this.

Dotted with the second vector
be, which is all this.

OK. Now what we want to do
now is use our

distributivity rule to be
able to workout to expand

these brackets and workout
the individual dot products.

And we can do that using the
rules that we know that in turn

each one of these vectors.

In the first bracket,
multiplies each of these in

the SEC bracket and just the
normal algebraic way. So

first of all we want a one I
dotted with B1 I those terms.

Then we want a one I.

Dotted with B2J

and finally a one
I dotted with B3K.

So that takes care of this first

vector here. And we moved to
the second vector. So now we

want plus A2, J, dotted with
each of these three.

So it's a 2 J dotted with B1I.

A2 J dotted with B2J.

And a two J dotted with B3K.

And finally, this loss vector in
the first bracket dotted with

each of these vectors in the SEC
bracket will be a 3K.

Dotted with B1I

plus a 3K dotted with B2J.

And finally, a 3K
dotted with B3K.

Now this looks horrendous. With
all these nine terms in here,

but a lot of this is going to
cancel out now. You'll remember

that any vectors which are
perpendicular to each other.

Have a dot product which is 0.

Now this vector.

Is in the eye direction and this

vector. Is in the J direction,
so a one I is perpendicular to

be 2 J.

So where we have the item dotted
with the J term, this must be 0.

Similarly, any vector in the
direction of I any multiple of I

must be perpendicular to any
multiple of K.

So this dot product is
also zero.

And similarly for the A2JB1I,
those vectors are perpendicular.

Their dot product is 0.

A2J dot B3K. A perpendicular

that's zero. Similarly, that
will be 0 and that will be 0.

And we'll be left with
three non zero terms.

What about this term here?

A1 I be one I now this is a
multiple of I and this is a

multiple of I. So these two

vectors are parallel. The angle
between them is 0, so when we

work out the dot product will
want the length of the first

one, which is a one.

The length of the second one,

which is. He wants.

And the cosine of the angle
between the two. The angle

between the two. Is zero and the
cosine of 0 is one, so we get a

one B 1 * 1.

So this first term, whilst it
looked quite complicated, just

simplifies to a 1B one.

Similarly, these two vectors
here. This is a vector in the

direction of J and this is a
vector in the direction of Jay.

So the angle between the two of
them is 0, the cosine of 0 is

one, so when we workout this dot

product we want. The length of
the first one, which is a 2. The

length of the second one which

is B2. Multiplied by the cosine
of 0. Cosine of 0 is one, so we

just get a 2B2.

And finally, same argument
applies here. This is a

multiple of K. This is a
multiple of K, so the vectors

are parallel. The angle between
them is zero and the cosine of

0 is one. So we get the length
of the first being A3 length of

the second being B3 and the
cosine of 0 cosine of 0 is one,

so the whole lot. All of this
complicated stuff just

simplifies down at the end of
the day to this very important

result here.

And what this result says is
if you want to find the scalar

product of two vectors A&B
which are given in this

cartesian form, like that, all
we need to do is multiply the

I components together. You see
the A1B one is the eye

components multiplied
together.

Multiply the J components

together A2B2. Multiply the K
components together A3B3.

And then finally add up these

results. So the dot product.

Is the sum.

Of the products of the
corresponding components.

And you'll notice that the
result we get here on the

right hand side doesn't have
any eyes or Jays orkez in the

answer anymore. This is purely
a number. This is purely a

scalar, as we'd expect when we
calculate a scalar product.

Point out also that this
expression on the right, which

is the sum of the products of
the corresponding components is

also sometimes referred to as
the inner product.

Of A&B, so on occasions you may
hear some staff or reading some

textbooks that this is called
the inner product of Bambi.

OK, so we have a formula
now for the dot product.

It's a one B 1 +
8, two B2 plus A3B3.

Can very important results and
this will allow us to calculate

the scalar product of two
vectors given in cartesian form.

Let me give you an example.

Suppose the first vector a is
this one 4I Plus 3J minus.

Sorry plus 7K.

And Suppose B is the vector two

I. +5 J.

Plus 4K.

Suppose we want to calculate the
scalar product A dot B.

Well, this formula tells us that
we multiply the corresponding I

components together A1B one.
This is a one for this is B1

which is 2. So we multiply the
four by the two.

We multiply the corresponding J
components together. A2B2 well,

this is a two and this is B2, so
we want 3 * 5.

And then we multiply the
corresponding K components

together. A3B3, which is 7 * 4.

When we do that and we add up
all the results and will get 8

+ 3, five, 15, Seven, 428. And
if we work those out we shall

get 2815 and eight to add up
to 51. So the dot product of

A&B is the scalar 51.

Now, sometimes, instead of
writing it all out like this, we

can use column vector notation
and some people find it a bit

easier to work with column
vectors. So just let me show you

how we do the same calculation.
If we had these vectors given

not in this form, but as column
vectors, suppose we had a

written as 437. And
be written AS254.

Then the a dotted with B.

Would be 437 dotted
with 254.

Now to multiply corresponding
components together is quite

easy to see what they are now
because we want 4 * 2, which is

8. 3 * 5 which is 15 and 7 *
4, which is 28. And if we had

those that will get 51 like we
did over here. So some people

might find it a bit easier just
to write them as column vectors

and then just read off the
corresponding components,

multiply them together, and add
them up to get the result.

Now it's so important that you
know how to calculate a scalar

product that I'm going to give
you one more example, just to

make sure that you understand
the process.

And will go straight into column
vector notation. Suppose the

vector A is the vector minus 6.

3 - 11.

That's minus six I plus 3J minus

11 K. And let's suppose
the vector B is the

vector 12:04 that's twelve
IOJ plus 4K.

So the dot product A dot B will
be minus six 3 - 11.

Dotted with twelve 04.

And in the column vector
notation, it's so easy to work

this out. We want minus 6 * 12,
which is minus 72.

3 * 0 is 0.

And minus 11 * 4 is minus 44.

And if we add these up, we get.

Minus 116

so the dot product of A&B is
just the sum.

Of the products of the
corresponding components.

I want to move on now to look at
some applications of the scalar

product and the first
application is one that I

mentioned very early on and it's
to do with testing whether or

not two given vectors are

perpendicular or not. Now you
remember from the definition of

the scalar product of two
vectors, 8 be that is the

modulus of the first one times
the modulus of the second one

times the cosine of the angle in
between the two vectors.

Now, if we find that
when we work this out

that the result is 0.

Then either the modulus of a
must be 0, or the modulus of be

must be 0. Or Alternatively,
cosine theater must be 0.

Now, if we know that the two
given vectors do not have a zero

length, In other words, the
modulus of a can't be 0, and the

modulus of B can't be 0.

Then the only conclusion we can
draw is that cosine theater must

be 0, and from that we did use
that theater must be 90 degrees.

In other words, if we take the
dot product of two vectors and

we find we get the answer 0.

Then, provided the two factors
were nonzero vectors, we can

deduce the amb must be
perpendicular vectors, so.

If A&B.

A nonzero vectors.

Search that.

Hey, don't be when we work it
out. Turns out to be 0.

Then A&B.

Are perpendicular.

There at right angles.

Another word we sometimes use is
orthogonal. You may hear that,

particularly in more advanced
work, two vectors at right

angles to vectors which are
perpendicular are sometimes said

to be orthogonal.

So we have a test here we take
the two vectors, we find their

dot product and if the answer
that we get zero then we deduce

that the two vectors must be
perpendicular. So that's a very

easy test we can apply to see if
two vectors are Purple ****.

Let me give you an example.

Suppose we have the Vector A,
which is the vector three 2 -

1 and the Vector B which is
1 - 2 - 1.

Now I think you would agree that
just by looking at these vectors

three, I +2 J Minus K.

And one I minus two J minus K.
You'd have no idea just from

looking at them written down
like that. Whether or not these

vectors where at right angles to

each other. We can apply this
dot product test and workout a

baby to see what happens. So
let's workout a dot B.

I want the dot product of three
2 - 1 with 1 - 2 - 1.

And we'll get 3 ones or three.

Two times minus 2 - 4.

And minus one times minus one is
plus one. So we've got three and

one which is 4 - 4 which is 0.
So you'll see that when we work

out the dot product of these two
nonzero vectors, the answer that

we get is 0.

So we can deduce from this
result that this vector a must

be at right angles to this
vector B and that's something

that's not obvious just from
looking at it. So we've got a

very useful test there to test
for. The two vectors are

perpendicular or not.

Now another important
application of the scalar

product is to finding the angle
between two given vectors. Let

me remind you of the definition
of the scalar product again

because we will need that the
scalar product of A&B is found

by taking the length of the
first vector times the length of

the second vector times the
cosine of the angle in between

the two vectors so.

If we get a vector, if we get
two vectors in Cartesian form.

And we calculate their dot

product. And if we know how to
calculate the modulus of each of

those two vectors, then the only
thing in this expression that we

don't know is cosine Theta, so
will be able to use this

definition to calculate cosine
theater the cosine of the angle

between the two vectors, from
which we can deduce the angle

itself. OK, let's rearrange

this. If we divide both sides by
the modular surveying, the

modulus of B will be able to get
cosine theater is a dot B

divided by the modulus of a
modulus of B and that's the

formula that we're going to use
in a minute to find cosine,

Theta and hence theater.

Now we've done a lot of work
already in this unit on

calculating the dot product. Let
me just remind you a little bit

about how you find the modulus
of a vector when it's given in

cartesian form. If we have a
Vector A, which is a one

I plus A2J plus a 3K.

Then the modulus of this
vector is found by squaring

the individual components,
adding them.

And finally, taking the square
root of the result. So this is a

formula that we're going to use
for finding the modulus of a

vector in Cartesian form, and
this has been covered in an

early you unit. If you need to
look back at that.

Let's look at a specific
example where we want to find

the angle between two vectors.

So the problem is
find the angle.

Between the two vectors, I'm
going to choose are A which is

4I Plus 3J plus 7K.

And B.

Which is 2 I plus 5J Plus
4K, so two vectors given in

cartesian form and the
problem is to find the angle

between them.

And we have this result that we
just reduced the cosine of the

required angle. Is the dot
product 8B divided by the length

of a times the length of be? So
that's the formula that we're

going to use.

OK, a dot B now. In fact a dot B
has already been evaluated in an

earlier example, but I'll just
remind you of that a dot B. If

we use the column vector
notation will be 437.

Dotted with 254

which is 428-3515 and Seven 428.
And if you work that out, you

find you get 51.

So all we need to do is
calculate now the modular

survey in the modulus of B and
then we've got everything we

need to know to put in the
right hand side of this

formula.

OK, the modular survey.

Now the modulus of a is found by
taking the square root of the

squares of these components
added up. So we want 4 squared.

3 squared

7 squared.

Which is going to be
the square root of 16.

330977's of 49 and if you work
that out, you'll find that

that's the square root of 74.

Similarly, the modulus of B will
be the square root of 2 squared.

+5 squared, +4 squared.

Which is the square root of 4
+ 25 + 16 and 25 +

4 + 16 is 45.

So we've got all the
ingredients we need now to

pop into this formula.

Cosine theater will be a dotted
with B, which we found was 51

divided by the modular survey,
which is the square root of 74.

The modulus of B, which is the
square root of 45 now will need

a Calculator to work this out.

We want 51.

Divided by the square
root of 74.

Divided by the square root.

45 Which is not .8838
to 4 decimal places. That's the

cosine of the angle between the
two vectors A&B.

If we find the inverse cosine.

Of .8838.

Which is 27.90
degrees.

So there we've seen how we can
use the scalar product.

To find the angle between
two vectors when they're

given in Cartesian form.

Now there's one more application
I'd like to tell you about, and

it's concerned with finding the
components of one vector in the

direction of a second vector.
Let me give you an example.

Suppose we have a vector A.

Let me have a second vector.

B.

And I've drawn these vectors so
that their tails coincide and.

As usual, the angle between the
two vectors is theater.

Let's call this .0.

And what I'm going to do is
I'm going to drop a

perpendicular from this
point, which I'll call B

down.

To meet the vector A.

Let's call this point here a.

And we can regard the vector OB.

As the sum of the vector
from O2 Capital A and then

the vector from A to B. In
other words, the vector OB.

Is the sum of OA.

Plus a bee.

Now this vector here, which
we've called OK, that vector is

said to be the component of B in
the direction of a. You'll see

we can regard be as being made
up of two components, one in the

direction of A and one which is
at right angles to that. So this

component, oh A.

Is the component of B
in the direction of a.

And what we're going to do is
we're going to see how we can

use the scalar product to
find this component.

Let's call this length here the
length from O to a. Let's call

that little. And then let's
focus our attention on this

right angled triangle.

Oba.

So there's a right angle
in there.

Now, using our knowledge of
trigonometry, we can write down

the cosine of this angle is
adjacent over hypotenuse, so the

cosine of Theta.

Is this length L
the adjacent side?

Over the hypotenuse, now
the hypotenuse is the

length of this side.

And this is the length of
this side is the modulus of

vector be.

Now, In other words, if we
rearrange this L, the component

of B in the direction of a week
and right as the modulus of be

times, the cosine Theta.

Now we've already got an
expression for the cosine of

this angle in the previous work
that we've done, we already know

that the cosine of feta can be
written as the dot product for

amb. Divided by the modular
survey times the modulus of be,

that's the result we just had.

So if you look at this
expression now, you'll see

with the modulus B in the
numerator and a modulus of be

in the denominator, they
cancel.

So we can write this.

As a not be over
the modulus of A.

Let me write that down
again on the next page.

So this remember L is the
component of be in the direction

of a. We can write it like this.
I'm going to Rearrange this a

little bit to write it as a
divided by the modulus of A.

Dotted with effective be.

And because of the result we
got very early on the

commutativity of the dot
product. We can write this as

B dot A divided by the modulus
of A.

Now when you take any vector and
you divide it by its modulus,

what you get is a unit vector in
the direction of that vector,

and we write a unit vector as
the vector with a little hat on

like that. So this is a unit
vector in the direction of A.

So this is an important result
we've got. We've got L, which

is the component of be in the
direction of a can be found by

taking the dot product of be
with a unit vector in the

direction of A.

So suppose we wish to find the
component of Vector B which is 3

I plus J Plus 4K in the
direction of the Vector A, which

is I minus J Plus K.

What we want to do is
evaluate B dotted with a

divided by the modular
survey.

Now we can do that as follows.
We can workout the dot product B

dot A and then divide by the
modulus of a. So B Dot A is

going to be 3 * 1, which is 3.

Plus one times minus one which
is minus 1 + 4 * 1 which is 4 or

divided by the modulus of a
which is the square root of 1

squared plus minus one squared
plus one squared, which is the

square root of 3.

If we work that out, will get 3
- 1 is 2 + 4 six six over Route

3. This means that the component
of B in the direction of a is 6.

Over Route 3. We might want to
write that in an alternative

form just to tidy it up so we
don't have the square root in

the denominator and we can do
that by multiplying top and

bottom by the square root of 3,
which will produce Route 3 times

route 3 in the denominator,
which is 3 and three into six

goes twice. So that
would just simplify to

two square root of 3.

So in this unit we've introduced
the scalar product.

Learn how to calculate it and
looked at some of its

properties and applications.