WEBVTT

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

00:00:06.214 --> 00:00:10.204
combining two vectors. This
method is called the scalar

00:00:10.204 --> 00:00:14.076
product. And it's called the
scalar product because when we

00:00:14.076 --> 00:00:18.332
calculate it, the result that we
get or the answer is a scalar as

00:00:18.332 --> 00:00:19.548
opposed to a vector.

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

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That's cool that vector A.

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

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

00:00:41.386 --> 00:00:43.804
tales of the two vectors
coincide.

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Now that they coincide, we can

00:00:47.788 --> 00:00:52.090
measure this angle. Between the
two vectors and I'm going to

00:00:52.090 --> 00:00:53.470
label the angle theater.

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So there are two vectors.

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

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

00:01:05.013 --> 00:01:06.000
the first vector.

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Which is the modulus of a.

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We multiply it by the length.

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

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

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vectors. We multiplied by the
cosine of Theta.

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

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the first vector times the
length of the second vector

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multiplied by the cosine of the
angle in between the two

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vectors. And we have a
notation for the scalar

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

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And this is the definition.
Sometimes this scalar product is

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

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

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

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product of two vectors.

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Let's suppose our vector a here.

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Has got length for units.

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

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suppose vector B has length
5 units.

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

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the two vectors is 60 degrees.

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

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dot B. It's a vector there.

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A dot B will be.

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

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Which is 4.

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Multiplied by the modulus
or the length of the

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second vector and the
length of the second

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vector is 5.

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

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want the cosine of 60 degrees.

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

00:03:25.066 --> 00:03:28.888
out on your Calculator, but it's
a common angle an the cosine of

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

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

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

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

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

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

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

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

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Let me illustrate some other
features of this scalar product

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by doing the calculation the
opposite way round, supposing

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

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through and see what happens.

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

00:04:23.998 --> 00:04:25.822
baby. I've now got dot A.

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

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the dot product of DNA.

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

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

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

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Which is now 4.

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

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angle still 60 degrees.

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

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

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

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

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

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choose. If we workout a dot B.

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

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

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and we call this property

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commutativity. We say that the
dot product or the scalar

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product is commutative.

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Now, as well as the
commutativity property, I

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want to tell you about
another property which is

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known as distributivity.

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

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The distributivity property
tells us to expand these

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brackets in the normal sort of
algebraic way that you would

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

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Dotted with B.

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Add it to.

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A dotted with C.

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This also works the other
way round, so suppose we

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have B Plus C.

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

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the result that will get is B
dotted with a.

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Attitude.

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See dotted with a.

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And these properties are said to
be the distributivity.

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Rules.

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I will need those rules along
with the commutativity rules

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later on in this unit.

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Now there's another property
very important property of the

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

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it's the scalar product of two
perpendicular vectors.

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

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

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

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

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the vectors are perpendicular.

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

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

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the second one times the
cosine of the angle between

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

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when the vectors are
perpendicular.

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

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modulus of A.

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

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

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from the cosine of 90 degrees.

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Now you either should know
or you can easily check on

00:08:46.888 --> 00:08:49.668
your Calculator that the
cosine of 90 degrees is 0.

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So these two vectors, these two

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moduli? Whatever they are
multiplied by zero. So the

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answer that will get will be 0.

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

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which are perpendicular, their
scalar product is 0.

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So for two perpendicular
vectors a.

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The.

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Important result which
you should learn that

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their dot product their
scalar product is 0.

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

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

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

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

00:09:53.333 --> 00:09:57.272
be perpendicular and will use
this as a test later on in this

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unit to test whether or not to

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given vector. Are
perpendicular or not?

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

00:10:13.880 --> 00:10:17.480
product of two vectors when
they're given in Cartesian form.

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Let me remind you about
cartesian form when we're

00:10:20.720 --> 00:10:23.960
dealing with vectors in
Cartesian form, where we express

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

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dealing with vectors of this

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form 3I. Minus two J.

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Plus 7K.

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Expect to be might be something
like minus 5I.

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

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like this with eyes and Jays and

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

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

00:10:59.921 --> 00:11:02.457
to calculate the dot
product. The scalar product

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vectors like these.

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

00:11:08.331 --> 00:11:11.841
results. Let me remind you a
little bit about IJ&K.

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In three dimensions, where we
have an X axis.

00:11:20.410 --> 00:11:23.310
AY axis. And as Ed Access.

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We did note unit vectors along
the X axis.

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

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notice that the vectors IJ&K.

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

00:11:43.730 --> 00:11:47.890
I&J is 90 degrees. The angle
between Jane K is 90 degrees and

00:11:47.890 --> 00:11:51.730
the angle between I&K is 90
degrees and all of these vectors

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

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

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

00:12:11.600 --> 00:12:14.625
product, which says that we
write down the length of the

00:12:14.625 --> 00:12:18.475
first vector and the length of I
being a unit vector is just one.

00:12:20.880 --> 00:12:24.389
The length of J being a unit
vector is also one.

00:12:25.800 --> 00:12:29.687
And we want the cosine of the
angle between the two and the

00:12:29.687 --> 00:12:33.275
angle between the two is 90
degrees, so we want the cosine

00:12:33.275 --> 00:12:38.500
of 90 degrees. And again, the
cosine of 90 degrees is 0, so

00:12:38.500 --> 00:12:42.964
we find that I got Jay is
simply zero. We could have

00:12:42.964 --> 00:12:45.940
deduced that from the
previous result that I

00:12:45.940 --> 00:12:49.660
obtained for you, which was
that for any two vectors

00:12:49.660 --> 00:12:52.636
which are perpendicular,
their dot product is 0.

00:12:54.900 --> 00:13:00.234
So I got Jays 0. Now the same
argument tells us also that Jay

00:13:00.234 --> 00:13:02.520
dotted with K must be 0.

00:13:07.400 --> 00:13:09.296
Be cause J&K are
perpendicular vectors

00:13:09.296 --> 00:13:12.456
there at right angles and
the angle between them is

00:13:12.456 --> 00:13:15.300
90 degrees and the cosine
of 90 is 0.

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

00:13:25.900 --> 00:13:28.958
What about dotting the vector
with itself? Let's just have a

00:13:28.958 --> 00:13:31.738
look at that. Suppose we wanted
I dotted with I.

00:13:34.330 --> 00:13:36.877
Well, the length of the
first vector is one.

00:13:38.360 --> 00:13:42.248
The length of the second vector
is still one, and if we're

00:13:42.248 --> 00:13:45.488
dotting a vector with itself,
the angle between the two

00:13:45.488 --> 00:13:49.700
vectors must be 0. So this time
we want the cosine of 0.

00:13:50.680 --> 00:13:53.859
And the cosine of 0. The cosine
of note is one.

00:13:54.540 --> 00:13:59.556
So we have 1 * 1 * 1, which
is just one.

00:14:00.170 --> 00:14:03.686
So if we're dotting the vector I
with itself, we get one.

00:14:04.690 --> 00:14:09.201
A similar argument will tell us
that if we got the vector Jay

00:14:09.201 --> 00:14:14.059
with itself, we also get one,
and if we drop the vector K with

00:14:14.059 --> 00:14:15.447
itself, we get one.

00:14:16.400 --> 00:14:19.344
Let me summarize those
important results here and

00:14:19.344 --> 00:14:23.392
I've results that you should
remember that if you took the

00:14:23.392 --> 00:14:27.808
vector I and dotted it with
itself, that would be the same

00:14:27.808 --> 00:14:29.648
as dotting Jay with itself.

00:14:31.060 --> 00:14:35.785
And it's the same as Dot Inc. A
with itself, and in any case we

00:14:35.785 --> 00:14:37.045
get the answer 1.

00:14:38.850 --> 00:14:46.007
If we don't, I with J we get
zero if we got I with K if we

00:14:46.007 --> 00:14:51.480
get zero and if we got J with K
we also get 0.

00:14:52.480 --> 00:14:55.312
And all those results are
particularly important, and

00:14:55.312 --> 00:14:58.852
once that you should become
familiar with as you work

00:14:58.852 --> 00:15:00.976
through the exercises
accompanying this unit.

00:15:05.890 --> 00:15:10.674
Now I want to use those results
about the dot products of those

00:15:10.674 --> 00:15:15.458
unit vectors to find a way to
calculate the dot product of two

00:15:15.458 --> 00:15:19.138
arbitrary vectors that are given
in Cartesian form. So let's

00:15:19.138 --> 00:15:22.450
choose two arbitrary vectors.
Let's suppose the first one.

00:15:23.750 --> 00:15:26.060
Is a one.

00:15:27.430 --> 00:15:28.010
I.

00:15:29.830 --> 00:15:35.120
A2, J and
a 3K.

00:15:36.520 --> 00:15:38.062
So this is now an arbitrary

00:15:38.062 --> 00:15:44.094
vector. A where a one A2 and a
three are the three cartesian

00:15:44.094 --> 00:15:46.154
components of the vector A.

00:15:47.270 --> 00:15:52.926
And let's suppose vector B
similarly is B1I.

00:15:54.500 --> 00:16:00.593
Plus B2J Plus V3K and
again B1B2B3 arbitrary numbers,

00:16:00.593 --> 00:16:06.686
so this is an arbitrary
vector in three dimensions.

00:16:08.260 --> 00:16:12.280
What I want to do is workout the
dot product of A&B.

00:16:13.510 --> 00:16:17.434
So let's work it through a
dot B equals.

00:16:18.900 --> 00:16:21.268
It's the first vector
which is all this.

00:16:31.180 --> 00:16:36.060
Dotted with the second vector
be, which is all this.

00:16:46.100 --> 00:16:51.170
OK. Now what we want to do
now is use our

00:16:51.170 --> 00:16:54.329
distributivity rule to be
able to workout to expand

00:16:54.329 --> 00:16:57.137
these brackets and workout
the individual dot products.

00:16:58.310 --> 00:17:02.524
And we can do that using the
rules that we know that in turn

00:17:02.524 --> 00:17:04.029
each one of these vectors.

00:17:04.850 --> 00:17:07.775
In the first bracket,
multiplies each of these in

00:17:07.775 --> 00:17:11.025
the SEC bracket and just the
normal algebraic way. So

00:17:11.025 --> 00:17:15.575
first of all we want a one I
dotted with B1 I those terms.

00:17:24.480 --> 00:17:26.196
Then we want a one I.

00:17:26.880 --> 00:17:28.800
Dotted with B2J

00:17:37.020 --> 00:17:41.388
and finally a one
I dotted with B3K.

00:17:47.490 --> 00:17:50.290
So that takes care of this first

00:17:50.290 --> 00:17:54.730
vector here. And we moved to
the second vector. So now we

00:17:54.730 --> 00:17:57.470
want plus A2, J, dotted with
each of these three.

00:18:01.200 --> 00:18:04.024
So it's a 2 J dotted with B1I.

00:18:09.120 --> 00:18:11.450
A2 J dotted with B2J.

00:18:17.180 --> 00:18:19.686
And a two J dotted with B3K.

00:18:27.000 --> 00:18:30.443
And finally, this loss vector in
the first bracket dotted with

00:18:30.443 --> 00:18:34.199
each of these vectors in the SEC
bracket will be a 3K.

00:18:37.350 --> 00:18:38.940
Dotted with B1I

00:18:40.860 --> 00:18:43.500
plus a 3K dotted with B2J.

00:18:48.500 --> 00:18:51.524
And finally, a 3K
dotted with B3K.

00:18:56.540 --> 00:19:00.016
Now this looks horrendous. With
all these nine terms in here,

00:19:00.016 --> 00:19:04.124
but a lot of this is going to
cancel out now. You'll remember

00:19:04.124 --> 00:19:06.968
that any vectors which are
perpendicular to each other.

00:19:08.000 --> 00:19:10.065
Have a dot product which is 0.

00:19:11.590 --> 00:19:12.640
Now this vector.

00:19:14.320 --> 00:19:16.742
Is in the eye direction and this

00:19:16.742 --> 00:19:21.792
vector. Is in the J direction,
so a one I is perpendicular to

00:19:21.792 --> 00:19:22.830
be 2 J.

00:19:23.890 --> 00:19:28.420
So where we have the item dotted
with the J term, this must be 0.

00:19:31.840 --> 00:19:36.076
Similarly, any vector in the
direction of I any multiple of I

00:19:36.076 --> 00:19:38.900
must be perpendicular to any
multiple of K.

00:19:39.900 --> 00:19:42.147
So this dot product is
also zero.

00:19:43.890 --> 00:19:47.850
And similarly for the A2JB1I,
those vectors are perpendicular.

00:19:47.850 --> 00:19:50.050
Their dot product is 0.

00:19:52.620 --> 00:19:55.955
A2J dot B3K. A perpendicular

00:19:55.955 --> 00:20:01.200
that's zero. Similarly, that
will be 0 and that will be 0.

00:20:01.940 --> 00:20:05.117
And we'll be left with
three non zero terms.

00:20:07.310 --> 00:20:08.610
What about this term here?

00:20:10.170 --> 00:20:15.386
A1 I be one I now this is a
multiple of I and this is a

00:20:15.386 --> 00:20:17.342
multiple of I. So these two

00:20:17.342 --> 00:20:22.251
vectors are parallel. The angle
between them is 0, so when we

00:20:22.251 --> 00:20:26.319
work out the dot product will
want the length of the first

00:20:26.319 --> 00:20:28.014
one, which is a one.

00:20:28.810 --> 00:20:30.310
The length of the second one,

00:20:30.310 --> 00:20:31.510
which is. He wants.

00:20:33.170 --> 00:20:36.052
And the cosine of the angle
between the two. The angle

00:20:36.052 --> 00:20:41.649
between the two. Is zero and the
cosine of 0 is one, so we get a

00:20:41.649 --> 00:20:43.264
one B 1 * 1.

00:20:44.340 --> 00:20:47.740
So this first term, whilst it
looked quite complicated, just

00:20:47.740 --> 00:20:49.440
simplifies to a 1B one.

00:20:52.090 --> 00:20:54.884
Similarly, these two vectors
here. This is a vector in the

00:20:54.884 --> 00:20:58.186
direction of J and this is a
vector in the direction of Jay.

00:20:58.830 --> 00:21:03.000
So the angle between the two of
them is 0, the cosine of 0 is

00:21:03.000 --> 00:21:04.946
one, so when we workout this dot

00:21:04.946 --> 00:21:09.453
product we want. The length of
the first one, which is a 2. The

00:21:09.453 --> 00:21:10.911
length of the second one which

00:21:10.911 --> 00:21:16.373
is B2. Multiplied by the cosine
of 0. Cosine of 0 is one, so we

00:21:16.373 --> 00:21:17.657
just get a 2B2.

00:21:19.150 --> 00:21:22.102
And finally, same argument
applies here. This is a

00:21:22.102 --> 00:21:26.038
multiple of K. This is a
multiple of K, so the vectors

00:21:26.038 --> 00:21:29.974
are parallel. The angle between
them is zero and the cosine of

00:21:29.974 --> 00:21:34.894
0 is one. So we get the length
of the first being A3 length of

00:21:34.894 --> 00:21:39.486
the second being B3 and the
cosine of 0 cosine of 0 is one,

00:21:39.486 --> 00:21:42.766
so the whole lot. All of this
complicated stuff just

00:21:42.766 --> 00:21:46.702
simplifies down at the end of
the day to this very important

00:21:46.702 --> 00:21:47.358
result here.

00:21:49.810 --> 00:21:55.036
And what this result says is
if you want to find the scalar

00:21:55.036 --> 00:21:59.056
product of two vectors A&B
which are given in this

00:21:59.056 --> 00:22:03.880
cartesian form, like that, all
we need to do is multiply the

00:22:03.880 --> 00:22:08.302
I components together. You see
the A1B one is the eye

00:22:08.302 --> 00:22:09.508
components multiplied
together.

00:22:11.410 --> 00:22:13.374
Multiply the J components

00:22:13.374 --> 00:22:18.458
together A2B2. Multiply the K
components together A3B3.

00:22:19.370 --> 00:22:21.860
And then finally add up these

00:22:21.860 --> 00:22:24.400
results. So the dot product.

00:22:25.420 --> 00:22:26.488
Is the sum.

00:22:27.170 --> 00:22:30.579
Of the products of the
corresponding components.

00:22:32.220 --> 00:22:35.751
And you'll notice that the
result we get here on the

00:22:35.751 --> 00:22:39.603
right hand side doesn't have
any eyes or Jays orkez in the

00:22:39.603 --> 00:22:43.134
answer anymore. This is purely
a number. This is purely a

00:22:43.134 --> 00:22:46.344
scalar, as we'd expect when we
calculate a scalar product.

00:22:48.460 --> 00:22:52.660
Point out also that this
expression on the right, which

00:22:52.660 --> 00:22:57.280
is the sum of the products of
the corresponding components is

00:22:57.280 --> 00:23:00.640
also sometimes referred to as
the inner product.

00:23:02.330 --> 00:23:07.530
Of A&B, so on occasions you may
hear some staff or reading some

00:23:07.530 --> 00:23:11.530
textbooks that this is called
the inner product of Bambi.

00:23:19.520 --> 00:23:22.424
OK, so we have a formula
now for the dot product.

00:23:24.800 --> 00:23:31.807
It's a one B 1 +
8, two B2 plus A3B3.

00:23:35.210 --> 00:23:38.928
Can very important results and
this will allow us to calculate

00:23:38.928 --> 00:23:42.308
the scalar product of two
vectors given in cartesian form.

00:23:42.308 --> 00:23:44.336
Let me give you an example.

00:23:49.030 --> 00:23:56.950
Suppose the first vector a is
this one 4I Plus 3J minus.

00:23:56.950 --> 00:23:58.930
Sorry plus 7K.

00:24:00.620 --> 00:24:03.952
And Suppose B is the vector two

00:24:03.952 --> 00:24:06.480
I. +5 J.

00:24:07.980 --> 00:24:08.910
Plus 4K.

00:24:11.300 --> 00:24:16.459
Suppose we want to calculate the
scalar product A dot B.

00:24:18.390 --> 00:24:22.636
Well, this formula tells us that
we multiply the corresponding I

00:24:22.636 --> 00:24:27.268
components together A1B one.
This is a one for this is B1

00:24:27.268 --> 00:24:31.514
which is 2. So we multiply the
four by the two.

00:24:35.290 --> 00:24:38.530
We multiply the corresponding J
components together. A2B2 well,

00:24:38.530 --> 00:24:43.570
this is a two and this is B2, so
we want 3 * 5.

00:24:47.080 --> 00:24:50.280
And then we multiply the
corresponding K components

00:24:50.280 --> 00:24:53.080
together. A3B3, which is 7 * 4.

00:24:56.050 --> 00:25:03.415
When we do that and we add up
all the results and will get 8

00:25:03.415 --> 00:25:10.289
+ 3, five, 15, Seven, 428. And
if we work those out we shall

00:25:10.289 --> 00:25:17.163
get 2815 and eight to add up
to 51. So the dot product of

00:25:17.163 --> 00:25:19.618
A&B is the scalar 51.

00:25:22.420 --> 00:25:25.940
Now, sometimes, instead of
writing it all out like this, we

00:25:25.940 --> 00:25:29.780
can use column vector notation
and some people find it a bit

00:25:29.780 --> 00:25:33.620
easier to work with column
vectors. So just let me show you

00:25:33.620 --> 00:25:37.460
how we do the same calculation.
If we had these vectors given

00:25:37.460 --> 00:25:41.300
not in this form, but as column
vectors, suppose we had a

00:25:41.300 --> 00:25:47.258
written as 437. And
be written AS254.

00:25:49.020 --> 00:25:51.306
Then the a dotted with B.

00:25:52.170 --> 00:25:57.786
Would be 437 dotted
with 254.

00:26:00.350 --> 00:26:02.886
Now to multiply corresponding
components together is quite

00:26:02.886 --> 00:26:07.641
easy to see what they are now
because we want 4 * 2, which is

00:26:07.641 --> 00:26:13.820
8. 3 * 5 which is 15 and 7 *
4, which is 28. And if we had

00:26:13.820 --> 00:26:17.850
those that will get 51 like we
did over here. So some people

00:26:17.850 --> 00:26:21.880
might find it a bit easier just
to write them as column vectors

00:26:21.880 --> 00:26:24.360
and then just read off the
corresponding components,

00:26:24.360 --> 00:26:27.770
multiply them together, and add
them up to get the result.

00:26:32.200 --> 00:26:36.088
Now it's so important that you
know how to calculate a scalar

00:26:36.088 --> 00:26:39.976
product that I'm going to give
you one more example, just to

00:26:39.976 --> 00:26:42.244
make sure that you understand
the process.

00:26:43.420 --> 00:26:48.060
And will go straight into column
vector notation. Suppose the

00:26:48.060 --> 00:26:51.308
vector A is the vector minus 6.

00:26:51.870 --> 00:26:54.420
3 - 11.

00:26:55.710 --> 00:26:59.014
That's minus six I plus 3J minus

00:26:59.014 --> 00:27:06.532
11 K. And let's suppose
the vector B is the

00:27:06.532 --> 00:27:11.600
vector 12:04 that's twelve
IOJ plus 4K.

00:27:14.670 --> 00:27:20.354
So the dot product A dot B will
be minus six 3 - 11.

00:27:21.520 --> 00:27:24.348
Dotted with twelve 04.

00:27:26.050 --> 00:27:29.504
And in the column vector
notation, it's so easy to work

00:27:29.504 --> 00:27:33.272
this out. We want minus 6 * 12,
which is minus 72.

00:27:35.410 --> 00:27:37.680
3 * 0 is 0.

00:27:40.100 --> 00:27:43.604
And minus 11 * 4 is minus 44.

00:27:44.980 --> 00:27:46.764
And if we add these up, we get.

00:27:48.180 --> 00:27:51.210
Minus 116

00:27:53.080 --> 00:27:57.030
so the dot product of A&B is
just the sum.

00:27:57.850 --> 00:28:00.986
Of the products of the
corresponding components.

00:28:06.430 --> 00:28:11.022
I want to move on now to look at
some applications of the scalar

00:28:11.022 --> 00:28:13.974
product and the first
application is one that I

00:28:13.974 --> 00:28:17.910
mentioned very early on and it's
to do with testing whether or

00:28:17.910 --> 00:28:19.550
not two given vectors are

00:28:19.550 --> 00:28:23.799
perpendicular or not. Now you
remember from the definition of

00:28:23.799 --> 00:28:27.506
the scalar product of two
vectors, 8 be that is the

00:28:27.506 --> 00:28:31.550
modulus of the first one times
the modulus of the second one

00:28:31.550 --> 00:28:35.257
times the cosine of the angle in
between the two vectors.

00:28:36.280 --> 00:28:39.460
Now, if we find that
when we work this out

00:28:39.460 --> 00:28:41.050
that the result is 0.

00:28:42.330 --> 00:28:47.048
Then either the modulus of a
must be 0, or the modulus of be

00:28:47.048 --> 00:28:50.418
must be 0. Or Alternatively,
cosine theater must be 0.

00:28:51.540 --> 00:28:55.740
Now, if we know that the two
given vectors do not have a zero

00:28:55.740 --> 00:28:59.640
length, In other words, the
modulus of a can't be 0, and the

00:28:59.640 --> 00:29:01.440
modulus of B can't be 0.

00:29:02.070 --> 00:29:06.714
Then the only conclusion we can
draw is that cosine theater must

00:29:06.714 --> 00:29:12.132
be 0, and from that we did use
that theater must be 90 degrees.

00:29:13.290 --> 00:29:16.865
In other words, if we take the
dot product of two vectors and

00:29:16.865 --> 00:29:18.790
we find we get the answer 0.

00:29:19.800 --> 00:29:24.510
Then, provided the two factors
were nonzero vectors, we can

00:29:24.510 --> 00:29:28.278
deduce the amb must be
perpendicular vectors, so.

00:29:30.360 --> 00:29:32.110
If A&B.

00:29:34.090 --> 00:29:35.839
A nonzero vectors.

00:29:41.660 --> 00:29:42.550
Search that.

00:29:46.000 --> 00:29:49.575
Hey, don't be when we work it
out. Turns out to be 0.

00:29:50.420 --> 00:29:53.640
Then A&B.

00:29:55.780 --> 00:29:56.710
Are perpendicular.

00:30:02.530 --> 00:30:03.630
There at right angles.

00:30:04.300 --> 00:30:08.216
Another word we sometimes use is
orthogonal. You may hear that,

00:30:08.216 --> 00:30:11.420
particularly in more advanced
work, two vectors at right

00:30:11.420 --> 00:30:14.624
angles to vectors which are
perpendicular are sometimes said

00:30:14.624 --> 00:30:15.692
to be orthogonal.

00:30:16.660 --> 00:30:20.566
So we have a test here we take
the two vectors, we find their

00:30:20.566 --> 00:30:24.193
dot product and if the answer
that we get zero then we deduce

00:30:24.193 --> 00:30:27.262
that the two vectors must be
perpendicular. So that's a very

00:30:27.262 --> 00:30:30.889
easy test we can apply to see if
two vectors are Purple ****.

00:30:33.560 --> 00:30:34.946
Let me give you an example.

00:30:39.840 --> 00:30:46.522
Suppose we have the Vector A,
which is the vector three 2 -

00:30:46.522 --> 00:30:52.690
1 and the Vector B which is
1 - 2 - 1.

00:30:54.580 --> 00:30:58.701
Now I think you would agree that
just by looking at these vectors

00:30:58.701 --> 00:31:00.603
three, I +2 J Minus K.

00:31:01.530 --> 00:31:05.618
And one I minus two J minus K.
You'd have no idea just from

00:31:05.618 --> 00:31:08.830
looking at them written down
like that. Whether or not these

00:31:08.830 --> 00:31:10.582
vectors where at right angles to

00:31:10.582 --> 00:31:15.800
each other. We can apply this
dot product test and workout a

00:31:15.800 --> 00:31:19.870
baby to see what happens. So
let's workout a dot B.

00:31:21.390 --> 00:31:27.166
I want the dot product of three
2 - 1 with 1 - 2 - 1.

00:31:28.550 --> 00:31:30.020
And we'll get 3 ones or three.

00:31:30.950 --> 00:31:33.380
Two times minus 2 - 4.

00:31:33.940 --> 00:31:38.168
And minus one times minus one is
plus one. So we've got three and

00:31:38.168 --> 00:31:43.000
one which is 4 - 4 which is 0.
So you'll see that when we work

00:31:43.000 --> 00:31:46.624
out the dot product of these two
nonzero vectors, the answer that

00:31:46.624 --> 00:31:47.832
we get is 0.

00:31:48.660 --> 00:31:52.296
So we can deduce from this
result that this vector a must

00:31:52.296 --> 00:31:55.629
be at right angles to this
vector B and that's something

00:31:55.629 --> 00:31:59.265
that's not obvious just from
looking at it. So we've got a

00:31:59.265 --> 00:32:02.598
very useful test there to test
for. The two vectors are

00:32:02.598 --> 00:32:03.507
perpendicular or not.

00:32:09.320 --> 00:32:11.980
Now another important
application of the scalar

00:32:11.980 --> 00:32:16.160
product is to finding the angle
between two given vectors. Let

00:32:16.160 --> 00:32:20.340
me remind you of the definition
of the scalar product again

00:32:20.340 --> 00:32:24.900
because we will need that the
scalar product of A&B is found

00:32:24.900 --> 00:32:29.460
by taking the length of the
first vector times the length of

00:32:29.460 --> 00:32:33.640
the second vector times the
cosine of the angle in between

00:32:33.640 --> 00:32:35.160
the two vectors so.

00:32:35.270 --> 00:32:38.286
If we get a vector, if we get
two vectors in Cartesian form.

00:32:39.190 --> 00:32:40.665
And we calculate their dot

00:32:40.665 --> 00:32:45.238
product. And if we know how to
calculate the modulus of each of

00:32:45.238 --> 00:32:48.826
those two vectors, then the only
thing in this expression that we

00:32:48.826 --> 00:32:52.414
don't know is cosine Theta, so
will be able to use this

00:32:52.414 --> 00:32:55.404
definition to calculate cosine
theater the cosine of the angle

00:32:55.404 --> 00:32:58.693
between the two vectors, from
which we can deduce the angle

00:32:58.693 --> 00:33:01.476
itself. OK, let's rearrange

00:33:01.476 --> 00:33:06.540
this. If we divide both sides by
the modular surveying, the

00:33:06.540 --> 00:33:11.860
modulus of B will be able to get
cosine theater is a dot B

00:33:11.860 --> 00:33:16.420
divided by the modulus of a
modulus of B and that's the

00:33:16.420 --> 00:33:20.980
formula that we're going to use
in a minute to find cosine,

00:33:20.980 --> 00:33:22.500
Theta and hence theater.

00:33:23.940 --> 00:33:27.252
Now we've done a lot of work
already in this unit on

00:33:27.252 --> 00:33:30.564
calculating the dot product. Let
me just remind you a little bit

00:33:30.564 --> 00:33:34.152
about how you find the modulus
of a vector when it's given in

00:33:34.152 --> 00:33:41.640
cartesian form. If we have a
Vector A, which is a one

00:33:41.640 --> 00:33:45.270
I plus A2J plus a 3K.

00:33:46.470 --> 00:33:51.370
Then the modulus of this
vector is found by squaring

00:33:51.370 --> 00:33:53.820
the individual components,
adding them.

00:33:55.890 --> 00:33:59.894
And finally, taking the square
root of the result. So this is a

00:33:59.894 --> 00:34:03.590
formula that we're going to use
for finding the modulus of a

00:34:03.590 --> 00:34:06.978
vector in Cartesian form, and
this has been covered in an

00:34:06.978 --> 00:34:10.366
early you unit. If you need to
look back at that.

00:34:14.830 --> 00:34:17.701
Let's look at a specific
example where we want to find

00:34:17.701 --> 00:34:19.006
the angle between two vectors.

00:34:23.760 --> 00:34:26.140
So the problem is
find the angle.

00:34:34.840 --> 00:34:41.128
Between the two vectors, I'm
going to choose are A which is

00:34:41.128 --> 00:34:43.748
4I Plus 3J plus 7K.

00:34:45.320 --> 00:34:46.410
And B.

00:34:47.740 --> 00:34:54.149
Which is 2 I plus 5J Plus
4K, so two vectors given in

00:34:54.149 --> 00:34:59.079
cartesian form and the
problem is to find the angle

00:34:59.079 --> 00:35:00.065
between them.

00:35:01.300 --> 00:35:06.370
And we have this result that we
just reduced the cosine of the

00:35:06.370 --> 00:35:10.660
required angle. Is the dot
product 8B divided by the length

00:35:10.660 --> 00:35:15.730
of a times the length of be? So
that's the formula that we're

00:35:15.730 --> 00:35:16.900
going to use.

00:35:18.470 --> 00:35:23.926
OK, a dot B now. In fact a dot B
has already been evaluated in an

00:35:23.926 --> 00:35:28.359
earlier example, but I'll just
remind you of that a dot B. If

00:35:28.359 --> 00:35:31.428
we use the column vector
notation will be 437.

00:35:32.460 --> 00:35:34.359
Dotted with 254

00:35:37.210 --> 00:35:44.347
which is 428-3515 and Seven 428.
And if you work that out, you

00:35:44.347 --> 00:35:46.543
find you get 51.

00:35:48.130 --> 00:35:50.715
So all we need to do is
calculate now the modular

00:35:50.715 --> 00:35:53.535
survey in the modulus of B and
then we've got everything we

00:35:53.535 --> 00:35:56.355
need to know to put in the
right hand side of this

00:35:56.355 --> 00:35:56.590
formula.

00:35:58.010 --> 00:35:59.518
OK, the modular survey.

00:36:02.030 --> 00:36:06.650
Now the modulus of a is found by
taking the square root of the

00:36:06.650 --> 00:36:10.280
squares of these components
added up. So we want 4 squared.

00:36:11.470 --> 00:36:12.570
3 squared

00:36:13.620 --> 00:36:14.700
7 squared.

00:36:16.700 --> 00:36:19.730
Which is going to be
the square root of 16.

00:36:21.030 --> 00:36:26.670
330977's of 49 and if you work
that out, you'll find that

00:36:26.670 --> 00:36:29.490
that's the square root of 74.

00:36:31.860 --> 00:36:37.827
Similarly, the modulus of B will
be the square root of 2 squared.

00:36:39.350 --> 00:36:42.298
+5 squared, +4 squared.

00:36:44.400 --> 00:36:51.862
Which is the square root of 4
+ 25 + 16 and 25 +

00:36:51.862 --> 00:36:54.527
4 + 16 is 45.

00:36:55.890 --> 00:36:58.740
So we've got all the
ingredients we need now to

00:36:58.740 --> 00:36:59.880
pop into this formula.

00:37:01.190 --> 00:37:06.572
Cosine theater will be a dotted
with B, which we found was 51

00:37:06.572 --> 00:37:11.540
divided by the modular survey,
which is the square root of 74.

00:37:12.810 --> 00:37:16.982
The modulus of B, which is the
square root of 45 now will need

00:37:16.982 --> 00:37:18.770
a Calculator to work this out.

00:37:21.650 --> 00:37:22.949
We want 51.

00:37:24.980 --> 00:37:27.745
Divided by the square
root of 74.

00:37:29.090 --> 00:37:30.390
Divided by the square root.

00:37:31.370 --> 00:37:38.560
45 Which is not .8838
to 4 decimal places. That's the

00:37:38.560 --> 00:37:43.609
cosine of the angle between the
two vectors A&B.

00:37:44.440 --> 00:37:46.306
If we find the inverse cosine.

00:37:49.340 --> 00:37:51.610
Of .8838.

00:37:54.160 --> 00:37:58.228
Which is 27.90
degrees.

00:38:00.980 --> 00:38:04.566
So there we've seen how we can
use the scalar product.

00:38:06.040 --> 00:38:09.883
To find the angle between
two vectors when they're

00:38:09.883 --> 00:38:11.591
given in Cartesian form.

00:38:17.270 --> 00:38:21.314
Now there's one more application
I'd like to tell you about, and

00:38:21.314 --> 00:38:25.021
it's concerned with finding the
components of one vector in the

00:38:25.021 --> 00:38:28.728
direction of a second vector.
Let me give you an example.

00:38:30.860 --> 00:38:33.308
Suppose we have a vector A.

00:38:35.170 --> 00:38:36.556
Let me have a second vector.

00:38:38.300 --> 00:38:38.990
B.

00:38:42.280 --> 00:38:46.988
And I've drawn these vectors so
that their tails coincide and.

00:38:47.580 --> 00:38:51.280
As usual, the angle between the
two vectors is theater.

00:38:52.440 --> 00:38:54.260
Let's call this .0.

00:38:54.850 --> 00:38:58.546
And what I'm going to do is
I'm going to drop a

00:38:58.546 --> 00:39:01.010
perpendicular from this
point, which I'll call B

00:39:01.010 --> 00:39:01.318
down.

00:39:04.330 --> 00:39:06.100
To meet the vector A.

00:39:09.510 --> 00:39:11.640
Let's call this point here a.

00:39:13.100 --> 00:39:15.347
And we can regard the vector OB.

00:39:16.810 --> 00:39:21.046
As the sum of the vector
from O2 Capital A and then

00:39:21.046 --> 00:39:25.282
the vector from A to B. In
other words, the vector OB.

00:39:27.190 --> 00:39:28.940
Is the sum of OA.

00:39:31.680 --> 00:39:32.988
Plus a bee.

00:39:36.550 --> 00:39:40.598
Now this vector here, which
we've called OK, that vector is

00:39:40.598 --> 00:39:45.750
said to be the component of B in
the direction of a. You'll see

00:39:45.750 --> 00:39:50.902
we can regard be as being made
up of two components, one in the

00:39:50.902 --> 00:39:56.054
direction of A and one which is
at right angles to that. So this

00:39:56.054 --> 00:39:57.158
component, oh A.

00:39:57.960 --> 00:40:00.910
Is the component of B
in the direction of a.

00:40:18.340 --> 00:40:21.980
And what we're going to do is
we're going to see how we can

00:40:21.980 --> 00:40:24.060
use the scalar product to
find this component.

00:40:26.070 --> 00:40:30.282
Let's call this length here the
length from O to a. Let's call

00:40:30.282 --> 00:40:34.930
that little. And then let's
focus our attention on this

00:40:34.930 --> 00:40:36.070
right angled triangle.

00:40:36.830 --> 00:40:38.300
Oba.

00:40:39.470 --> 00:40:41.514
So there's a right angle
in there.

00:40:43.320 --> 00:40:47.390
Now, using our knowledge of
trigonometry, we can write down

00:40:47.390 --> 00:40:51.867
the cosine of this angle is
adjacent over hypotenuse, so the

00:40:51.867 --> 00:40:53.088
cosine of Theta.

00:40:54.640 --> 00:40:57.874
Is this length L
the adjacent side?

00:40:59.350 --> 00:41:01.878
Over the hypotenuse, now
the hypotenuse is the

00:41:01.878 --> 00:41:03.142
length of this side.

00:41:04.650 --> 00:41:07.410
And this is the length of
this side is the modulus of

00:41:07.410 --> 00:41:07.870
vector be.

00:41:12.020 --> 00:41:16.299
Now, In other words, if we
rearrange this L, the component

00:41:16.299 --> 00:41:22.134
of B in the direction of a week
and right as the modulus of be

00:41:22.134 --> 00:41:23.690
times, the cosine Theta.

00:41:27.750 --> 00:41:31.290
Now we've already got an
expression for the cosine of

00:41:31.290 --> 00:41:35.538
this angle in the previous work
that we've done, we already know

00:41:35.538 --> 00:41:40.140
that the cosine of feta can be
written as the dot product for

00:41:40.140 --> 00:41:45.070
amb. Divided by the modular
survey times the modulus of be,

00:41:45.070 --> 00:41:47.344
that's the result we just had.

00:41:48.470 --> 00:41:51.200
So if you look at this
expression now, you'll see

00:41:51.200 --> 00:41:54.476
with the modulus B in the
numerator and a modulus of be

00:41:54.476 --> 00:41:55.841
in the denominator, they
cancel.

00:41:56.900 --> 00:41:58.420
So we can write this.

00:41:59.170 --> 00:42:03.535
As a not be over
the modulus of A.

00:42:06.760 --> 00:42:08.730
Let me write that down
again on the next page.

00:42:20.970 --> 00:42:24.546
So this remember L is the
component of be in the direction

00:42:24.546 --> 00:42:28.718
of a. We can write it like this.
I'm going to Rearrange this a

00:42:28.718 --> 00:42:32.592
little bit to write it as a
divided by the modulus of A.

00:42:34.000 --> 00:42:35.908
Dotted with effective be.

00:42:37.240 --> 00:42:40.771
And because of the result we
got very early on the

00:42:40.771 --> 00:42:43.981
commutativity of the dot
product. We can write this as

00:42:43.981 --> 00:42:46.870
B dot A divided by the modulus
of A.

00:42:49.400 --> 00:42:53.313
Now when you take any vector and
you divide it by its modulus,

00:42:53.313 --> 00:42:57.226
what you get is a unit vector in
the direction of that vector,

00:42:57.226 --> 00:43:01.440
and we write a unit vector as
the vector with a little hat on

00:43:01.440 --> 00:43:05.353
like that. So this is a unit
vector in the direction of A.

00:43:14.770 --> 00:43:18.502
So this is an important result
we've got. We've got L, which

00:43:18.502 --> 00:43:22.856
is the component of be in the
direction of a can be found by

00:43:22.856 --> 00:43:26.588
taking the dot product of be
with a unit vector in the

00:43:26.588 --> 00:43:27.521
direction of A.

00:43:31.340 --> 00:43:37.794
So suppose we wish to find the
component of Vector B which is 3

00:43:37.794 --> 00:43:43.787
I plus J Plus 4K in the
direction of the Vector A, which

00:43:43.787 --> 00:43:46.553
is I minus J Plus K.

00:43:48.180 --> 00:43:53.306
What we want to do is
evaluate B dotted with a

00:43:53.306 --> 00:43:55.636
divided by the modular
survey.

00:43:59.190 --> 00:44:03.026
Now we can do that as follows.
We can workout the dot product B

00:44:03.026 --> 00:44:07.136
dot A and then divide by the
modulus of a. So B Dot A is

00:44:07.136 --> 00:44:09.602
going to be 3 * 1, which is 3.

00:44:10.190 --> 00:44:15.970
Plus one times minus one which
is minus 1 + 4 * 1 which is 4 or

00:44:15.970 --> 00:44:20.390
divided by the modulus of a
which is the square root of 1

00:44:20.390 --> 00:44:24.130
squared plus minus one squared
plus one squared, which is the

00:44:24.130 --> 00:44:25.490
square root of 3.

00:44:27.570 --> 00:44:32.988
If we work that out, will get 3
- 1 is 2 + 4 six six over Route

00:44:32.988 --> 00:44:37.503
3. This means that the component
of B in the direction of a is 6.

00:44:37.503 --> 00:44:41.115
Over Route 3. We might want to
write that in an alternative

00:44:41.115 --> 00:44:45.329
form just to tidy it up so we
don't have the square root in

00:44:45.329 --> 00:44:48.640
the denominator and we can do
that by multiplying top and

00:44:48.640 --> 00:44:52.553
bottom by the square root of 3,
which will produce Route 3 times

00:44:52.553 --> 00:44:56.165
route 3 in the denominator,
which is 3 and three into six

00:44:56.165 --> 00:44:58.884
goes twice. So that
would just simplify to

00:44:58.884 --> 00:45:00.404
two square root of 3.

00:45:02.560 --> 00:45:06.169
So in this unit we've introduced
the scalar product.

00:45:06.760 --> 00:45:10.225
Learn how to calculate it and
looked at some of its

00:45:10.225 --> 00:45:11.170
properties and applications.