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.