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

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

of combining two vectors.

This method is called the vector

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

combine the two vectors in this
way, the result that will get is

another vector. OK, let's start
with the two vectors.

So suppose we have a vector A.

And another vector.

Be. And we have
drawn these vectors A&B so that

their tails coincide.

And so that we can label
the angle between A&B's

theater like that.

So we start with two vectors and
we're going to do some

calculations with these two
vectors to generate what's

called the vector product.

And the vector product is
defined to be the length of the

first vector. That's the length

of A. Multiplied by the length
of the second vector, the length

of be. Multiplied this time by
the sign of the angle between a

envy. Length of the first one,
length of the second one, and

the sign of the angle between

the two vectors. Now that's all
well and good, but this isn't a

vector. This is a scalar. This
is just a number times a number

times a number. And if we're
going to get a result which is a

vector we've got to give some

direction to this. If we look
back at the diagram, will notice

that if we choose any two
vectors, these two vectors lie

in a plane. They form a plane in
which they both lie.

When we calculate the vector
product of these two vectors, we

define the direction of the
vector product to be the

direction which is perpendicular
to the plane containing A&B. So

if we have a vector which is
perpendicular to a man to be and

to enter the plane containing an
be, this gives us the direction

of the vector product.

And in fact, there are two
possible directions which are

perpendicular to this plane,
because In addition to the one

that I've drawn upwards like
this, There's also one.

That's downwards like that, so
we have two possible directions

which are perpendicular to the
plane containing A&B.

And we have a convention for
deciding which direction we

should choose. Now, this
convention is variously called

the right hand screw rule or the
right hand thumb rule, so I'm

going to show you both of these

different conventions. First of
all, the right hand screw rule.

If you take a conventional
screwdriver and right handed

screw, and if you imagine
turning the screwdriver handle

so that we turn from the
direction of a round to the

direction of be. So I'm turning
my screwdriver In this sense.

That way around like that.

Imagine the direction in which
the screw would advance. So in

turning from the sensor data,
the sense of B.

The scroll advance.

Upwards. So it's this upwards
direction here, which we take as

the direction we require for
evaluating this vector product.

Let me denote a unit vector in
this direction by N.

So if we want to calculate the
vector product of amb, this is

going to be its magnitude.

And the direction we want is
one in the direction of this

unit vector N. So I put a unit
vector N there to give the

direction to this quantity
here.

There's another way, which
sometimes people use to define

the direction of the vector
product, which is equivalent,

and it's called the right hand
thumb rule. Let me explain this

as well. Imagine you take your
fingers of your right hand and

you curl them in the sense going
round from a round towards be.

So you curl your fingers.

In the direction from a round
to be then the thumb points

in the required direction of
the vector product.

Yet another way of thinking

about this. Still with the right
hand is to point the first

finger in the direction of A.

The middle finger in the

direction of B. And then the
thumb points in the required

direction of the vector product

of A&B. So using either the
right hand screw rule or the

right hand thumb rule, we can
deduce that when we want to find

the vector product of these two
vectors A&B, the direction that

we require is the one that's
moving upwards on this diagram

here, not the one moving

downwards. So the vector product
is defined as the length of the

first times the length of the
second times the sign of the

angle between the two times this

unit vector N. Which is defined
in a sense defined by the right

hand thumb rule or the right

hand screw rule. Now we have a
notation for the vector product

and we write the vector product
of A&B, like this A and we use a

time sign or across.

And so sometimes instead of
calling this the vector product,

we sometimes call this the cross
product and throughout the rest

of this unit, sometimes you'll
hear me refer to this either as

the cross product product, all
the vector product.

And I'll use these two words

interchangeably. So that's the
definition will need. I'll also

mention that some authors and
some lecturers will use a

different notation again, and
you might see a cross be written

using this wedge symbol like
this, and that's equally

acceptable and often people will
use the wedge as well to define

the vector product.

It's very important that you put
this symbol in either the wedge

or the cross because you don't
want to just write down to

vectors like that when you might
mean the vector product or you

in fact might mean a scalar
product. Or you might mean

something else, so don't use
that sort of notation. Make sure

you're very explicit about when
you want to use a vector product

by putting the time sign or the

wedge sign in. Let's look at
what happens now. When we

calculate the vector product of
A&B. But we do it in a different

order, so we look at B cross a.

Now, as before, we want the
modulus of the first factor,

which is the modulus of be.

Multiplied by the modulus of the
second vector, the modular

survey. We want the sign of the
angle between B&A, which is

still the sign of angle theater.

And we want to direction.

Now again using our right hand
screw or right hand thumb rule

we can try to find the sense we
require so that as we turn from

B round to a, this time from
the first round of the second

vector, reoccuring my fingers
in the sense from be round to

a. You'll see now that the
thumb points downwards.

So this time we want a vector
which is a unit vector pointing

downwards instead of upwards.

Now unit vector downwards must
be minus N hot.

If the output vector was an hat.

So this time the required
direction of be cross a is

minus an hat.

If we bring the minus sign out
to the front, you'll see that we

get modulus of be modulus of a
sine theater. An hat with the

minus sign at the front now, and
if you examine this vector with

the vector we had before, when
we calculated a cross, B will

see that the magnitudes of these
two vectors are the same,

because we've got a modulus of a
modulus of being sign theater in

each. But the directions are now
different, because where is the

direction of this one? Was an
hat the direction now is minus

and hat. And we see that
this quantity in here.

Is the same as we had here.

So in fact, what we've found is
that B Cross A is the negative

of a crossbite. This is very

important. When we interchange
the order of the two vectors.

We get a different answer be
cross a is in fact the negative

of across be. So it's not true
that a cross B is the same as be

across A. Across be is not
equal to be across A.

And in fact, what is true is
that B cross a is the negative

of a cross be.

So we say that the vector
product is not commutative.

So what that means in practice
is that when you've got to find

the vector product of two
vectors, you must be very clear

about the order in which you
want to carry out the operation.

Now there's a second property of
the vector product. I want to

explain to you, and it's called
the distributivity of the vector

product over addition. What does
that mean? It means that if we

have a vector A and we want to
find the cross product with the

sum of two vectors B Plus C.

We can evaluate this or remove
the brackets in the way that you

would normally expect algebraic
expressions to be evaluated. In

other words, the cross product
distributes itself over this

edition. In other words, that
means that we workout a crossed

with B. And then because of this
addition here, we add that to a

crossed with C.

So this is the distributivity
rule which allows you to remove

brackets in a natural way.

It's also works through the way
around, so if we had B Plus C

first. And we want to cross that

with a. We do this in a way you

would expect. Be cross a.

Plus

secrecy.
And together those rules

are called the distributivity

rules. I will use those in a
little bit later, little bit

later on in this unit.

Another property I want to
explain to you is what happens

when the two vectors that we're
interested in our parallel. So

let's look at what happens when
we want to find the vector

product of parallel vectors.

Suppose now then we have a
Vector A and another vector B

where A&B are parallel vectors.
Is that parallel and we make the

tales of these two vectors
coincide? Then the angle between

them will be 0. So when the
vectors are parallel theater is

0. So when we come to
work out across VB, the modulus

of a modulus of B, the sign,
this time of 0.

And the sign of 0 is 0.

So when the two vectors are
parallel, the magnitude of this

vector turns out to be not. So
in fact what we get is the zero

vector. So for two parallel

vectors. The vector product is
the zero vector.

We now want to start to look at
how we can calculate the vector

product when the two vectors are
given in Cartesian form. Before

we do that, I'd like to develop
some results which will be

particularly important. In this
diagram I've shown a 3

dimensional coordinate system,
so we can see an X

axis or Y axis and it

said access. And superimposed on
this system I've got a unit

vector I in the X direction unit
vector J in the Y direction and

unit vector K and the zed

direction. And what I want to do
is explore what happens when we

cross these unit vectors so we
workout I cross J or J Cross K

etc. And let's see what happens.

Suppose we want I crossed

with J. Well, by definition, the
vector product of iron Jay will

be the modulus of the first

vector. And we want the modulus
of this vector I now this is a

unit vector remember, so its

modulus is one. Times the
modulus of the second vector,

and again the modulus of J is
the modulus of a unit vector. So

again, it's one.

Times the sign of the angle

between. I&J, that's the sign of

90 degrees. And then we've got
to give it a direction, and the

direction is that defined by the
right hand screw rule or right

hand thumb rule. So as we
imagine, turning from Iran to J.

Imagine curling the fingers
around from I to J. The thumb

points in the upward direction
points in the K direction. So

when we work out across J, the
direction of the result will be

Kay Kay being a unit vector in

the direction. Which is
perpendicular to the plane

containing I&J. Now, this
simplifies a great deal because

the sign of 90 side of 90
degrees is one. So we've got 1 *

1 * 1, which is 1K.

So we've got this important
result that I cross. Jay is K.

Let's look at what happens now
when we work out. Jake Ross I.

When we work out Jake Ross, I
were doing the vector product

in the opposite order to which
we did it when we calculated

across J and we know from what
we've just done that we get a

vector of the same magnitude
of the opposite sense,

opposite direction. So when we
work out Jake Ross, I in fact

will get minus K, which is
another important result.

Similarly, if we work out, say,
for example J Cross K, let's

look at that one.

Again, you want the modulus of
the first one, which is one the

modulus of the second one, which
is one the sign of the angle

between J&K. Which is the sign
of 90 degrees. And again you

want to direction defined by the
right hand screw rule and J

Cross K being where they are
here. If you kill your fingers

in the sense around from the
direction of J round to the

direction of K and imagine which
way your thumb will move, your

thumb will move in the eye

direction. So Jake Ross K equals

I. 1 * 1 * 1 gives you
just the eye.

So this is another important
result. J Cross K is I.

And equivalently, if we reverse,
the order will get cake. Ross

Jay is minus I from the result
we had before.

So we've dealt with I crossing
with JJ, crossing with K. What

about I crossing with K?

If we workout I Cross K again.
Modulus of the first one is one

modulus of the second one is one
and I encounter at 90 degrees.

So with the sign of 90 degrees
and we want a direction again

and again with the right hand
screw rule I cross K will give

you a direction which is in this
time in the sense of minus J we

just look at that for a minute.
If you imagine to curling your

fingers in the sense of round
from my towards K.

Then the thumb will point in
the direction of minus J, so

we've got I Cross K is minus
J, or equivalently K Cross I

equals J.

So again important results.

Now, it's important that you
can remember how you calculate

vector product of the eyes in
the Jays. In the case that I'm

going to suggest a way that
you might remember this if you

write down the IJ&K in a
cyclic order like that, and

imagine moving around this
cycle in a clockwise sense.

And if you want to workout, I
cross Jay. The result will be

the next vector round K.

If you want to
workout J Cross K.

The result will be the next
vector round when we move around

clockwise, which is I.

And finally came across I.

Same argument.

Next vector round is Jay, so
that's that's an easy way of

remembering the vector products
of the eyes and Jason case. And

clearly when you reverse the
order of any of these, you

introduce a minus sign.
Alternatively, if you wanted for

example K Cross J, you'd realize
that to move from kata ju moving

anticlockwise, so K Cross Jay is
minus I, that's the way that I

find helpful to remember these

results. We're now in a position
to calculate the vector product

of two vectors given in

Cartesian form. Suppose we start
with the Vector A and let's

suppose this is a general 3
dimensional vector, a one I plus

A2J plus a 3K where A1A2A3 or
any numbers we choose.

Suppose our second vector B,
again arbitrary is be one. I

be 2 J and B3K.

And we set about trying to
calculate the vector product

across be. So we
want the first one A1I

Plus A2J plus a 3K.

I'm going to cross
it with B1IB2J and

B3K. Now this is a little bit
laborious and it's going to take

a little bit of time to develop
it at the end of the day, will

end up with a formula which is
relatively simple for

calculating the vector product.
So we start to remove these

brackets in the way that we've
learned about previously. We use

the distributive law to say that
the first component here a one

I. And then the 2nd and then
the third distributes themselves

over the second vector here. So
will get a one. I crossed with

all of this second vector.

The One I plus
B2J plus B3K.

Then added to the second vector

here. A2J Crossed with.

This whole vector here.

The One I
plus B2J plus

B3K. And finally, the third
one here, a 3K.

Crossed with the whole
of this vector B1

I add B2J ad

D3K. So at that stage we've used
the distributive law wants to

expand this first bracket.

We can use it again because now
we've got a vector here crossed

with three vectors added
together and we can use the

distributive law to distribute
this. A one I cross product

across each of these three terms

here. Then again, to distribute
this across these three terms

and to distribute this across
these three terms. So if we do

that, we'll get a one. I crossed

with B1I. Added to a
one I cross B2J.

Do you want I cross B2J?

Added to a one I.

Crossed with B3K.

So that's taken care of
removing the brackets from

this first term here.

Let's move to the second term.
We've got a 2 J crossed with

B1I. A2 J
crossed with

B2J. And
a two J

crossed with B3K.

And that's taking care
of this term.

And finally we use the
distributive law once more to

remove the brackets over. Here
will get a 3K cross be one I.

Plus a
3K cross

be 2

J. And finally,
a 3K cross

be 3K. So we get all
these nine terms. In fact, here

now it's not as bad as it looks,
because some of this is going to

cancel out, and in particular
one of the things you may

remember we said was that if you
have two parallel vectors.

Their vector product is 0.

Now these two vectors A1I and B1
I a parallel because both of

them are pointing in the
direction of Vector I. So these

are these are parallel and so
the vector product is 0, so that

will become zero.

For the same reason, the A2 J
Crosby to Jay is 0 because A2 J

is parallel to be 2 J.

And a 3K Crosby 3K is 0 because
a 3K and B3K apparel vectors so

they disappear. So that's
reduced it to six terms.

Now, what about this term here?
Let's look at a one I cross be 2

J. If you work out the vector
product of these, we've got a

vector in the direction of. I
crossed with a vector in the

direction of Jay, and we've seen
already that if you have an I

cross OJ the result is K. So
when we workout a one across be

2, J will write down the length
of the first, the length of the

2nd. And then the direction is
going to be such that it's at

right angles to I and TJ in a
sense defined by the right

hand screw rule. And that
sense is K. So when we

simplify this first term here,
it'll just simplify to A1B2K.

What about this one here?

This direct this factor here a
one is in the direction of I.

This ones in the direction of K
and we've already seen that if

we workout I cross K let me
remind you of our little diagram

we had. I JK this cycle of
vectors here. If we want a

vector in the direction of
across with vector in the

direction of K were coming
anticlockwise around this

diagram and I Cross K is going
to be minus J.

So when we come to workout this
term, we want the length of the

first term A1 length of the
second one which is B3. But I

Cross K. Is going to be minus J.

So that start with

that. What about this one?
Again, length of the first one

is A2. Length of the second one
is B1, and a Jake Ross I.

For the diagram again Jake Ross
I moving anticlockwise around

this circle, J Cross Eye is
going to be minus K.

And also this one will have a 2
J Crosby 3K length of the first

one is A2 length of the second
one is B3 and Jake Ross K.

Clockwise now Jake Ross K is I.

We dealt with that one and the
last two terms, a 3K cross be

one. I will be a 3B One and
K Cross. I came across. I will

be J. And last of all,
a 3K cross B2J will be a

3B2K Cross JK cross. Jay going
anticlockwise will be minus I.

And these six times if we study
them now, you realize that two

of the terms of involved K2 of
the terms involved Jay and two

of the terms involved I. So we
can collect those like terms

together and in terms of eyes,
there will be a 2B3I

And and minus a 3B2I. So
just the items will give you

this term here.

Just the Jay terms will be a

3B one. Minus

A1B3? There the Jay terms.

And the Kay terms, there's

an A1B2. Minus an

A2B one. And those are
the key terms.

So All in all, with now reduced
this complicated calculation

down to one which just gives us
a formula for calculating a

cross be in terms of the
components of the original

vectors, and that's an important
formula that will now illustrate

in the following example.

OK, so to illustrate this
formula with an example, let me

write down the formula again
across be is given by a

two B 3 - 8 three

B2. Plus a
3B one minus

a one V3J.

Plus A1 B 2 - 8, two
B1K. So that's the formula will

use. Let's look at a specific

example. And let's choose
A to be the Vector

4I Plus 3J plus 7K.
And let's suppose that B

is the vector to I
plus 5J plus 4K.

So to use the formula we
need to identify A1A2A3B1B2B3

the components, so a one will

be 4. A2 will be 3 and
a three will be 7B. One will be

two, B2 will be 5 and be three,
will be 4 and all we need to do

now is to take these numbers and
substitute them in the

appropriate place in the
formula. So will do that and see

what we get across be. We want a

2B3. Which is 3 * 4.

Which is 12

subtract A3B2. Which is
7 * 5, which is 35 and that

will give us the I component of

the answer. Plus a 3B
One which is 7 *

2 which is 14. Subtract
A1B3 which is 4 *

4 which is 16.

And that will give us the J
component of the answer.

And finally. A1B2?

Which is 4 * 5, which is 20.
Subtract a 2B one which is 3

* 2, which is 6 and that will
give us the key component of

the answer.

So just tidying this up 12.
Subtract 35 is minus 23 I.

14 subtract 16 is minus

2 J. And 20
subtract 6 is plus 14K.

That's the result of calculating
the vector product of these two

vectors A&B and you'll notice
that the answer we get is indeed

another vector. Now that example
that we've just seen shows that

it's a little bit cumbersome to
try to workout of vector

product, and for those of you
familiar with mathematical

objects, called determinants,
there's another way which we can

use to calculate the vector
product and I'll illustrate it.

Now, if you've never seen a
determinant before, it doesn't

really matter because you should
still get the Gist of what's

going on here. If we want to
calculate a cross be.

We can evaluate this as a
determinant, which is an object

with two straight lines down the
size like this along the first

line will write the three unit
vectors I Jane K.

On the second line, will write
the components of the first

vector, the first vector being a
as components A1A2 and A3.

And on the last line, the line
below will write the cover 3

components of the vector be
which is B1B 2B3.

Now, as I say, when you evaluate
the determinant, you do it like

this, and if you've never seen
it before, it doesn't matter.

Just watch what I do and we'll
see how to do it.

Imagine first of all, that we
cover up the row and column with

the I in it.

This first entry here corrupt
their own column and look at

what's left. We've got four
entries left and what we do is

we calculate the product of the
entries from the top left to the

bottom right. And subtract the
product of the entries from the

top right to the bottom left.

In other words,
we workout a 2B3.

Subtract A3B2.
That's 82B3, subtract

A3B2. So that Operation A2
B 3 - 8 three B2 is what will

give us the eye components of
the vector product.

Then we come to the J component.

And again, we cover up the row
and the column with a J in it,

and we do the same thing. We
calculate the product A1B3.

Subtract A3B, One.

And that will give us a one B 3
minus a 3B One J. But when we

work a determinant out, the
convention is that we change the

sign of this middle term here.

Finally, we moved to the last
entry here on the 1st Row.

And we cover up their own
column with a K in and do

the same thing again,
A1B2 subtract a 2B one.

A1B2 subtract 8 two B1 and that
will give us the key component

and this formula is equivalent
to the formula that we just

developed earlier on. The only
difference is there's a minus

sign here, but when you apply
the minus sign to these terms in

here, this will swap them around
and you'll get the same formula

as we have before.

So let's repeat the previous
example, doing it using these

determinants we had that a was
the vector 4I Plus 3J plus 7K,

and B was the vector to I
plus 5J Plus 4K, so will find

the vector product. But this
time will use the determinant.

Always first row right down
the unit vectors IJ&K.

Always in the 2nd row, the three
components of the first vector,

which is A the three components
being 4, three and Seven.

And the last line, the three
components of the second vector.

25 and four.

And then we evaluate this.
As I said before, imagine

crossing out the row and
the column with the IN.

And calculating the product 3
* 4 - 7 * 5, which is 3, four

12 - 7, five 35 and that will
give you the I component of

the answer.

Cross out their own column
with the Jays in.

4 * 4 - 7
* 2, which is four

416-7214. That will give you the
J component and as always with

determinants, we change the sign
of this middle term.

And finally cross out the row
and column with a K in.

We want 4 * 5.

Which is 20. Subtract 3 * 2
which is 6. So we've got 20

subtract 6 and that will give
you the cake component of the

answer and just to tidy it all
up, 12 subtract 35 will give you

minus 23 I. 16 subtract 14 is 2
with the minus sign. There is

minus 2 J.

And 20 subtract 6 is 14K and
that's the answer we got before

this method that we've used to
expand the determinant is called

expansion along the first row
and those of you that know

determinants will find this very
straightforward and those of you

that haven't. All you need to
know about determinants is what

we've just done here.

We now want to look at some
applications of the vector

product. And the first
application I want to

introduce you to is how to
find a vector which is

perpendicular to two given
vectors.

And the easiest way to
illustrate this is by using an

example. So let's suppose are
two given vectors. Are these

supposing the vector A is I plus
3J minus 2K?

The second given vector B is
5. I minus 3K.

Now you remember when we defined
the vector product of the two of

two vectors A&B, the direction
of the result was perpendicular

both to a.

And to be, and indeed the plane
which contains A and be. So if

we want to find a vector which
is perpendicular to these two

given vectors or we have to do
is find the vector product

across be. So we do that a
cross B and again will use the

determinants, firstrow being the
unit vectors IJ&K.

The 2nd row being the components
of A the first vector, which are

1, three and minus 2.

And the last drove being the
components of the second vector,

which is B and those components
are 5 zero because there are no

JS in here.

And minus three from the case.

So let's workout this
determinant. So when we work it

out, we cross out the row and a
column containing I.

And we calculate the products of
these entries that are left

three times, minus three
subtract minus two times. Not.

So we want three times minus
three, which is minus 9 subtract

minus two times. Not. So we're
subtracting zero, and that will

give you the I component of the

result. Then we want to move to
the J component, cross out the

role in the column with the Jays
in, and again evaluate the

product's one times. Minus three
is minus 3, subtract minus 2 *

5, so we're subtracting minus
10, which is the same as adding

10. And you remember we change
the sign of the middle term.

Finally, last of all the

K component. Cross out the row
in the column with a K in.

And the products that are left
are 1 * 0, which is 0. Subtract

3 five 15, so it's 0 subtract
15. And if we tidy it, what

we've got, they'll be minus nine

I. This 10 subtract 3 is
7 so it will be minus Seven

J. Minus 15.

And this vector that we've found
here is perpendicular to both A

and to be.

And we've solved the problem.

Sometimes you asked to find a
unit vector which is

perpendicular to two given
vectors. So if this problem had

been find a unit vector
perpendicular to a into B, or we

have to do is calculate a Crosby
and then find a unit vector in

this direction. Now to find a
unit vector in the direction of

any given vector or we have to
do is divide the vector by its

modulus. It's a general result,
the unit vector in the direction

of any vector is found by taking
the vector and dividing it by

its modulus. So if we want a
unit vector in the direction of

a cross be.

All we have to do is divide
minus nine. I minus Seven J

minus 15 K by the modulus of
that vector and the modulus of

this vector is found by finding
the square root of minus 9

squared. Add 2 - 7 squared.

Added 2 - 15 squared and if you
do that calculation you'll find

out that this number at the
bottom is the square root of

355, so I can write my unit
vector in this form one over the

square root of 355.

Minus 9 - 7 J minus 15
K, so that's now a unit vector.

Which is perpendicular to the
two given factors.

A second application that I want
to introduce you to is a

geometrical one. We can use the
vector product to calculate the

area of a parallelogram.

Let's suppose
we have

a parallelogram.

And let
me denote.

A vector along one of the sides

as be. And along this side here

as see. And this angle in the
parallelogram here is theater.

Now this is a parallelogram so
that sides parallel to this side

and this sides parallel to that
side. The area of a

parallelogram is the length of

the base. Times the
perpendicular height. Let

me put this perpendicular
height in here.

So there's a perpendicular
and let's call this

perpendicular height age.

Now if we focus our attention on
this right angle triangle in

here, we can do a bit of
trigonometry in here to

calculate this perpendicular

height H. In particular, if we
find the sign of Theta, remember

that the sign is the opposite
over the hypotenuse. We can

write down that sign Theta.

Is the opposite side, which is
H, the perpendicular height

divided by the hypotenuse.
That's the length of this side.

And the length of this side is
just the length of this vector,

see, so that's the modulus of C.

If we rearrange this formula, we
can get a formula for the

perpendicular height age and we
can write it as modulus of C

sign theater. We're now in a
position to write down the area

of the parallelogram. The area
of the parallelogram is the

length of the base times the
perpendicular height and the

length of the base is just the
modulus of this vector. Be now

it's just modulus of be.

What's the length of the base
when we multiply it by the

perpendicular height, the
perpendicular height is the

modulus of C sign theater.

If you look at what we've got

here now. We've got the modulus
of a vector, the modulus of

another vector times the sign of
the angle in between the two

vectors, and this is just the
definition of the modulus of the

vector product be crossy, so
that is just the modulus of the

cross, see. So this is an
important result. If we ever

want to find the area of a
parallelogram and we know that

two of the sides are represented
by vector being vector C or we

have to do to find the area of
the parallelogram is find the

vector product be crossy and
take the modulus of the answer

that we get? That's a very
straightforward way of finding

the area of a parallelogram.

The final application I want to
look at is to finding the volume

of the parallelepiped now
parallelepiped such as that

shown here is A6 faced solid
where each of the faces is a

parallelogram. And the opposite
faces are identical

parallelograms. Now, if we want
the volume of this

parallelepiped, we want to find
the area of the base and

multiply it by the perpendicular
height. So let's put that in.

Extend the top here.

So that we can see what the
perpendicular height is.

And let's call that
perpendicular height H. Now in

fact, this perpendicular height
is the component of a which is

in the direction which is normal
to the base.

Now we've seen that a direction
which is normal to the base can

be obtained by finding the
vector product be crossy.

Remember when we find be cross C
we get a vector which is normal,

so the component of a in the
direction which is normal to the

base is given by a dotted with a
unit vector in this

perpendicular direction, which
is a unit vector in the

direction be across see so this

formula. Will give us this
perpendicular height.

Now we want to multiply this
perpendicular height by the area

of the base. We've already seen
that the base area in the

previous application was just
the modulus of be cross, see.

So In other words, the volume of
the parallelepiped, let's call

that V is going to be a dot B
Cross SI unit vector multiplied

by the modulus of be cross, see.

Now remember when you find the
unit vector, one way of doing it

is to find the cross product and
divide by the length.

So you'll see when we recognize
it in that form, you'll see

there's a bee crossy modulus
here, Annabi, Crossy modulus

here, and those will cancel.

And what we're left with is that
if we want to find the volume of

the parallelepiped, all we have
to do is evaluate the dot

product of A with the vector be
cross. See now that may or may

not give you a positive or
negative answer. So what we do

normally is say that if we want
the volume we want the modulus

of a dotted with be Cross C and
that is the formula for the

volume of the parallelepiped.

We look at one final example
which illustrates the previous

formula that the volume of the
parallelepiped is given by the

modulus of a dotted with the
vector product of BNC.

Let's suppose that a is
the vector 3I Plus 2J

Plus K. B is the
vector, two I plus J

Plus K. And see
is the vector I plus

2J plus 4K.

So those three vectors represent
the three edges of the

parallelepiped. OK, first of
all, to apply this formula we

want to workout the vector
product of B&C.

And I'll use the determinants
again, be crossy first row. As

always I JK.

The 2nd row we want the three
components of the first vector,

which are 211.

And then the last row we want
the three components of the

second vector, which are 1, two

and four. And then we proceed to
evaluate the determinant

crossing out the row and column
with the eyes in will get ones,

four is 4. Subtract ones too is
2, four subtract 2 is 2, and

that will give you the number of
eyes in the solution to I.

Then we come to the Jays. We
want 248 subtract 1 is one is

one, so it's 8. Subtract 1 is 7
and then we change the signs of

the middle term. So we'll get
minus Seven J and finally for

the case two tubes of 4 - 1 is
one is one 4 - 1 is 3. So will

end up with plus 3.

So that's the vector product of

BNC. And we now need to find the
scalar product the dot product

of A with this result that we've

just obtained. And I'll use the
column vector notation to do

this because it's easy to
identify the terms we need to

multiply together the vector a
as a column vector is 321.

We're going to dot it with be
cross, see which we've just

found is 2 - 7 and three.

And you'll remember that to
calculate this dot product we

multiply corresponding
components together and then add

up the results. So we want 3

twos at 6. Two times minus 7
- 14 and 1. Three is

3. So we've got 9 subtract 14,
which is minus 5.

Finally, to find the volume of
the parallelepiped, we find the

modulus of this answer, so V
will be simply 5, so five is its

volume. And that's an
application of both scalar

product and the vector product.