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PROFESSOR: Hi.
Today, we finish up the last of the basic vector
operations.
Today, what we're going to discuss is something called
the vector product or the cross product.
And it's a new type of multiplication, whereby we now
multiply two vectors to obtain a vector.
You see, the dot product multiplied two vectors, but it
gave a scalar.
This is why the dot product was called a scalar product.
Now, the idea is, again, we can go to the physical world
to seek motivation.
And my only problem with that is that some of the examples
that motivated this are not quite as elementary as the
work equals force times distance formula of last time.
The cross product does come up in sort of trivial
applications, in terms of moments, angular momentum, in
terms of mechanics, and in more advanced situations, in
terms of the Maxwell equations where, you may remember, in
physics courses, going through this routine of the force and
the flux and the intensity and what have you.
And at any rate, rather than waste time-- and I mean, I use
the word waste here judiciously, because, for the
purpose of our course, it is not nearly as crucial to
motivate why we invented the cross product physically as it
is to define it mathematically and to use it structurally.
You see, what I want to do is simply get to the crux of the
situation, discuss the concept of a cross product, and simply
point out that if A and B are vectors, by A cross B--
written this way, A cross B--
we mean a vector.
Now, here's the kicker again.
As soon as I say vector, that means to define it, I have to
tell you three ingredients, namely magnitude,
direction, and sense.
Right?
Now, for the magnitude of A cross B, it's simply the
magnitude of A times the magnitude of B times the
magnitude of the sine of the angle between A and B. And I'm
going to come back and comment on that in a few moments,
about the difference between the sine here versus the
cosine of the previous lecture.
Again, physically, there's a motivation for this.
As far as we're concerned, all we have to do is say, this is
what the magnitude is defined to be.
This is the game.
We define it this way.
Next thing I have to do is tell you its direction.
And the direction of A cross B is, by definition,
perpendicular to both A and B. All right?
Another way of saying that is in the event that A and B are
not parallel, then the vectors A and B determine a plane.
And what we're saying is that A cross B is in a vector
perpendicular to the plane determined by
A and B. All right?
By the way, if A and B are parallel, they don't determine
the plane, because they coincide.
They have only one direction.
But if A and B are parallel, the angle between them is 0 or
180, depending on whether their sense
is the same or not.
Once the angle is either 0 or 180, the sine of
that angle is 0.
Once one of the factors here is 0, the whole product is 0.
So in the event that A and B do not determine the plane,
the magnitude of A cross B is 0, which means that A cross B
is the zero vector.
And that's a pretty trivial situation.
So we just won't even worry about that.
And thirdly, I must now give you the sense of A cross B.
And the sense is the usual right-hand rule, in other
words, the direction in which a right-handed thread screw
would turn as A is rotated into B--
and now I have to be careful because of this sine
situation--
through the smaller angle.
You see, look it.
Notice that this sense is going to depend on what-- if
you rotate A into B, you get the opposite sense than if you
rotated B into A.
And not only that, but if you rotate A into B through the
larger of the two angles, notice that rotating A into B
through one angle-- say the smaller one--
gives you a different sense than rotating A into B through
the larger of the two angles.
So there's a certain ambiguity that comes up here.
And the ambiguity occurs because of the fact that we're
using the sine rather than the cosine.
In other words, if you reverse the sense of an angle, if you
change its sine, the algebraic sine, the trigonometric sine,
s-i-n changes.
OK?
And similarly, the sine of an angle is the negative of the
sine of 360 degrees minus that angle.
But that, at any rate, is the definition
of the cross product.
Given two vectors, their cross product is what?
A vector whose magnitude is the product of the two
magnitudes times the magnitude of the sine of the angle
between the two vectors.
The cross product is perpendicular to each of the
two vectors.
And it's sense is the right-hand rule as the first
is rotated into the second through the
smaller of the two angles.
This is important because, you see, what happens is that as
important as the cross product is from a physical point of
view, arithmetically, it's a nuisance.
We can take very few liberties with the cross product.
It turns out that almost every rule that we would like to be
true about numerical multiplication is false for
vector cross products.
For example, among other things, A cross B is not equal
to B cross A.
In fact, A cross B is the negative of B cross A. And the
reason for this is simply, as you check this thing through,
A cross B and B cross A certainly
have the same magnitude.
They have the same direction, because each is perpendicular
to both A and B.
But notice that the sense is different.
Namely, A cross B tells us to rotate A into B through the
smaller of the two angles.
And in our particular diagram that is what, the
counterclockwise direction?
On the other hand, B cross A says, rotate B into A through
the smaller of the two angles.
And if we do that, notice that we get the clockwise direction
so that B cross A and A cross B have the same magnitude.
They have the same direction.
But they have the opposite sense.
That is, A cross B is the negative of B cross A.
And what that means, you see, in computations is that every
time you change the order of A and B in a cross product, you
must remember to change the sign.
In fact, what's particularly embarrassing is if you forgot
to do this.
If this particular term is going to be added on to
something, and you forget to change its sign, you're going
to be off by more than just the error of a sign.
In other words, if something was supposed to be plus 4, and
you have it as minus 4 or vice versa, the error is going to
be 8 if you add it rather than subtract it, you see.
So the error gets hidden in other computations.
But in other words, you can change the order of the
factors, but you must remember to change the sign.
Another interesting thing is that unlike in the dot
product, where A dot B dot C didn't even make sense, it is
true that A cross B cross C and A cross B
cross C do makes sense.
And by the way, what are these things?
They're vectors.
B cross C is a vector, which is perpendicular to both B and
C. It's perpend--
you see.
And A is a vector.
And the cross product of two vectors is, again,
a vector, et cetera.
These are two vectors.
But in general, these two vectors do
not have to be equal.
In fact, as I shall show you in a moment, these two vectors
don't even have to be in the same plane,
let alone being equal.
They don't even have to be in parallel planes.
And again, without making too much of a fuss over this,
let's take a look and see if we can see why.
Let's look at A cross B cross C. By definition, let's just
even look at the direction.
Let's not even worry about magnitudes.
What is the direction of A cross (B cross C)?
It must be what?
It must be, on the one hand, perpendicular to A and on the
other hand perpendicular to B cross C. OK?
That's the definition of being a cross product.
On the other hand, what about B cross C as a vector itself?
B cross C is perpendicular to both B and C. So
what do you have here?
You have that whatever A cross (B cross C) is, it's
perpendicular to B cross C, which, in turn, is
perpendicular to both B and C.
That sounds like a tongue twister.
But if you decipher it, all it says is that A cross (B cross
C) is perpendicular to the perpendicular to B and C. And
I call that deciphering it.
What is the perpendicular to the perpendicular?
You say let's just write it this way.
Here are two vectors in the plane.
Here's a perpendicular.
I now take a perpendicular to that.
And that's going to be parallel to
the original plane.
This is written up in the text.
It's written up as one of our learning exercises.
But just for a quick run-through now, all I want
you to see is that A cross (B cross C) is parallel to the
plane determined by B and C.
In fact, the easy way to memorize this is look at the
vector that's in parentheses.
And the cross product is what?
It's parallel to the plane determined by the two vectors
in parentheses.
The only time these two vectors, by the way, won't
determine a plane is if they're parallel.
But in that case, the cross product would
have been zero anyway.
In other words, to summarize what I've written on the
blackboards here, A cross (B cross C) is a vector which
lies in the plane determined by B and C.
After all, if the vector is parallel to the plane, you can
assume it lies in the plane, because you can shift it
parallel to itself.
And correspondingly, (A cross B) cross C is a vector in the
plane determined by A and B. In particular, there's no
reason why the AB plane and the BC plane have to be the
same plane.
No law says that our three vectors, A, B, and C, all lie
on the same plane.
So in particular, this is why the associative property
doesn't hold for a cross product.
Because as soon as you shift the parentheses, you actually
shift the plane in which the vector exists.
OK?
But again, I just want to go through these highlights to
make sure that you don't miss them.
But these are done in more detail in the notes.
Before I start blasting the cross product too much, let me
point out that certain properties that we like in
numerical arithmetic are present in the cross product.
For example, there is a--
I say this, there's a little structure.
The distributive property holds to the cross product.
In other words, if the vector A is crossed with the sum of
the two vectors, B and C--
and by the way, again notice, if B and C are vectors, B plus
C is a vector.
I'll put a v down here to indicate vector.
A is a vector.
And the cross product of two vectors is a vector.
Notice that, again, with the cross product, all of the
things that you're working with happen to be vectors.
But my claim is that A crossed with (B plus C) is equal to A
crossed with B plus the product of A crossed with C.
That's a very nice structural property.
Again, as usual, Cartesian coordinates have a
lot to offer us.
If we switch to Cartesian coordinates, again, as usual,
writing A as a1 i plus a2 j plus a3 k, and B as b1 i plus
b2 j plus b3 k, and multiplying term by term, we
observe a rather interesting situation.
You see, notice things like this.
When you cross i with i, j with j, and k with k, you're
always going to get the zero vector.
And the reason for that is, we already saw that since the
sine of the angle between two parallel vectors is 0, the
magnitude of the cross product is 0, that means that the
cross product is the zero vector.
The question comes up, what happens when you cross i with
j or j with i, i with k, k with j, et cetera?
I think it's easy to see that when you cross i and j and j
and i, you're going to get either k or minus k.
The right-handed rule is set up very, very conveniently,
again, to help us with a memory device.
The easiest way to remember how these things go is i, j,
and k were set up to form a right-handed rule, a
right-handed system.
Write i, j, and k sequentially.
See, i, j, k, i, j, k, et cetera.
Call the normal order the positive direction and the
opposite order the negative direction.
For example, when you want i cross j, just read this list
as you go along: i followed by j gives k, which is in the
right order-- see, i, j, k--
so i cross j is k.
On the other hand, j cross i means what?
j followed by i.
j followed by i is this way.
That also yields k.
But you're now going in the negative direction.
To summarize these results, i cross j is the same as minus j
cross i, which is k.
k cross i is the negative of i cross k, which is j.
i cross i, j cross j, and k cross k are all 0.
And notice the use here of zero vector, because we're
dealing with vectors.
And again, leaving the details to you.
They're very simple to come up with.
Just carry out the multiplication.
And what you find is that A cross B is
given by the following.
The i component of the vector is a2b3 minus a3b2.
The j component is a3b1 minus a1b3.
And the k component is a1b2 minus a2b1.
And if you want a convenient way of memorizing this, again,
write 1, 2, 3 in their correct sequential order.
And what you're saying is, each term in A cross B
involves what?
An a factor followed by a b factor.
The subscripts involve the
component, which isn't present.
See, for the first component, i, the subscripts are 2 and 3.
For the second component, j, the second component, the
subscripts are 3 and 1.
And for the third component, k, the subscripts are 1 and 2.
And notice that the way the sines are determined is that
if the subscripts occur in the correct sequential order, it's
positive, otherwise negative.
For example, 2 followed by 3 is the correct order.
But 3 followed by 2 is the negative order.
Notice, in this sequence, 1 follows 3, whereas 1 to 3 is
in the negative order.
See?
Just the same, this works out very, very nicely.
And the reason I went through this-- it may seem like an
awful lot of work to go through that to show you how
logical this thing is--
is that something comes up in the textbook, which is not
bad, but which I find as frightening if you have not
had experience with something called determinants.
You see, the author, Professor Thomas, elects to introduce a
convenient device for memorizing this result.
And he writes it as a 3 by 3 determinant.
Now, let me just write down the 3 by 3 determinant, and
then we'll talk about it later.
I put convenient in quotation marks to emphasize to you that
this is convenient only if you already know determinants.
If you don't know determinants, this is not only
not convenient, but it's frightening.
We're always frightened of things that we don't
understand too well.
And I call it a memory device, because you do not have to
have determinants to remember this particular result.
In fact, if one were going to list, in order, the reasons
for inventing determinants, and the man was reading off of
very, very long list, and you were getting bored, as soon as
he came to the reason that it was convenient to list the
cross product, you could heave a sigh of relief, because you
would know that he's near the end of the list.
In other words, to make a long story short, of all the
reasons for inventing determinants, using them for a
cross product representation is perhaps the least
significant.
But you see, in Professor Thomas's approach,
determinants are assumed to be known.
And if you do know them, it's a good notation.
If you don't, forget it.
The trouble is, I tell you to forget it, and you
won't forget it.
It always works that way.
When you tell the student not to forget
something, he forgets.
When you say, please ignore it and forget
it, he doesn't forget.
So since you may be a little bit rusty on determinants, let
me give you a very, very brief review of determinants.
In fact, it's not only very brief, it's a pseudo-review,
because I will give you no logic behind this at all.
I'm just going to tell you the recipe.
First of all, by a 2 by 2 determinant, you mean a square
array of numbers consisting of two rows and two columns.
You use absolute value signs to indicate the determinant.
And by definition, the 2 by 2 determinant, a, b, c, d is the
product of the two terms, a and d, minus the product of
the two terms, b and c.
In other words, if you multiply the upper left by the
lower right and subtract from that the upper right times the
lower left, that is, by definition, the value of the 2
by 2 determinant, just by definition.
By the way, in block seven of our course, we're going to do
this in much more detail.
For now I just want to give you enough of a hint, so that
if you are frightened by determinants, you can see what
Professor Thomas is talking about in the text here.
Now, to expand a 3 by 3 determinant, one uses
a little sign code.
Starting at the upper left-hand corner, you put a
little plus sign.
Then you alternate plus, minus, plus, et cetera.
The plus simply is a code to tell you, leave the sign of
the term alone.
And the minus tells you, change the sign of the term.
What that means is this.
By definition, this 3 by 3 determinant
is obtained as follows--
take x1 as it appears and strike out the row and column
in which x1 appears and multiply x1 by the remaining 2
by 2 determinant.
Then what you do is you factor out x2 with a minus sign,
strike out the row and column in which x2 appears, and
multiply x2 by the remaining 2 by 2 determinant that's left.
Similarly, take out x3 with the sign that it appears with,
strike out the row and column in which it appears, and
multiply by the remaining 2 by 2 determinant.
Summarized, to expand this 3 by 3 determinant, it's x1
times the determinant, y2, y3, z2, z3 minus x2 times the
determinant, y1, y3, z1, z3, plus x3 times the determinant,
y1, y2, z1, z2.
Now, that happens to have very practical application.
As I say, for this particular situation, I just want to do
it with you to have you see how it works.
Let me come back to what we were doing before, and show
you how, if we use this definition, we can
mechanically write down A cross B in a hurry.
What did the rule say?
It says, to expand the 3 by 3 determinant, take out i, as it
appears, and multiply by the 2 by 2 determinant that's left,
when you strike out the row and column that i appears in.
OK?
Then it says what?
Take out j with a minus sign, strike out the row and column,
multiply by the remaining 2 by 2 determinant.
Then it says, factor out k and strike out the row and column
that k appears in.
Multiply by the remaining 2 by 2 determinant.
Well lookit, by our rule of multiplying a 2 by 2
determinant, what is this thing here?
This is nothing more than a2 b3 minus a3 b2.
I'm not going to go through this, because time is short
and you can carry this operation out by yourself.
What I would like you to see, however, is that this term
here is precisely the same as the term we have here.
In other words, if I carry out this mechanical recipe, I am
going to be able to get the same answer that I got
logically, only without having to worry about keeping track
of things in my head.
In other words, if I do know determinants, it is very
convenient to write down the cross product of two vectors
using Cartesian coordinates.
And just to illustrate that, let me
give you a few examples.
I call this example one.
Find the vector perpendicular to the plane, determined by
the three points, (1, 2, 3); (5, 9, 4); and (7, 6, 8).
Now, try to visualize this problem if
you didn't know vectors.
Try to visualize drawing this thing to scale, trying to
imagine what slope the plane has and how you find a vector,
or a line perpendicular to the plane.
And by the way, I don't even need the word vector in here.
Later I could say, find a direction.
But we'll worry about that more next time.
The thing that I want to emphasize here is, I just very
quickly draw myself a little convenient device.
I don't draw it to scale.
I assume that the points A, B, and C are not in the same
straight line.
And if they are, it'll become very interesting that I simply
won't get--
I'll get 0 for a cross product to tell me that they don't
form two non-parallel vectors.
But I mark off A, B, and C. And the first thing I do, as
always, is I vectorize--
I draw the lines AB and AC, add I vectorize them.
That forms for me, as usual, the vector AB
and the vector AC.
Remembering how I form a vector from two points in
Cartesian coordinates, again, subtracting the coordinates of
the point that I'm going--
from the point that I'm going to, I subtract away the
coordinates of a point that I'm leaving.
In other words, 5 minus 1, 9 minus 2, 4 minus 3.
The vector AB, very simply, is 4i plus 7j plus k.
And similarly, the vector AC, very conveniently, is 6i plus
4j plus 5k.
You see, 7 minus 1, 6 minus 2, 8 minus 3.
Now, I better stop checking these things.
Because, with my luck, I've probably subtracted wrong, and
you might not notice it if we just go through it rapidly.
But look at how quickly I've got AB and AC now.
Now, here's the beauty.
What do I want?
I want a vector perpendicular to the plane determined by A,
B, and C.
In particular, if I can find a vector perpendicular to both
AB and AC, namely if a vector is perpendicular to two lines
in a plane, it's perpendicular to the plane itself.
If I can find a vector perpendicular to both AB and
AC, I'm home free.
That's the vector I'm looking for, or at least a vector that
I'm looking for.
Do I know, conveniently, a vector perpendicular to two
given vectors?
And the answer had darn well better hinge somehow on the
title of today's lecture, being the cross product.
Yes, by definition, if I cross AB and AC, whatever vector I
get is going to be perpendicular to the plane
determined by A, B, and C.
And now what I'm saying is that the beauty of Cartesian
coordinates is that I can now very quickly write down what
AB cross AC is as a determinant.
Namely, I write down i, j, k.
Then I write down the components of AB.
Then I write down the components of AC.
See, (4, 7, 1); (6, 4, 5).
And now I do what?
I strike out the row and column in which i appears.
And the remaining 2 by 2 determinant is what?
7 times 5 minus 1 times 4-- in other words, 35 minus 4.
I take out minus j, strike out the row and
column in which j appears.
The remaining 2 by 2 determinant is 4 times 5,
minus 1 times 6-- in other words, 20 minus 6.
I strike out the row and column in which k appears.
The remaining 2 by 2 determinant is 4 times 4
minus 7 times 6.
That's 16 minus 42.
And now simplifying, what do I have?
I have 31i minus 14j minus 26k.
In other words, the vector 31i minus 14j minus 26k is the
vector that I'm looking for.
It's a vector perpendicular to the plane.
And what I would like you to see is how convenient this
vector arithmetic is to studying space geometry.
And in the next lesson I'm going to say some more about
space geometry.
But for now I thought maybe another nice example before we
finish up, and let it go at that.
Let me just do one more example for you.
And to simplify the computation, I might as well
use the vectors I've already used.
Let's take the same three points we had before, (1, 2,
3); (5, 9, 4); (7, 6, 8).
And we're going to ask, what is the area of the triangle
determined by these three points?
And the subtlety that I wanted to bring out here was an
excuse to show you how the magnitude of the cross product
of two vectors can be interpreted.
Namely, look at two vectors.
All right?
Call them x and y.
Draw them into scale from a common point.
And let theta be the angle formed
between these two vectors.
Consider the parallelogram formed by these two vectors.
The area of a parallelogram is what?
It's the base times the height.
Now, look at.
The base of this parallelogram is the magnitude of y.
And the height of this parallelogram--
remember, this magnitude is x.
So this height is x sine theta.
This is the area of the parallelogram.
But notice that this is also, by definition, the magnitude
of x cross y.
In other words, notice that this is, by definition, the
magnitude of x cross y. x cross y is a vector whose
magnitude is this.
In other words, the magnitude of x cross y is the area of
the parallelogram determined by x and y, which gives you a
rather nice, interesting interpretation.
Imagine, for example, that x and y don't change in length.
Is it clear to you that as I change the angle between x and
y, the area of the parallelogram
formed by x and y changes?
In other words, one extreme case is when the angle between
x and y is 0, in which case the parallelogram has no area.
Another extreme case is when the angle between x and y is
90 degrees.
In other words, x and y are perpendicular to each other.
But if the lengths of x and y remain fixed, as I alter the
angle between them, I alter the area of the parallelogram
determined by them.
All right?
I just thought that's a cute little device whereby you can
visualize how the magnitude of the cross product of two
vectors varies.
Think of the two vectors at a common origin, and the
magnitude is affected by the area of that parallelogram as
the angle changes.
OK?
In other words, if you change the angle, the area of the
parallelogram changes.
But at any rate, getting back to our original problem, the
corollary to that is the following.
We already know what the magnitude of AB cross AC is
from example one.
But what does that mean now, geometrically?
From what we've just seen, the magnitude of AB cross AC is
the area of the parallelogram, which has AB and AC as edges.
Notice that that parallelogram is precisely one-half--
I should say that--I shouldn't say that-- the triangle that
we're looking for is precisely one-half the parallelogram.
In other words, notice that the area of the triangle is
simply 1/2 the magnitude of AB cross AC.
How do we find the magnitude of AB cross C?
We square each of its components, add them, and
extract the positive square root, as usual.
Without going through the details, it's
simply going to be what?
This particular amount.
And the rest is arithmetic's baby, so to speak.
All right?
Now, what I'm hoping is that by this stage of a game, you
now have a better feeling as to what the
cross product means.
Going back to the beginning of my lecture, notice that I
played down the physical applications.
Because one of the things that I wanted you to see is that
the cross product has tremendous application, just
as space geometry alone, quite apart from anything else.
Well, at any rate, what I'm going to do next time is-- now
that we've talked about lines and planes and what have you--
to get down to the Cartesian representation of what lines
and planes would look like.
They play a very vital role in the study of functions of
several variables.
We'll see more about that as the course
continues to unfold.
So until next time, goodbye.
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