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PROFESSOR: Hi.
Our unit to today concerns something
called the dot product.
Now it's very easy to just mechanically give a definition
of the dot product, but in keeping with the spirit both
of our game and of our correlation between the real
and the abstract world, let's keep in mind that when it came
time to define vector addition, we chose, as our
definition, a concept already used in the physical world--
namely, that of a resultant vector.
And now what we would like to do is to introduce a further
structure in our game of vectors.
And by the way, keep in mind that I do not need any more
physical interpretations to add more complexities to my
game, but that perhaps if we do this, it makes the subject
more meaningful, both to the applied person and to the
theoretical person.
The idea that I wanted to keep in mind for motivating today's
lesson was the old high school idea of work equals force
times distance in the elementary physics course.
Well rather than talk on like this, let's just take a look
and see what's going on.
Let's say the lecture is called "The Dot Product," and
our physical motivation is the recipe from elementary
physics, work equals force times distance.
Now this is a deceptive little formula.
First of all, what it really meant was, when we learned
this recipe, was that the force was taking place in the
same direction as the object was moving.
Now what happened next was the following situation.
We have an object, say, being moved by a force.
The object is being moved along a tabletop, say.
The force is represented now by a vector, an arrow.
And we're assuming enough friction here so that the
block moves along the table and doesn't rise.
And the question is, how much work is done on the object as
the object moves, say, from this point to this point?
Now without worrying about what motivates this thing
physically, the important thing was, is that people
observe that physically, the only thing that went into the
work was the component of the force in the direction of the
displacement.
In other words, this was the force that one found had to be
multiplied by the displacement.
In other words, to find the work being done, one took not
the magnitude of F, but rather the component of F in the
direction of s.
And in this diagram, that's just what?
It's the magnitude of F times cosine theta.
And that quantity, which was called the effective force,
was then multiplied by the displacement.
And to write that in more suggestive form, notice that
the work was not the magnitude of the force times the
magnitude of the displacement.
That was only true in the special case where the force
and the displacement were in the same direction.
But that rather the work is what?
It's the magnitude of the force times the magnitude of
the displacement times the cosine of the angle between
the force and the displacement.
And notice by the way, the very special cases, that if F
and s happen to be parallel and have the same sense, then
the angle between F and s is 0.
The cosine of 0 is 1, in which case that you have the work is
the magnitude of force times the displacement.
The other extreme case magnitude-wise is if s and F
happen to be perpendicular, in which case the angle, of
course, is 90 degrees.
The cosine of a 90-degree angle is 0.
In which case, what?
If the force was at right angles to the displacement,
the work was 0.
At any rate then, whether we understand this physical
motivation or not is irrelevant.
The important point is if we want to keep this structure or
this particular motivation in a structural form, we now
generalize this as follows.
We simply say, let A and B be any two vectors, arrows.
And we will define A dot B to be the magnitude of A times
the magnitude of B times the cosine of the angle between A
and B. And I write this this way to indicate an ordering.
In other words, don't think of A and B as being A and B.
Think of A as denoting the first and B as denoting the
second vector.
And what we're saying is the first dotted with the second
is the magnitude of the first, times the magnitude of the
second, times the cosine of the angle as you rotate the
first vector into the second.
And the beauty of having a cosine over here is
the fact that what?
If you reverse the angle of rotation--
in other words, from B into A--
notice that you change the sign of the angle, but the
cosine of theta is the same as cosine minus theta, so no harm
is done this particular way.
On the other hand, had we been dealing with sine of the
angle, as we will in our next lecture, this will make a
difference.
But be this as it may, we define A dot B to be the
magnitude of A times the magnitude of B times the
cosine of the angle between A and B. All right?
The two extreme cases being what?
If A and B are perpendicular, the dot product is 0.
If A and B are parallel, the dot product is equal to the
magnitude of the product of the two magnitudes.
Now the only difficult thing here, or what we sometimes
call undesirable--
and maybe this is what separates the new
three-dimensional geometry from the old
three-dimensional geometry--
is that the cosine of an angle is particularly difficult to
keep track of, especially when the lines are in
three-dimensional space.
How do you measure an angle this way?
You see, in the plane, it's simple.
You draw the thing to scale.
But in three space, this can be rather difficult.
So what we would like to do is to eliminate
this particular term.
We would like to find an expression for A dot B that
doesn't involve the cosine of an angle, at least directly.
And to do this, we simply draw a little diagram.
And notice that even if A and B are three-dimensional
vectors, since A and B are lines, if they are not
parallel, if they emanate at a common point, they form a
plane of their own.
Let's call that the plane of the blackboard.
The third side of the triangle is either A minus B, or B
minus A, depending on where you put the arrowhead here.
But we've already discussed that idea.
And the interesting point here is notice that A minus B very
subtly includes the angle theta.
In other words, imagine the magnitudes of
A and B to be fixed.
And now fan out A and B. As you fan out A and B, what
you're doing is what?
Just changing this angle.
As you change this angle, notice that A minus B changes.
Namely, as these fan out, the vector that joins the two
arrowheads here becomes a different vector, both in
magnitude and direction.
In other words, whether it looks it or not, one of the
beauties of our vector notation is that the cosine of
the angle theta is indirectly included in A minus B.
But now you see, once we have this triangle here, notice
that the law of cosines tells us how to relate the third
side of a triangle in terms of two sides and
the included angle.
OK?
And in fact, when we use the Law of Cosines, notice that
one of the terms is going to be what?
The product of the magnitudes of two sides times the cosine
of the included angle.
And that, roughly speaking, is just what we mean by A dot B.
So without any further ado, what we do now is we just
write the Law of Cosines down here, which says what?
The magnitude of A minus B squared is equal to the
magnitude of A squared plus the magnitude of B squared
minus twice the magnitude of A times the magnitude of B times
the cosine of the angle between A and B. And then, you
see, we simply recognize that this term here is, by
definition, A dot B. We can now take this equation and
solve for A dot B-- which, by the way, this is very, very
important to notice.
I should have pointed this out sooner.
But A dot B is a number.
It's a product of two magnitudes times the cosine of
an angle, which is a number.
This is a numerical equation.
We can therefore solve for A dot B. And we
wind up with what?
That A dot B is the magnitude of A minus B squared minus the
magnitude of A squared minus the magnitude of B squared,
all divided by 2.
And the beauty now is that we have expressed A dot B solely
in terms of magnitudes.
And notice especially in Cartesian coordinates--
and I'll do that next-- but in terms of Cartesian
coordinates, notice that magnitudes are particularly
simple to find.
We just subtract corresponding components
and square, et cetera.
But the important point is that even without Cartesian
coordinates, this particular result expressed as A dot B in
terms of the magnitudes of A, B, and A minus B, and is a
result which is independent of any coordinate system.
However--
and this is done very simply in the text,
reinforced in our exercises.
If you elect to write A, B and A minus B in Cartesian
coordinates and use this particularly straightforward
recipe, what we wind up with is a rather elegant result--
elegant in terms of simplicity, at least.
And that is-- remember in Cartesian coordinates, we
would write A as a1 i plus a2 j plus a3 k.
B would be b1 i plus b2 j plus b3 k.
And then the beauty is what?
That A dot B turns out to be very simply and conveniently
a1 b1 plus a2 b2 plus a3 b3.
In other words, that to dot A and B, if the vectors are
written in Cartesian coordinates--
and this is crucial.
If this is not done in Cartesian coordinates, you can
get into a heck of a mess.
And I have deliberately made an exercise on this unit, get
you into that mess, if you fall into
that particular trap.
But if we have Cartesian coordinates, it turns out that
to dot two vectors, you simply do what?
Multiply the two i components together.
Multiply the two j components together.
Multiply the two k components together, and add.
All right?
By the way, to show you why this works from a structural
point of view without belaboring this point right
now, notice that if you were to multiply in the usual sense
of the word "multiplication," form the dot product here, you
would expect to get nine terms.
In other words, each of the terms in A multiplies each of
the three terms in B. So that altogether you would expect
nine terms.
The thing that's rather interesting here is that
notice that i dot i, j dot j, and k dot k all
happen to be 1.
Because after all, the magnitudes of these vectors
are each 1.
The angle between our i and i is 0, j and j is 0.
The angle between k and k is 0.
So that i dot i, j dot j, and k dot k are all 1.
Whereas on the other hand, when you take mixed terms,
notice that because i and j, i and k, and j and k are all at
right angles, i dot j, j dot k, i dot k are
all going to be 0.
And that therefore, those other six terms will drop out.
In other words, structurally what's happening here is the
fact that the three vectors that we're using here all
happen to have unit length.
And they happen to be mutually perpendicular.
If they were not perpendicular, these mixed
terms would appear in here.
In other words, in general, when you dot two vectors in
three space, depending on the coordinate system, you can
expect up to nine terms in your answer.
But the beauty is that as long as we have Cartesian
coordinates, there happens to be a particularly simple,
beautiful recipe to compute A dot B.
Now keep in mind, the A dot B that we're talking about here
is the same one that we defined before.
It's the magnitude of A times the magnitude of B times the
cosine of the angle between A and B. All we're saying is
that if we use Cartesian coordinates, we can compute it
almost as fast as we can read.
And let me show you that in terms of some examples.
My first example is the following.
Let's imagine that we have three points in Cartesian
three space.
A is the point (1,2,3), B is the point (2,4,1), and C is
the point (3,0,4).
We draw the straight lines AB and AC, and we would like to
find the angle BAC--
in other words, the angle theta.
The first thing that we do--
and this is one of the beauties of how vectors are
used in geometry--
is that we vectorize the lines A and B. We put
arrow heads on them.
That immediately makes them vectors.
We already know from last time how to read the
vectors AB and AC.
Namely, AB is the vector i plus 2j.
See-- just subtract component by component.
2 minus 1 is 1.
4 minus 2 is 2.
1 minus 3 is minus 3, et cetera.
So that the vector AB is the vector i plus 2j minus 2k.
And the vector AC, working in a similar way, is 2i
minus 2j plus k.
Now the beauty is that we can compute these magnitudes very,
very quickly by recipe.
And we've just learned the recipe for finding
A dot B in a hurry.
I mean, well in this case, I don't mean A dot B. I mean the
vector AB dotted with the vector AC.
And going through the computational details here, we
square the components of AB, extract the positive square
root, and we find very easily that the magnitude of AB is 3.
And the hardest part of these problems for me, as a teacher,
is to find ones where I find the sum of 3 squares coming
out to be a whole number.
So I always use the vector 1, 2, 2, because that's a nice
vector that way.
Similarly, the vector AC also happens to have magnitude 3.
OK?
And to find AB dot AC, what do we do?
Let's just come back here and make sure we know
what we're doing now.
We simply dot component by component.
It's (1 x 2) + (2x -2) + (-2 x 1).
In other words, AB dot AC is ( 2- 4 -2), which is minus 4.
Now using our recipe, we see what?
That cosine theta is AB dot AC divided by the product of the
magnitudes of AB and AC, from which we very quickly conclude
that the cosine of theta is minus 4/9.
And if you're still mixed up as to what that minus sign
means, just by way of a quick review of the inverse
trigonometric functions, you locate the point (-4, 9) in
the xy-plane, and your angle theta is this particular angle
here, which means that in terms of principle values, if
you look up the angle in the tables whose cosine is 4/9,
that will give you this angle here.
Subtract that from 180.
And that's the angle that you're looking for.
But the beauty is what?
That you can now find an angle between two lines in space
without having to geometrically worry about what
the angle looks like.
The algebra in Cartesian coordinates takes
care of this by itself.
The same thing happens when you're looking for projections
in three-dimensional space.
Suppose you have a force and a displacement in
three-dimensional space, and you want to project a force
onto a line, a direction.
And our next example shows how the dot product can be used to
find projections.
Namely here's a vector A, here's a vector B. And I would
like to project the vector A onto the vector B. And I would
like to find what the length of that projection is.
Well, from elementary trigonometry, I know that the
length of this projection is just the magnitude of A times
the cosine of theta.
And by the way, notice if theta were be greater than
90-degrees, cosine theta would be negative.
And the minus sign would not affect the length.
It would simply tell us that the projection was in the
opposite sense of B. That's all that would mean.
But here's the interesting point.
If you look at the magnitude of A times the cosine of
theta, it almost looks like a dot product.
After all, theta is the angle between A and B. And if the
magnitude of B were in here, this would just be A dot B.
But the magnitude of B isn't in here.
Of course if the magnitude of B happened to be 1,
that would be fine.
But the magnitude of B might not be 1.
And the most honest way to make it 1 is to divide B by
its magnitude.
And what will that give you?
If you divide any vector by its magnitude, that
automatically gives you a unit vector having the same
direction and sense as the vector that you started with.
In other words, let u sub B, which is B divided by its
magnitude, be the unit vector in the direction--
and by the way, here direction includes sense in the
direction of B--
and notice that the unit vector in the direction of B
has the same direction as B itself.
Therefore, to find the angle between u sub B and A is the
same as finding the angle between B and A.
In other words, the kicker now seems to be that I take this
length, which is the magnitude of A times cosine theta and
rewrite that as follows.
It's the magnitude of A--
and now remembering that u sub B has unit length, I just
throw that in as a factor--
and theta, being the angle between A and B, is also the
angle between A and u sub B. But this, by definition, is A
dot u sub B.
You see, in other words, to find the projection of A onto
B, all you have to do is dot A with the unit vector in the
direction of B. In fact, to summarize that without the u
sub B in there, all I'm saying is given two vectors A and B,
if you dot A and B and then divide by the magnitude of B,
that will be the projection of A in the direction of B. OK?
The projection of A in the direction of B. And of course,
if you want it the other way around, you have to reverse
the roles of A and B. The beauty of this unit in
Cartesian coordinates is how easy it is to compute A dot B
in Cartesian coordinates.
Oh, another example that you might be interested in that I
think is very interesting, and that's the special case where
A and B already happen to be unit vectors.
If A and B already happen to be unit vectors, then if we
use our recipe for the formula for A dot B, we observe that,
in this case, by definition of unit vectors, both the
magnitudes of A and B are 1.
And we find that A dot B is then the cosine of the angle
between A and B. Which means that if A and B happen to be
unit vectors, as soon as you dot them, you have
automatically found the cosine of the angle between the two
vectors, which suggests a rather
general type of approach.
Given any two vectors, divide each by the magnitude, right?
That gives you unit vectors.
Dot them, and that gives you the cosine of the angle
between them.
You see?
In particular, and here's an interesting thing--
You know, I don't know if it's that funny, it just
struck me as funny.
Last night at supper as we were sitting down to eat, my
four-year-old looked at me and said, "Dad, did they have
baked potatoes when you was a little boy?" And you get the
feeling sometimes that people think that the modern world
really changed the old in certain basic ways that didn't
happen at all.
And one of the interesting points is that long before
vector geometry was invented, people were doing
three-dimensional geometry using non-vector methods.
And one technique that happened to be used were
things called directional cosines.
Namely, suppose you were given a line in space.
OK?
As a vector, if you wish to look at it that way-- or if
you didn't want to look at it as a vector, imagine the line
parallel to the given line that goes through the origin.
As soon as I know what that line looks like, I can compute
the angle that it makes with the positive x-axis.
I can compute the angle that it makes with
the positive y-axis.
I can compute the angle that it makes with
the positive z-axis.
And those three angles uniquely determine the
position of the line in space, the direction of the line.
OK?
And those were called the directional angles.
You understand that?
That was the three-dimensional analog of slope.
In other words, to find the slope of a line in
three-dimensional space, draw the line parallel to that line
that goes through the origin, and measure each of the three
angles that that line makes with the
positive x-, y-, and z-axes.
And what the beauty was of the dot product was it just gave
us a simpler way of doing that.
Namely, if A is any vector, I divide A by its magnitude.
That gives me the unit vector in the direction of A.
If I now dot the unit vector in the direction of A with i--
and after all, what is i?
i is the unit vector in the direction of
the positive x-axis.
Since these are both unit vectors, this would be the
cosine of the angle between i and A. And that's just what?
The cosine of the angle that A makes with
the positive x-axis--
traditionally, that angle was called alpha.
So u A dot i is simply cosine alpha.
Correspondingly, u A dot j is the cosine of the angle
between A and j.
That's the cosine of the angle that A makes with
the positive y-axis.
That angle was called beta.
This is cosine beta.
And u sub A dot k is cosine gamma where gamma is the angle
that A makes with the positive k direction.
And these, being the cosines of the angles, these were
called the directional cosines, and they yielded the
slope of lines.
But one must not believe that one needed vectors before he
could do three-dimensional geometry.
What did happen was that vector techniques greatly
simplified many of the aspects of three-dimensional geometry.
Well let's leave this part for a moment and close for today
by coming back to our game idea.
Remember that ultimately, all we will ever use, once we get
started with our game, all we will ever use are the
structural properties.
Now I've gone through these in the notes.
I've gone through them--
well you go through them with me in the text or with
yourselves in the text.
Let me just point out certain properties of the dot product
that are shared by regular arithmetic as well.
For example, A dot B equals B dot A. The dot product is
commutative.
Why is that?
Well think of what A dot B is.
It's the magnitude of A times the magnitude of B times the
cosine of the angle between A and B. But
that's the same as what?
The magnitude of B times the magnitude of A. After all,
numbers, we know are commutative when
you multiply them.
And the cosine of the angle between A and B is the same as
the cosine of the angle between B and A. So these are
equal numerical quantities.
Again, without going through the proof here, it turns out
that if you want to dot a vector with the sum of two
given vectors, the distributive property holds.
Namely, A dot B plus C is A dot B plus A dot C. By the
way, if you did want to prove this, all you would have to do
if you couldn't see it geometrically,
is to argue as follows.
You say, you know?
The easiest way to add and dot vectors is in Cartesian
coordinates.
So let me prove that this result is true in Cartesian
coordinates.
Carry out the details, and if it works in Cartesian
coordinates, since the result doesn't depend on the
coordinate system, the result must be true, regardless of
the coordinate system.
But it's a very simple exercise to actually write A,
B, and C in terms of i, j, k components, compute both sides
of this expression, and show that
they're numerically equal.
I say "numerically" because it's crucial to notice that
both expressions on either side of the
equal sign are numbers.
B plus C is a vector.
A is a vector.
When you dot two vectors, you get a number.
Finally, a scalar multiple of a vector dotted with another
vector has the property that you can leave the scalar
multiple outside.
In other words, you can first dot the two vectors and then
multiply by the scalar.
In other words, in a way, you don't have to worry about
voice inflection when you have a scalar multiple.
And we will talk more about these as we go along.
However, what's very crucial is to notice that the dot
product does have some difficulties not associated
with ordinary multiplication.
So I say, beware.
For example, somebody might say, I wonder if the dot
product is associative.
I wonder if A dot B dotted with C is the same as A dotted
with B dotted with C. And this is nonsensical.
I wanted to do this with great dramatic gesture, but I
probably would have broken two fingers
against the board here.
Let's just cross this out, so that you won't be inclined to
remember that.
This not only is false, it's stupid.
And the reason that it's stupid is that it's
nonsensical, that these things don't make sense.
Namely, the dot product has been defined to be an
operation between two vectors that yields a number.
Notice that as soon as you dot A and B, you no
longer have a vector.
You have a number.
And you cannot dot a number with a vector.
In other words, neither (A dot B) dot C nor A dot (B dot C)
is defined.
Because you see, a number is never dotted with a vector.
All right?
And finally, a closing note, as we've already seen, if A is
perpendicular to B, then the cosine of the angle between A
and B is 0.
And that says that if A is perpendicular to B,
then A dot B is 0.
In fact, this is one of the most common usages on the
elementary level, of the dot product, is to prove that two
vectors are perpendicular.
But whereas that's a nice property, what causes great
hardship here is to notice the following, in particular, that
if A dot B is 0, we cannot conclude that either A is the
0 vector or B is the 0 vector.
For example, i dot j is 0.
But neither i nor j is the 0 vector.
You see, with ordinary arithmetic, we had the
cancellation rule, things that said if the product of two
numbers is 0, at least one of the factors must be 0.
With the dot product, this need not be true.
The thing I want you to get from this lesson more than
anything else, other than the applications that you can get
from the book and from the exercises, is to learn to get
a feeling for the structure.
Don't be upset that certain vector properties are
different than arithmetic properties, and certain ones
are the same.
Notice in terms of the game, we take our rules as they may
apply and just carry them out towards inescapable
conclusions.
But I think that will become clearer as you read the text
and do the exercises.
And until next time when we'll talk about a new vector
product, let's just say, so long.
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