So let's start right away with
stuff that we will need to see before we can go on to more
advanced things. So, hopefully yesterday in
recitation, you heard a bit about vectors.
How many of you actually knew about vectors before that?
OK, that's the vast majority. If you are not one of those
people, well, hopefully you'll learn about
vectors right now. I'm sorry that the learning
curve will be a bit steeper for the first week.
But hopefully, you'll adjust fine.
If you have trouble with vectors, do go to your
recitation instructor's office hours for extra practice if you
feel the need to. You will see it's pretty easy.
So, just to remind you, a vector is a quantity that has
both a direction and a magnitude of length. So -- So, concretely the way
you draw a vector is by some arrow, like that,
OK? And so, it has a length,
and it's pointing in some direction.
And, so, now, the way that we compute things
with vectors, typically, as we introduce a
coordinate system. So, if we are in the plane,
x-y-axis, if we are in space, x-y-z axis.
So, usually I will try to draw my x-y-z axis consistently to
look like this. And then, I can represent my
vector in terms of its components along the coordinate
axis. So, that means when I have this
row, I can ask, how much does it go in the x
direction? How much does it go in the y
direction? How much does it go in the z
direction? And, so, let's call this a
vector A. So, it's more convention.
When we have a vector quantity, we put an arrow on top to
remind us that it's a vector. If it's in the textbook,
then sometimes it's in bold because it's easier to typeset.
If you've tried in your favorite word processor,
bold is easy and vectors are not easy.
So, the vector you can try to decompose terms of unit vectors
directed along the coordinate axis.
So, the convention is there is a vector that we call
***amp***lt;i***amp***gt; hat that points along the x
axis and has length one. There's a vector called
***amp***lt;j***amp***gt; hat that does the same along
the y axis, and the
***amp***lt;k***amp***gt; hat that does the same along
the z axis. And, so, we can express any
vector in terms of its components.
So, the other notation is ***amp***lt;a1,
a2, a3 ***amp***gt; between these square brackets.
Well, in angular brackets. So, the length of a vector we
denote by, if you want, it's the same notation as the
absolute value. So, that's going to be a
number, as we say, now, a scalar quantity.
OK, so, a scalar quantity is a usual numerical quantity as
opposed to a vector quantity. And, its direction is sometimes
called dir A, and that can be obtained just
by scaling the vector down to unit length,
for example, by dividing it by its length.
So -- Well, there's a lot of notation to be learned.
So, for example, if I have two points,
P and Q, then I can draw a vector from P to Q.
And, that vector is called vector PQ, OK?
So, maybe we'll call it A. But, a vector doesn't really
have, necessarily, a starting point and an ending
point. OK, so if I decide to start
here and I go by the same distance in the same direction,
this is also vector A. It's the same thing.
So, a lot of vectors we'll draw starting at the origin,
but we don't have to. So, let's just check and see
how things went in recitation. So, let's say that I give you
the vector ***amp***lt;3,2,1***amp***gt;.
And so, what do you think about the length of this vector?
OK, I see an answer forming. So, a lot of you are answering
the same thing. Maybe it shouldn't spoil it for
those who haven't given it yet. OK, I think the overwhelming
vote is in favor of answer number two.
I see some sixes, I don't know. That's a perfectly good answer,
too, but hopefully in a few minutes it won't be I don't know
anymore. So, let's see.
How do we find -- -- the length of a vector three,
two, one? Well, so, this vector,
A, it comes towards us along the x axis by three units.
It goes to the right along the y axis by two units,
and then it goes up by one unit along the z axis.
OK, so, it's pointing towards here.
That's pretty hard to draw. So, how do we get its length?
Well, maybe we can start with something easier,
the length of the vector in the plane.
So, observe that A is obtained from a vector,
B, in the plane. Say, B equals three (i) hat
plus two (j) hat. And then, we just have to,
still, go up by one unit, OK?
So, let me try to draw a picture in this vertical plane
that contains A and B. If I draw it in the vertical
plane, so, that's the Z axis,
that's not any particular axis, then my vector B will go here,
and my vector A will go above it.
And here, that's one unit. And, here I have a right angle.
So, I can use the Pythagorean theorem to find that length A^2
equals length B^2 plus one. Now, we are reduced to finding
the length of B. The length of B,
we can again find using the Pythagorean theorem in the XY
plane because here we have the right angle.
Here we have three units, and here we have two units.
OK, so, if you do the calculations,
you will see that, well, length of B is square
root of (3^2 2^2), that's 13.
So, the square root of 13 -- -- and length of A is square root
of length B^2 plus one (square it if you want) which is going
to be square root of 13 plus one is the square root of 14,
hence, answer number two which almost all of you gave.
OK, so the general formula, if you follow it with it,
in general if we have a vector with components a1,
a2, a3, then the length of A is the
square root of a1^2 plus a2^2 plus a3^2.
OK, any questions about that? Yes?
Yes. So, in general,
we indeed can consider vectors in abstract spaces that have any
number of coordinates. And that you have more
components. In this class,
we'll mostly see vectors with two or three components because
they are easier to draw, and because a lot of the math
that we'll see works exactly the same way whether you have three
variables or a million variables.
If we had a factor with more components, then we would have a
lot of trouble drawing it. But we could still define its
length in the same way, by summing the squares of the
components. So, I'm sorry to say that here,
multi-variable, multi will mean mostly two or
three. But, be assured that it works
just the same way if you have 10,000 variables.
Just, calculations are longer. OK, more questions?
So, what else can we do with vectors?
Well, another thing that I'm sure you know how to do with
vectors is to add them to scale them.
So, vector addition, so, if you have two vectors,
A and B, then you can form, their sum, A plus B.
How do we do that? Well, first,
I should tell you, vectors, they have this double
life. They are, at the same time,
geometric objects that we can draw like this in pictures,
and there are also computational objects that we
can represent by numbers. So, every question about
vectors will have two answers, one geometric,
and one numerical. OK, so let's start with the
geometric. So, let's say that I have two
vectors, A and B, given to me.
And, let's say that I thought of drawing them at the same
place to start with. Well, to take the sum,
what I should do is actually move B so that it starts at the
end of A, at the head of A. OK, so this is, again, vector B.
So, observe, this actually forms,
now, a parallelogram, right?
So, this side is, again, vector A.
And now, if we take the diagonal of that parallelogram,
this is what we call A plus B, OK, so, the idea being that to
move along A plus B, it's the same as to move first
along A and then along B, or, along B, then along A.
A plus B equals B plus A. OK, now, if we do it
numerically, then all you do is you just add
the first component of A with the first component of B,
the second with the second, and the third with the third.
OK, say that A was ***amp***lt;a1,
a2, a3***amp***gt; B was ***amp***lt;b1,
b2, b3***amp***gt;, then you just add this way.
OK, so it's pretty straightforward.
So, for example, I said that my vector over
there, its components are three, two, one.
But, I also wrote it as 3i 2j k. What does that mean?
OK, so I need to tell you first about multiplying by a scalar.
So, this is about addition. So, multiplication by a scalar,
it's very easy. If you have a vector,
A, then you can form a vector 2A just by making it go twice as
far in the same direction. Or, we can make half A more
modestly. We can even make minus A,
and so on. So now, you see,
if I do the calculation, 3i 2j k, well,
what does it mean? 3i is just going to go along
the x axis, but by distance of three instead of one.
And then, 2j goes two units along the y axis,
and k goes up by one unit. Well, if you add these
together, you will go from the origin, then along the x axis,
then parallel to the y axis, and then up.
And, you will end up, indeed, at the endpoint of a
vector. OK, any questions at this point?
Yes? Exactly.
To add vectors geometrically, you just put the head of the
first vector and the tail of the second vector in the same place.
And then, it's head to tail addition.
Any other questions? Yes?
That's correct. If you subtract two vectors,
that just means you add the opposite of a vector.
So, for example, if I wanted to do A minus B,
I would first go along A and then along minus B,
which would take me somewhere over there, OK?
So, A minus B, if you want,
would go from here to here. OK, so hopefully you've kind of
seen that stuff either before in your lives, or at least
yesterday. So, I'm going to use that as an
excuse to move quickly forward. So, now we are going to learn a
few more operations about vectors.
And, these operations will be useful to us when we start
trying to do a bit of geometry. So, of course,
you've all done some geometry. But, we are going to see that
geometry can be done using vectors.
And, in many ways, it's the right language for
that, and in particular when we learn
about functions we really will want to use vectors more than,
maybe, the other kind of geometry that you've seen
before. I mean, of course,
it's just a language in a way. I mean, we are just
reformulating things that you have seen, you already know
since childhood. But, you will see that notation
somehow helps to make it more straightforward.
So, what is dot product? Well, dot product as a way of
multiplying two vectors to get a number, a scalar.
And, well, let me start by giving you a definition in terms
of components. What we do, let's say that we
have a vector, A, with components a1,
a2, a3, vector B with components b1,
b2, b3. Well, we multiply the first
components by the first components, the second by the
second, the third by the third. If you have N components,
you keep going. And, you sum all of these
together. OK, and important:
this is a scalar. OK, you do not get a vector.
You get a number. I know it sounds completely
obvious from the definition here,
but in the middle of the action when you're going to do
complicated problems, it's sometimes easy to forget.
So, that's the definition. What is it good for?
Why would we ever want to do that?
That's kind of a strange operation.
So, probably to see what it's good for, I should first tell
you what it is geometrically. OK, so what does it do
geometrically? Well, what you do when you
multiply two vectors in this way,
I claim the answer is equal to the length of A times the length
of B times the cosine of the angle between them.
So, I have my vector, A, and if I have my vector, B,
and I have some angle between them,
I multiply the length of A times the length of B times the
cosine of that angle. So, that looks like a very
artificial operation. I mean, why would want to do
that complicated multiplication? Well, the basic answer is it
tells us at the same time about lengths and about angles.
And, the extra bonus thing is that it's very easy to compute
if you have components, see, that formula is actually
pretty easy. So, OK, maybe I should first
tell you, how do we get this from that?
Because, you know, in math, one tries to justify
everything to prove theorems. So, if you want,
that's the theorem. That's the first theorem in
18.02. So, how do we prove the theorem?
How do we check that this is, indeed, correct using this
definition? So, in more common language,
what does this geometric definition mean?
Well, the first thing it means, before we multiply two vectors,
let's start multiplying a vector with itself.
That's probably easier. So, if we multiply a vector,
A, with itself, using this dot product,
so, by the way, I should point out,
we put this dot here. That's why it's called dot
product. So, what this tells us is we
should get the same thing as multiplying the length of A with
itself, so, squared, times the cosine of the angle.
But now, the cosine of an angle, of zero,
cosine of zero you all know is one.
OK, so that's going to be length A^2.
Well, doesn't stand a chance of being true?
Well, let's see. If we do AdotA using this
formula, we will get a1^2 a2^2 a3^2.
That is, indeed, the square of the length.
So, check. That works.
OK, now, what about two different vectors?
Can we understand what this says, and how it relates to
that? So, let's say that I have two
different vectors, A and B, and I want to try to
understand what's going on. So, my claim is that we are
going to be able to understand the relation between this and
that in terms of the law of cosines.
So, the law of cosines is something that tells you about
the length of the third side in the triangle like this in terms
of these two sides, and the angle here.
OK, so the law of cosines, which hopefully you have seen
before, says that, so let me give a name to this
side. Let's call this side C,
and as a vector, C is A minus B.
It's minus B plus A. So, it's getting a bit
cluttered here. So, the law of cosines says
that the length of the third side in this triangle is equal
to length A2 plus length B2. Well, if I stopped here,
that would be Pythagoras, but I don't have a right angle.
So, I have a third term which is twice length A,
length B, cosine theta, OK?
Has everyone seen this formula sometime?
I hear some yeah's. I hear some no's.
Well, it's a fact about, I mean, you probably haven't
seen it with vectors, but it's a fact about the side
lengths in a triangle. And, well, let's say,
if you haven't seen it before, then this is going to be a
proof of the law of cosines if you believe this.
Otherwise, it's the other way around.
So, let's try to see how this relates to what I'm saying about
the dot product. So, I've been saying that
length C^2, that's the same thing as CdotC,
OK? That, we have checked.
Now, CdotC, well, C is A minus B.
So, it's A minus B, dot product,
A minus B. Now, what do we want to do in a
situation like that? Well, we want to expand this
into a sum of four terms. Are we allowed to do that?
Well, we have this dot product that's a mysterious new
operation. We don't really know.
Well, the answer is yes, we can do it.
You can check from this definition that it behaves in
the usual way in terms of expanding, vectoring,
and so on. So, I can write that as AdotA
minus AdotB minus BdotA plus BdotB.
So, AdotA is length A^2. Let me jump ahead to the last
term. BdotB is length B^2,
and then these two terms, well, they're the same.
You can check from the definition that AdotB and BdotA
are the same thing. Well, you see that this term,
I mean, this is the only difference between these two
formulas for the length of C. So, if you believe in the law
of cosines, then it tells you that, yes, this a proof that
AdotB equals length A length B cosine theta.
Or, vice versa, if you've never seen the law of
cosines, you are willing to believe this.
Then, this is the proof of the law of cosines.
So, the law of cosines, or this interpretation,
are equivalent to each other. OK, any questions?
Yes? So, in the second one there
isn't a cosine theta because I'm just expanding a dot product.
OK, so I'm just writing C equals A minus B,
and then I'm expanding this algebraically.
And then, I get to an answer that has an A.B.
So then, if I wanted to express that without a dot product,
then I would have to introduce a cosine.
And, I would get the same as that, OK?
So, yeah, if you want, the next step to recall the law
of cosines would be plug in this formula for AdotB.
And then you would have a cosine.
OK, let's keep going. OK, so what is this good for?
Now that we have a definition, we should figure out what we
can do with it. So, what are the applications
of dot product? Well, will this discover new
applications of dot product throughout the entire
semester,but let me tell you at least about those that are
readily visible. So, one is to compute lengths
and angles, especially angles. So, let's do an example.
Let's say that, for example,
I have in space, I have a point,
P, which is at (1,0,0). I have a point,
Q, which is at (0,1,0). So, it's at distance one here,
one here. And, I have a third point,
R at (0,0,2), so it's at height two.
And, let's say that I'm curious, and I'm wondering what
is the angle here? So, here I have a triangle in
space connect P, Q, and R, and I'm wondering,
what is this angle here? OK, so, of course,
one solution is to build a model and then go and measure
the angle. But, we can do better than that.
We can just find the angle using dot product.
So, how would we do that? Well, so, if we look at this
formula, we see, so, let's say that we want to
find the angle here. Well, let's look at the formula
for PQdotPR. Well, we said it should be
length PQ times length PR times the cosine of the angle,
OK? Now, what do we know,
and what do we not know? Well, certainly at this point
we don't know the cosine of the angle.
That's what we would like to find.
The lengths, certainly we can compute.
We know how to find these lengths.
And, this dot product we know how to compute because we have
an easy formula here. OK, so we can compute
everything else and then find theta.
So, I'll tell you what we will do is we will find theta -- --
in this way. We'll take the dot product of
PQ with PR, and then we'll divide by the lengths. OK, so let's see.
So, we said cosine theta is PQdotPR over length PQ length
PR. So, let's try to figure out
what this vector, PQ,
well, to go from P to Q, I should go minus one unit
along the x direction plus one unit along the y direction.
And, I'm not moving in the z direction.
So, to go from P to Q, I have to move by
***amp***lt;-1,1,0***amp***gt;. To go from P to R,
I go -1 along the x axis and 2 along the z axis.
So, PR, I claim, is this. OK, then, the lengths of these
vectors, well,(-1)^2 (1)^2 (0)^2, square root,
and then same thing with the other one.
OK, so, the denominator will become the square root of 2,
and there's a square root of 5. What about the numerator?
Well, so, remember, to do the dot product,
we multiply this by this, and that by that,
that by that. And, we add.
Minus 1 times minus 1 makes 1 plus 1 times 0,
that's 0. Zero times 2 is 0 again.
So, we will get 1 over square root of 10.
That's the cosine of the angle. And, of course if we want the
actual angle, well, we have to take a
calculator, find the inverse cosine, and you'll find it's
about 71.5°. Actually, we'll be using mostly
radians, but for today, that's certainly more speaking.
OK, any questions about that? No?
OK, so in particular, I should point out one thing
that's really neat about the answer.
I mean, we got this number. We don't really know what it
means exactly because it mixes together the lengths and the
angle. But, one thing that's
interesting here, it's the sign of the answer,
the fact that we got a positive number.
So, if you think about it, the lengths are always
positive. So, the sign of a dot product
is the same as a sign of cosine theta.
So, in fact, the sign of AdotB is going to
be positive if the angle is less than 90°.
So, that means geometrically, my two vectors are going more
or less in the same direction. They make an acute angle.
It's going to be zero if the angle is exactly 90°,
OK, because that's when the cosine will be zero.
And, it will be negative if the angle is more than 90°.
So, that means they go, however, in opposite
directions. So, that's basically one way to
think about what dot product measures.
It measures how much the two vectors are going along each
other. OK, and that actually leads us
to the next application. So, let's see,
did I have a number one there? Yes.
So, if I had a number one, I must have number two.
The second application is to detect orthogonality.
It's to figure out when two things are perpendicular.
OK, so orthogonality is just a complicated word from Greek to
say things are perpendicular. So, let's just take an example.
Let's say I give you the equation x 2y 3z = 0.
OK, so that defines a certain set of points in space,
and what do you think the set of solutions look like if I give
you this equation? So far I see one,
two, three answers, OK.
So, I see various competing answers, but,
yeah, I see a lot of people voting for answer number four.
I see also some I don't knows, and some other things.
But, the majority vote seems to be a plane.
And, indeed that's the correct answer.
So, how do we see that it's a plane? So, I should say,
this is the equation of a plane.
So, there's many ways to see that, and I'm not going to give
you all of them. But, here's one way to think
about it. So, let's think geometrically
about how to express this condition in terms of vectors.
So, let's take the origin O, by convention is the point
(0,0,0). And, let's take a point,
P, that will satisfy this equation on it,
so, at coordinates x, y, z.
So, what does this condition here mean?
Well, it means the following thing.
So, let's take the vector, OP. OK, so vector OP,
of course, has components x, y, z.
Now, we can think of this as actually a dot product between
OP and a mysterious vector that won't remain mysterious for very
long, namely, the vector one,
two, three. OK, so, this condition is the
same as OP.A equals zero, right?
If I take the dot product OPdotA I get x times one plus y
times two plus z times three. But now, what does it mean that
the dot product between OP and A is zero?
Well, it means that OP and A are perpendicular.
OK, so I have this vector, A. I'm not going to be able to
draw it realistically. Let's say it goes this way.
Then, a point, P, solves this equation exactly
when the vector from O to P is perpendicular to A.
And, I claim that defines a plane.
For example, if it helps you to see it,
take a vertical vector. What does it mean to be
perpendicular to the vertical vector?
It means you are horizontal. It's the horizontal plane.
Here, it's a plane that passes through the origin and is
perpendicular to this vector, A.
OK, so what we get is a plane through the origin perpendicular
to A. And, in general,
what you should remember is that two vectors have a dot
product equal to zero if and only if that's equivalent to the
cosine of the angle between them is zero.
That means the angle is 90°. That means A and B are
perpendicular. So, we have a very fast way of
checking whether two vectors are perpendicular.
So, one additional application I think we'll see actually
tomorrow is to find the components of a vector along a
certain direction. So, I claim we can use this
intuition I gave about dot product telling us how much to
vectors go in the same direction to actually give a precise
meaning to the notion of component for vector,
not just along the x, y, or z axis,
but along any direction in space.
So, I think I should probably stop here.
But, I will see you tomorrow at 2:00 here, and we'll learn more
about that and about cross products.
