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  • 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.

So let's start right away with stuff that we will need to see

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Lec 1|MIT 18.02 多変数微分積分, 2007年秋学期 (Lec 1 | MIT 18.02 Multivariable Calculus, Fall 2007)

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    單柏祥 に公開 2021 年 01 月 14 日
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