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Perpendicular Lines
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0:02
Now that's great but ...
0:06
... we'd probably better compute something about it. Right? So that is ...
0:12
Well how would you find how to compute those walls? Well the wall and the direction of this flow. And I mentioned that they are perpendicular but how do you know? And can you see that the angle there is ninety degrees? Well it's close to ninety. But what if it was eighty nine degrees? You would never notice it. Right? So ... How to check that? And how to find those algebraically?
0:42
Well the idea is to understand better what does it mean ... ... to have direction of that wall? And what does it mean to have direction of the flow? What does it mean in terms of the formula? Algebraically. Because this is what we see, right? And this is what we have part-time computing at this moment. So how to compute this?
1:06
Well how would you characterize this wall thing? How would you explain that so that you explanation is hopefully computable?
1:18
So the wall is a line through the origin. And what's the property of that line? Student: [...]. So the vector you have ... ... at that point is away from the origin. Right? So ... The point you choose ... ... x, y has a vector assigned ... ... which is away from the origin. What does it mean to be away from the origin? How can we say that algebraically?
1:58
Well away from the origin is something about direction, right? So what we decide about this vector?
2:05
Er ... let's call it w. The vector assigned to that point. That vector is proportional to the vector x, y. Right? That's exactly what it means. So being away from the origin ... ... pointing out, away from the origin ...
2:26
... from the origin zero, zero means algebraically that this vector ...
2:36
[...] I denoted that by u comma v in the past. That vector is proportional ...
2:48
... to the vector x, y.
2:56
Now is that a good algebraic explanation? You [...] all know what it means to be proportional. Right? So let's explain that value. It means that the vector u comma v ... ... is equal to some lambda.
3:14
Or rather maybe some number l. Because that's now the same number as I used before. Some number l times
3:25
... the vector x, y. So that's exactly the condition we are looking for.
3:34
We are looking for all those points x, y ... ... where the vector assigned is equal to some number times x, y.
3:47
So algebraically we find ... ... all x, y such that ...
3:58
... the vector u, v at that point is equal to some constant l ... ... times the vector x, y ... ... where ... ... l is a constant.
4:16
OK. Well that's great. But what was u and v?
4:22
Is there any relation between u, v, l and x?
4:27
Well u and v are the coordinates of the vector determined by this matrix.
4:37
Right? So well let me take this matrix -- this specific matrix. Should I?
4:46
I probably ... Yeah, so let me take this specific example.
4:54
Now what should I write in terms of u and v? If I use this matrix?
5:02
Well this matrix means that ... ... my u and v are written in terms of x and y -- certain way, right? So u equals two x plus three y. Right? Those numbers are used as coefficients. And v equals three x minus two y.
5:28
So that means this matrix tells me that this side -- left-hand side -- should be ... ... it is actually ... ... two x plus three y comma ... ... three x minus two y. And that should be equal to l times ... Well let me bring l inside that vector. l times x comma l times y.
5:55
So I have a condition that I want to be satisfied ... ... and I want to solve for x and y.
6:04
Now what kind of problem do we see now? I have to find x and y -- two unknowns. And I have only one condition. And I also don't know this number l.
6:19
Isn't that kind of a problem? One condition, three unknowns. Well algebraically sort of a problem. But let's what it means for these two vectors to be equal?
6:33
What does it mean algebraically that this vector is equal to that vector? It means the first coordinate is equal to the first, and second to the second. So this one condition is actually ...
6:45
... that pair of conditions ...
6:49
... to be satisfied simultaneously. Two x plus three y should equal to l times x and at the same time ... ... tree x minus two y should be equal to l times y.
7:04
Well that's better. I have now two equations. Still three unknowns.
7:12
Well let me still try to solve it.
7:19
How do we solve this system?
7:26
Solve for one of the variables in the first, substitute in the second. Right? OK. So let me solve for ... ... y in the first.
7:37
So y if I leave it on this side ... ... everything goes to the right. So I have ... ... one third of ... ... l x minus two x.
7:54
And now I can factor x out. So it is l minus two divided by three times x.
8:04
And I can substitute that into the second equation. Now I have three x minus two y ...
8:14
... two thirds times l minus two x ... ... equals l times y. So it is ... ... one third times l times l minus two times x. All right. Am I making progress? Is the second equation easy to solve? Why is it easy to solve? What does it tell you?
8:45
You can divide by x. Are you allowed to divide by x? Can we always divide by x?
8:56
Unless x is ...
9:00
... zero. Right? You cannot divide by zero. So the second equation says that either x equals zero ... And in that case the equation is satisfied. Right? Zero equals zero -- easy. Or ...
9:16
... if x is not zero then you can divide by it. And you get something without x. Three minus two thirds times l minus two ... ... should be equal to l over three times l minus two.
9:35
Right? Now what kind of case is that -- x equals zero?
9:42
Well let's go back to the original question. I'm looking for all the points with coordinates x, y ... ... where the vector assigned is proportional ... ... to the point itself.
9:57
And the algebra suggests I should look at the point x equals zero. And also it says that if x equals zero then y equals zero. So it tells me one of the solutions is the origin zero, zero. If x equals zero then y equals zero. And I'm looking at zero, zero. [...] origin. Is that exciting? That the vector at the origin ... ... is proportional to the zero, zero vector? Well of course it is. So that's not exciting solution. So I'm going to disregard it.
10:42
And then I'm left with one equation on that number l. So let's look at that equation.
10:51
Let me expand all those things. So I'll multiply by three. It's nine minus ... ... two l ...
11:04
... plus four ... ... should be equal to l squared minus two l.
11:14
Oh, good things happen! Right? Don't you see ...
11:20
... something good happening here? Aren't you excited?
11:26
Yeah! Two l cancels! And now you don't need quadratic formula to solve it, right? l squared equals nine plus four -- thirteen. So now l equals ... ... plus minus square root of thirteen.
11:47
Student: [...].
11:59
You can try it. I [...].
12:06
So ...
12:09
All right. So let me put here a question mark.
12:17
I will come back to this later. The question is whether this was a coincidence ... ... or something we should expect in the future? But so far let's go further and enjoy this simplicity. Well the lambda is plus/minus square root of thirteen. Well ... If I go back to my question ... ... this is not even close to something I wanted. Right? I wanted x and y and I found lambda or l. How exciting! So what should I do now?
12:53
What should I do with this l?
12:57
Plug it back. Plug it right into this system, right? OK. So let's plug it back. Let me choose one of them -- l equals square root of thirteen -- positive one. I plug it back into this system. And what I see is ... ... two x plus three y equals ... ... square root of thirteen x. And also three x minus two y equals ... ... square root of thirteen y.
13:35
Oh, great! I have ...
13:39
... a system of two equations and only two unknowns. I should be able to solve it.
13:47
How do I solve it?
13:54
Divide by x ... Well I cannot divide by x here.
14:00
All right. I can bring things on the other side. So I bring two x to the other side. Three y equals ... ... x multiplied by square root of thirteen minus two.
14:21
And what about the second equation?
14:25
What does that tell? Well if I want to solve for y I can do that on the second equation. So if I bring y to the right ... ... I will have ...
14:40
... three x left ... ... divided by square root of thirteen plus two.
14:50
So I finally have ... ... y equals square root of thirteen minus two over three x. And the second says ... second equation says y equals ... ... three over square root of thirteen plus two x.
15:14
Isn't that a problem?
15:19
y equals that and simultaneously y equals that. How can that happen?
15:26
How can we possibly find x and y?
15:37
Are those inverse mutually? You know. Are those numbers related in any way? What's the relation?
15:51
What's the relation between square root of thirteen minus two over three and ... ... three over square root of thirteen plus two?
16:13
The relation is very simple. Those are the same numbers.
16:19
Let's multiply. Square root of thirteen minus two ...
16:27
... times square root of thirteen plus two.
16:33
Well is it equal to nine?
16:38
Isn't it?
16:40
We have difference of numbers times sum of numbers. Isn't it difference of squares? Square of that is thirteen. Square of two is four. That's nine.
16:58
All right. That's exciting. So what does that mean? Well [...] these two numbers are exactly the same. And the system is actually telling me exactly the same thing twice. Bla bla right?
17:17
Well I usually understand things from the first time. So let me keep just one statement -- y equals ... ... well let's speak this one -- square root of thirteen minus two over three x. So this is the only information I'm getting from the system.
17:38
So what is ... What is x and y that I want?
17:48
Well let's think about my expectation. Do I expect just one pair x and y? Do I expect ... ... by looking at this picture that there is only one point ... ... x, y where the vector is away from zero? Well of course not. [...]. So I should not expect to have a system of two equations producing just one ... ... point, one pair. And actually what that system produces is infinitely many pairs. It tells me that ... ... my x-coordinate and my y-coordinate should be related in exactly this way.
18:27
If I choose anything for x ... ... I can solve for y. For any choice of x ... ... on this picture ... ... I should find the point where ... ... the vector is away from zero ... ... having that x-coordinate. And that's exactly the way to find that point. So this is actually equation of a line, isn't it?
18:57
y equals ... ... m times x, right? So that's exactly the equation of that wall.
19:08
So this is the wall equation.
19:26
Is it?
19:29
Well yes because I chose l to be positive. Right? What does it mean physically that I choose l to be positive? It means that the vector I have is proportional to x, y and has the same direction. That's really away from the origin. So if I choose l positive ... ... it means this is really a wall where all vectors are away from the origin. But I still have another value -- negative square root of thirteen left.
20:02
So I can use that. And solve the system.
20:12
So if ... ... l equals negative square root of thirteen ... ... then I have a system of two equations. Two x plus three y equals negative square root of thirteen x. Three x minus two y equals negative square root of thirteen y. Now do we ... Should we expect something about this system? Looking back at the experience we have?
20:41
Well of course I can follow the same procedure -- I can solve for y in the first equation, I can solve for y in the second equation. Can you guess about ... ... those two things? Well those should be the same. So let me solve for y in this first. It's y equals ...
21:01
... something over three times x. And that something is ... ... negative square root of thirteen minus two. And well the second equation -- y equals ... ... if I bring it to the right ... ... I now have three over something times x. And that something is two minus square root of thirteen. Well again I have ... ... y equals, y equals -- again I have two equations of a line. Now are those equations of the same line? Well the only way to check is to check that these two numbers are the same.
21:39
Are those numbers the same?
21:42
Exactly the same for exactly the same reason. Those are actually very similar to these numbers -- these reciprocals. Or negative reciprocals. Negative reciprocals. Negative reciprocals. Does it tell you something?
21:59
This line is perpendicular to this one.
22:03
Remember I mentioned the ... ... [...] that the flow is perpendicular to the wall. Well that's exactly why. So the y equals ... ... negative square root of thirteen plus two over three x is the equation of the line. That is identifying the flow direction. So physically it's direction ...
22:33
... of the flow.
22:45
All right. So it is the same kind of computation ... ... should ... well can be done for any choice of those numbers. Right? Whenever you get a matrix ... ... write it as a sum of four. You look only at the two projections here at the liquid part. Again merge them together in one matrix. And then carry out this kind of computations. You will find those two lines. And there are some things that you have to expect about those. First is that ... ... you really have to get lines.
23:22
And those lines should be perpendicular to each other.
23:28
Now another thing ...
23:32
... happened right here.
23:35
In the process of solving for l. The linear part canceled. And this is exactly what you always have to expect ...
23:46
... in this process. This one is a check point if you like. If the linear part doesn't cancel, if your number l ... ... is a solution of quadratic equation ... ... which is not simply l squared equals something ... ... well you made mistake somewhere in the computations. So basically the algebraic consequence of that is that ... ... the two solutions for l -- and we always expect two solutions from quadratic equation, right? So the algebraic consequence is that those two solutions are always ...
24:22
... negative of each other.
24:26
Now what's the meaning of this? Let's think about physics behind this l. What's the meaning of l?
24:35
Well the meaning of l is actually ...
24:39
... the rate of expansion along this line.
24:45
Remember the meaning of this l. l is how much the vector is greater than is x, y vector. It's in the same spirit as we have here, you see? x comma y has the vector lambda x comma lambda y. And there ... ... x comma y has the vector l times x, l times y. So ... The meaning of l here is the rate of expansion along the wall. It is how fast everything is moving away from origin along this wall.
25:26
And the meaning of ... And that's the meaning of positive l. The meaning of negative l ... ... is the [...] coefficient along this line.
25:39
And that negative means that we are shrinking. But the central part here is that we are shrinking with exactly the same rate as we are expanding there.
25:56
And that should remind you of the fluid, of this incompressible stuff. Right? If you shrink it with the certain rate in one direction ... ... you expect it to expand with exactly the same rate ... ... not faster, not slower -- with exactly the same. So the fact that the solution of this equation ... ... the fact that the equation is very simple and solution is plus/minus something ... ... is exactly this incompressibility of liquid. It's algebraic interpretation of that.
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