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Parametric representations of lines

Parametric Representations of Lines in R2 and R3. Created by Sal Khan.

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  • leaf blue style avatar for user chandra.bhat
    How would you use this in real life?
    (113 votes)
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  • blobby green style avatar for user Sowmya Reddy
    How can a line be represented in 50 dimensions?What are its real-time applications?I mean, what is the need of a line represented in 50 dimensions.
    (37 votes)
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    • orange juice squid orange style avatar for user Jason Malan
      You can't graphically represent a line in 50 dimensions, because we only have three spacial dimensions to work with. A line in 50 dimensions would just be a representation of a set of values. Think of it, like this: In two dimensions I can solve for a specific point on a function or I can represent the function itself via an equation (i.e. a line). In three dimensions I can represent a point on a function or a line of a function or the function itself (a plane). So, in 50 dimensions a line would represent a range of specific values of a function.

      Real-world applications would involve 50 different variables, e.g.: quantity of flavours of ice-cream (Strawberry, Chocolate, Vanilla, Blueberry, ... etc.) and the cost of an amount of a specific flavour. Then a line could be a representation of an optimal set of values for the quantity of ice-cream for each flavour at the lowest cost.

      I hope that helps. (Please correct me if I'm wrong)
      (124 votes)
  • piceratops ultimate style avatar for user Dr. Math
    What is collinear?
    (28 votes)
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  • leaf green style avatar for user Ivan Nyemtsev
    (13m:43s) Can we instead of doing b-a do a-b?
    (32 votes)
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  • blobby green style avatar for user Dylan.Nertavich
    When determining the equation for L he says that it doesn't matter whether or not you were to use P1+t[P2-P1] or P2+t[P1-P2] or any other variation, or this is at least how I understood it. How is this possible if each variation provides you with different parametric equations? I attempted to apply this method to a problem from my calc txtbook and i found that the answer from the back of the book only corresponds with one of the forms of the equation.
    (17 votes)
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    • leafers ultimate style avatar for user GFauxPas
      Does your textbook say to find "a" set of parametric equations, or "the" set? Parametrizations for a given line are not unique, so your book should say "answers may vary", or something. Or, the problem would have to give you more information than the two points. If you look at t as a time parameter (you don't have to), you can think of different parametrizations of the same line as two particles going perhaps in opposite directions, or different speeds, but along the same path.
      (15 votes)
  • leaf green style avatar for user Chris Forsyth
    How would someone reading the set S = {c v | c ( R} know that it represents only the line in R^2 going through the point (0, 0)?

    Earlier on, we defined vectors such that v can be started anywhere on the Cartesian plane , not just the origin. Therefore, the set S should represent all lines in R^2 with the same slope as vector v .

    I know that at , Sal makes the clarification that this is only if we define v as the position vector version of v, which starts only at the origin. But when writing the set S, is there a way (a notation of some sort) to clarify that c v represents only the set of all scaled position vectors v , rather than just all v ? Because otherwise, adding the x vector later on would make literally no difference to S. You'd just be taking an infinite number of lines in R^2 and moving all of them by x, which would have no visible effect.
    (14 votes)
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    • primosaur ultimate style avatar for user Derek M.
      In Linear Algebra, we only consider a vector as an object referenced from the origin. In other words, we cannot move vectors wherever we want in Linear Algebra.
      However such a practice of allowing a vector v to be anywhere we want is acceptable in a class like Multivariable Calculus.
      (8 votes)
  • leaf green style avatar for user Nathan Clingan
    Wait a minute...could you actually describe the path of a fly with a parametric equation, if the path were not a straight line? Or was he just referring to straight lines?
    (11 votes)
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    • aqualine seed style avatar for user Bobak McCann
      I think the confusion with the non-straight lines (or at least my confusion with it) derived from the fact that in the video, sal was showing us parameterization with just a straight line. The functions in this video always produced straight lines because they were of the form mx + b. It does not have to stay in that form though. You could make a function like x= t^3 + 3, y= t^2 , z= t^.5 + 2t. Here is a graph of that function I just stated to visualize it: http://www.wolframalpha.com/input/?i=plot+%7Bx+%3Dt%5E3+%2B+3%2C+y%3Dt%5E2%2C+z%3Dt%5E.5+%2B+2t%7D
      (6 votes)
  • blobby green style avatar for user DarkAbyss175
    What is an example of higher dimensional problems in real life?
    (5 votes)
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    • leaf red style avatar for user Bob Fred
      probably one of the simplest cases is where you want to associate some sort of data with a 3D object, like a mountain and the elevation at every point. one way you might show this is by adding a 4th dimension (height) and represent it by coloring the mountain. AKA topography.
      (1 vote)
  • purple pi purple style avatar for user Trurl
    This may be a silly question, but at why B-A? Is not A+B the same vector? What am I missing?
    (4 votes)
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    • old spice man green style avatar for user newbarker
      A + B is not the same as B - A. You can see this clearly with the following example: Let A = [1,2], and B = [3,4]. B - A is [2,2] and A + B is [4,6].

      A + B is the vector you get by drawing A, then drawing B with B's tail at the head/tip/front of A. My suggestion is to draw some actual vectors on some axes and give it a go.

      A + B is easy to see, but B - A isn't as easy. Another way to think of it (hope this is not more confusing!) is that to get from A to B you need to add another vector. What is that vector? It is B - A. Let's see that in algebra:

      A + (B - A)
      = A + B - A
      = B
      (9 votes)
  • mr pink red style avatar for user Andy
    At , wouldn't all the vectors in set L have arrows that point to every point on the blue line, not actually being part of the blue line? The arrows start at the origin, they don't start on the point (2,4).
    (5 votes)
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Video transcript

Everything we've been doing in linear algebra so far, you might be thinking, it's kind of a more painful way of doing things that you already knew how to do. You've already dealt with vectors. I'm guessing that some of you all have already dealt with vectors in your calculus or your pre-calculus or your physics classes. But in this video I hope to show you something that you're going to do in linear algebra that you've never done before, and that it would have been very hard to do had you not been exposed to these videos. Well I'm going to start with, once again, a different way of doing something you already know how to do. So let me just define some vector here, instead of making them bold, I'll just draw it with the arrow on top. I'm going to define my vector to be-- I can do with the arrow on top or I can just make it super bold. I'm just going to define my vectors, it's going to be a vector in R2. Let's just say that my vector is the vector 2, 1. If I were to draw it in standard position, it looks like this. You go two to the right, up one, like that. That's my, right there, that is my vector v. Now, if I were to ask you, what are all of the possible vectors I can create? So let me define a set. Let me define a set, s, and it's equal to-- all of the vectors I can create, if I were to multiply v times some constant, so I multiply some constant, some scalar, times my vector v, and just to maybe be a little bit formal, I'll say such that c is a member of the real numbers. Now what would be a graphical representation of this set? Well, if we draw them all in standard position, c could be any real number. So if I were to multiply, c could be 2. If c is 2, let me do it this way. If I do 2 times our vector, I'm going to get the vector 4, 2. Let me draw that in standard position, 4, 2. It's right there. It's this vector right there. It's collinear with this first vector. It's along the same line, but just goes out further 2. Now I could've done another. I could have done 1.5 times our vector v. Let me do that in a different color. And maybe that would be, that would be what? That'd be 1.5 times 2, which is 3, 1.5. Where would that vector get me? I'd go one 1.5 and then I'd go 3, and then 1.5, I'd get right there. And I can multiply by anything. I can multiply 1.4999 times vector v, and get right over here. I could do minus 0.0001 times vector v. Let me write that down. I could do 0.001 times our vector v. And where were that put me? It would put me little super small vector right there. If I did minus 0.01, it would make a super small vector right there pointing in that direction. If I were to do minus 10, I would get a vector going in this direction that goes way like that. But you can imagine that if I were to plot all of the vectors in standard position, all of them that could be represented by any c in real numbers, I'll essentially get-- I'll end up drawing a bunch of vectors where their arrows are all lined up along this line right there, and all lined up in even in negative direction-- let me make sure I draw it properly-- along that line, like that. I think you get the idea. So it's a set of collinear vectors. So let me write that down. And if we view these vectors as position vectors, that this vector represents a point in space in R2-- this R2 is just our Cartesian coordinate plane right here in every direction-- if we view this vector as a position vector-- let me write that down-- if we view it as kind of a coordinate in R2, then this set, if we visually represent it as a bunch of position vectors, it'll be represented by this whole line over here. And I want to make that point clear because it's essentially a line, of slope 2. Right? Sorry, slope 1/2. Your rise is 1. Your rise is 1 for going over 2. But I don't want to go back to our Algebra 1 notation too much. But I want to make this point that this line of slope 2 that goes through the origin, this is if we draw all of the vectors in the set as in their standard form, or if we draw them all as position vectors. If I didn't make that clarification, or that qualification, I could have drawn these vectors anywhere. Right? Because this 4, 2 vector, I could have drawn over here. And then, to say that it's collinear probably wouldn't have made as much visual sense to you. But I think this collinearity of it makes more sense to you if you say, oh, let's draw them all in standard form. All of them start at the origin, and then their tails are at the origin, and their heads go essentially to the coordinate they represent. That's what I mean by their position vectors. They don't necessarily have to be position vectors, but for the visualization in this video, let's stick to that. Now I was only able to represent something that goes through the origin with this slope. So you can almost view that this vector kind of represented its slope. You almost want to view it as a slope vector, if you wanted to tie it in to what you learned in Algebra 1. What if we wanted to represent other lines that had that slope? What if we wanted to represent the the same line, or I guess a parallel line-- that goes through that point over there, the point 2 comma 4? Or if we're thinking in position vectors, we could say that point is represented by the vector, and we will call that x. It's represented by the vector x. And the vector x is equal to 2, 4. That point right there. What if I want to represent the line that's parallel to this that goes through that point 2, 4? So I want to represent this line right here. I'll draw it as parallel to this as I can. I think you get the idea, and it just keeps going like that in every direction. These two lines are parallel. How can I represent the set of all of these vectors, drawn in standard form, or all of the vectors, that if I were to draw them in standard form, would show this line? Well, you can think about it this way. If every one of the vectors that represented this line, if I start with any vector that was on this line, and I add my x vector to it, I'll show up at a corresponding point on this line that I want to be at. Right? Let's say I do negative 2 times my original, so minus 2 times my vector v, that equaled what? Minus 4, minus 2, so that's that vector there. But if I were to add x to it, if I were to add my x vector. So if I were to do minus 2 times my vector v, but I were to add x to it, so plus x. I'm adding this vector 2 comma 4 to it, so from here I'd go right 2 and up 4, so I'd go here. Or visually you could just say, heads to tails, so I would go right there. So I would end up at a corresponding point over there. So when I define my set, s, as the set of all points where I just multiply v times the scalar, I got this thing that went through the origin. But now let me define another set. Let me define a set l, maybe l for line, that's equal to the set of all of vectors where the vector x, I could do it bold or I'll just draw an arrow on it, plus some scalar-- I could use c, but let me use t, because I'm going to call this a parametrization of the line-- so plus some scalar, t times my vector v such that t could be any member of the real numbers. So what is this going to be? This is going to be this blue line. If I were to draw all of these vectors in standard position, I'm going to get my blue line. For example, if I do minus 2, this is minus 2, times my vector v, I get here. Then if I add x, I go there. So this vector right here that has its endpoint right there-- its endpoint sits on that line. I can do that with anything. If I take this vector, this is some scalar times my vector v, and I add x to it, I end up with this vector, whose endpoint, if I view it as a position vector, it's endpoint dictates some coordinate in the xy plane. So it will [UNINTELLIGIBLE] that point. So I can get to any of these vectors. This is a set of vectors right here, and all of these vectors are going to point-- they're essentially going to point to something-- when I draw them in standard form, if I draw them in standard form-- they're going to point to a point on that blue line. Now you might say, hey Sal, this was a really obtuse way of defining a line. I mean we do it in Algebra 1, where we just say, hey you know, y is equal to mx plus b. And we figure out the slope by figuring out the difference of two points, and then we do a little substitution. And this is stuff you learned in seventh or eighth grade. This was really straightforward. Why am I defining this obtuse set here and making you think in terms of sets and vectors and adding vectors? And the reason is, is because this is very general. This worked well in R2. So in R2, this was great. I mean, we just have to worry about x's and y's. But what about the situation, I mean notice, in your algebra class, your teacher never really told you much, at least in the ones I took, about how do you present lines in three dimensions? Maybe some classes go there, but they definitely didn't tell you how do you represent lines in four dimensions, or a hundred dimensions. And that's what this is going to do for us. Right here, I defined x and v as vectors in R2. They're two-dimensional vectors, but we can extend it to an arbitrary number of dimensions. So just to kind of hit the point home, let's do one more example in R2, where, it's kind of the classic algebra problem where you need to find the equation for the line. But here, we're going to call it the set definition for the line. Let's say we have two vectors. Let's say we have the vector a, which I'll define as-- let me just says it's 2, 1. So if I were draw it in standard form, it's 2, 1. That's my vector a, right there. And let's say I have vector b, let me define vector b. I'm going to define it as, I don't know, let me define it as 0, 3. So my vector b, 0-- I don't move to the right at all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could almost say that both of those vectors lie on-- I guess that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how to do that. We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. We have to be careful. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. Right? It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which is this line here, and we have to shift it. And we could shift it either by shifting it straight up, we could add vector b to it. So we could take this line right here, and add vector b to it. And so any point on here would have its corresponding point there. So when you add vector b, it essentially shifts it up. That would work. So we could, say, we could add vector b to it. And now all of these points for any arbitrary-- t is a member of the real numbers, will lie on this green line. Or the other option we could have done is we could have added vector a. Vector a would have taken any arbitrary point here and shifted it that way. Right? You would be adding vector a to it. But either way, you're going to get to the green line that we cared about, so you could have also defined it as the set of vector a plus this line, essentially, t times vector b minus a, where t is a member of the reals. So my definition of my line could be either of these things. The definition of my line could be this set, or it could be this set. And all of this seems all very abstract, but when you actually deal with the numbers, it actually becomes very simple. It becomes arguably simpler than what we did in Algebra 1. So this l, for these particular case of a and b, let's figure it out. My line is equal to-- let me just use the first example. It's vector b, so it's the vector 0, 3 plus t, times the vector b minus a. Well what's b minus a? 0 minus 2 is minus 2, 3, minus 1 is 2, for t is a member of the reals. Now, if this still seems kind of like a convoluted set definition for you, I could write it in terms that you might recognize better. If we want to plot points, if we call this the y-axis, and we call this the x-axis, and if we call this the x-coordinate, or maybe more properly that the x-coordinate and call this the y-coordinate, then we can set up an equation here. This actually is the x-slope. This is the x-coordinate, that's the y-coordinate. Or actually, even better, whatever-- actually, let me be very careful there. This is always going to end up becoming some vector, l1, l2. Right? This is a set of vectors, and any member of this set is going to look something like this. So this could be li. So, this is the x-coordinate, and this is the y-coordinate. And just to get this in a form that you recognize, so we're saying that l is the set of this vector x plus t times this vector b minus a here. If we wanted to write it in kind of a parametric form, we can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2 plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time, this was convoluted. You have to define these sets and all that. But now I'm going to show you something that you probably-- well, unless you have done this before, but I guess that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. It's going to be equal to-- we could just pick one of these guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1. 7 minus 4 is 3. So that thing is that vector. And so, our line can be described as a set of vectors, that if you were to plot it in standard position, it would be this set of position vectors. It would be P1, it would be-- let me do that in green-- it would be minus 1, 2, 7. I could've put P2 there, just as easily-- plus t times minus 1, minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is the z-axis. This is the x-axis, and let's say the y-axis. It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. And then finally, our z-coordinate is determined by that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in three dimensions, it has to be a parametric equation. Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically.