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Visualizing a projection onto a plane

Visualizing a projection onto a plane. Showing that the old and new definitions of projections aren't that different. Created by Sal Khan.

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  • leafers seed style avatar for user michaelthemba93
    Thank you for the video it was very helpfull.

    what I fail to understand is that is it mathematically correct to say that the projection of a vector V into another W is the length of the component of V in the direction of W?
    (2 votes)
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  • blobby green style avatar for user Yanyi
    Is the vector X parallel to plane V? I'm getting confused here..
    (2 votes)
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    • leaf blue style avatar for user Richard
      No. Vector x passes through the origin and heads up out of the plane V at perhaps a 40 degree angle as drawn. Don't feel bad. Understanding anything in 3-space from a drawing that of course has to be flat like your computer screen is difficult. In this case I should also add that what I called the origin should be thought of as the zero vector.
      (1 vote)
  • blobby green style avatar for user zhigangxu2007
    What do you mean by orthogonal to everything in L? I think there is only one thing (one vector) in L which the residual vector can be orthogonal to.
    (2 votes)
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  • leafers seed style avatar for user twilightsoutcast
    For a question relating orthogonal projection and linear regression, why does the orthogonal projection of the training data into the column space of X minimize the total error in the predictions of the model?
    (2 votes)
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  • blobby green style avatar for user dimi.proios
    sorry I might have a misunderstanding on a detail the schema of on the vector substraction.
    the Vector: x - projection_of_X_in_L
    shouldn't it start from the x beginning and end in the tip of -projection_of_X_in_L ?
    It seemd like it is starting from the tip of X and finishes on the tip of projection_of_X_in_L .
    What do I get wrong ? :/
    ps thanks for tha amazing videos
    (2 votes)
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    • ohnoes default style avatar for user Caleb Clark
      The diagram in the video is correct. Since it is x MINUS proj_L(x), this is why.

      If the vector were x PLUS proj_L(x), and the vectors were placed with the beginning of proj_L(x) at the end of the vector x (think of proj_L(x) as just another vector, maybe u), then the sum of the vectors would indeed be from the beginning of x to the end of the projection.

      Since we have x - proj_L(x), we could think about flipping the projection vector 180 degrees from its tip (to turn it into the negative version of itself, so we can add it to x), and then placing this at the top of the x- vector. Then moving from the start of x to the tip of the (flipped) projection, we see that this movement is identical to the yellow vector (x-proj_L(x)) in the diagram.

      In general, a trick that I use (but always good to be comfortable with the basic method) is if I have vectors a and b, both starting at a common origin (like we have in this diagram), then the vector b-a will move from the tip of a to the tip of b. (You can check that this is what we have in the diagram).

      Hope this helped!
      (1 vote)
  • blobby green style avatar for user steenharsted
    Thank you for the videos! When I watch them it all makes sense, but I am having some problems with translating this sense into actual problem solving.

    My problem (That I wish to both compute and understand the solution of) is i´m having four vectors in R3. I use two of the vectors to create a plane and the normal vector for that plane.
    Now I want to project the remaining two vectors onto my newly created plane and calculate the angle between the projected vectors.
    I feel like this should be possible using what I have learned in the videos so far, but I cant figure out what the coordinates of the projected lines are, and without the coordinates I dont know how to compute the angle between the projected vectors.

    I guess there is no easy explanation to my problem, but can you direct me to some videos where I can learn this and some assignments where I can practice what I learn?

    Thanks in advance.
    (1 vote)
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    • spunky sam red style avatar for user Bernard Field
      I think I have a way to find the projections onto the plane.
      You know the normal vector to the plane. Therefore, you know the orthogonal complement to the plane, which is a line. We already know how to project vectors onto a line.
      We also know that an arbitrary vector x can be expressed as a unique sum of a vector in a subspace v and a vector in its orthogonal complement w.
      You are given x. You can find w. Therefore, you can find v, the projection of x onto the plane.
      You should be able to work from there.
      (2 votes)
  • leafers seedling style avatar for user Sukriti
    Hello,
    Can someone explain me in 3D geometry most basic version?? For example, if say two points X and Y are given and are joined to form a line..(3 D line). Then how we know the length of the projection of the line segment XY in the plane, say x + y + z=d ?
    {Unfortunately, I could understand some part of the video}
    (1 vote)
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    • leaf green style avatar for user kubleeka
      What you've described is actually a relatively intensive problem, but let's give it a shot.

      We have two arbitrary points in space, (p₁, q₁, r₁) and (p₂, q₂, r₂), and an arbitrary plane, ax+by+cz=d. We want the distance between the projections of these points into this plane.

      A strategy might look like this:
      1) Find the normal vector to the plane
      2) Find equations of lines perpendicular to this plane through the given points.
      3) Find the intersections of these lines with our plane (these are the projected points)
      4) Compute the distance between them.
      (2 votes)
  • blobby green style avatar for user a.somjp
    this might sound pedantic, but we should say that x - projvx is also unique?
    (1 vote)
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  • piceratops seed style avatar for user iashvilinick
    Just curious: When we see a plane and an orthogonal line in R3, how can we tell if those are rowspace and nullspace of a 3x3 transformation, or of a 2x3 transformation?
    (1 vote)
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  • ohnoes default style avatar for user matrix
    In projection formula, (x.v)/(v.v), what if x is bigger than v?
    (1 vote)
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Video transcript

I'm going to do one more video where we compare old and new definitions of a projection. Our old definition of a projection onto some line, l, of the vector, x, is the vector in l, or that's a member of l, such that x minus that vector, minus the projection onto l of x, is orthogonal to l. So the visualization is, if you have your line l like this, that is your line l right there. And then you have some other vector x that we're take the projection of it on to l. So that's x. The projection of x onto l, this thing right here, is going to be some vector in l. Such that when I take the difference between x and that vector, it's going to be orthogonal to l. So it's going to be some vector in l. This was our old definition when we took the projection onto a line. Some vector in l. Maybe it's there. And if I take the difference between that and that, this difference vector's going to be orthogonal to everything in l. Just like that. So, this right here would be it's difference vector. That would be x minus the projection of x onto l. And then, of course, this vector right here. This is the one we were defining it. That was the projection onto l of x. Now, what's a different way that we could have written this? We could have written this exact same definition. We could have said it is the vector in l such that-- so we could say, let me write it here in purple. Is the vector v in l such that v-- let me write it this way-- such that x minus v, right? x minus the projection of l is orthogonal is equal to w, which is orthogonal to everything in l. Being orthogonal to l literally means being orthogonal to every vector in l. So I just rewrote it a little bit different, instead of just leaving it as a projection of x onto l. I said hey, that's some vector, v, in l such that x minus v is equal to some other vector, w, which is orthogonal to everything in l. Or we can rewrite that statement right there as x is equal to v plus w. So we can just say that the projection of x onto l is the unique vector v in l, such that x is equal to v plus w, where w is a unique vector-- I mean it is going to be unique vector-- in the orthogonal complement of l. Right? This is got to be orthogonal to everything in l. So that's going to be a member of the orthogonal complement of l. So this definition is actually completely consistent with our new subspace definition. And we could just extend it to arbitrary subspaces, not just lines. Let me help you visualize that. So let's say we're dealing with R3 right here. And I've got some subspace in R3. And let's say that subspace happens to be a plane. I'm going to make it a plane just so that it becomes clear that we don't have to take projections just onto lines. So this is my subspace v right there. Let me draw its orthogonal compliment. Let's say its orthogonal complement looks something like that. Let's say it's a line. And then it goes-- it intersects right there. Then it goes back. And, of course, it would have to intersect at the 0 vector. That's the only place where a subspace and its orthogonal complement overlap. And then it goes behind and you see it again. Obviously you wouldn't be able to you again because this plane would extend in every direction. But you get the idea. So this right here is the orthogonal complement of v, that line. Now, let's have some other arbitrary vector in R3 here. So let's say I have some vector that looks like that. Let's say that that is x. Now our new definition for the projection of x onto v is equal to the unique vector v. This is a vector v. That's a subspace v. The unique vector v, that is a member of v, such that x is equal to v plus w, where w is a unique member of the orthogonal complement of v. This is our new definition. So, if we say x is equal to some member of v and some member of its orthogonal complement-- we can visually understand that here. We could say, OK it's going to be equal to, on v, it'll be equal to that vector to right there. And then on v's orthogonal complement, you add that to it. So, if you were to shift it over, you would get that vector, just like that. This right here is v. That right there is v. And then this is vector that goes up like this, out of the plane, orthogonal to the plane, is w. You could see if you take v plus w, you're going to get x. And you could see that v is the projection onto the subspace capital v-- so this is a vector, v-- is the projection onto the subspace capital V of the vector x. So the analogy to a shadow still holds. If you imagine kind of a light source coming straight down onto our subspace, kind of orthogonal to our subspace, the projection onto our subspace is kind of the shadow of our vector x. Hopefully that help you visualize it a little better. But what we're doing here is we're going to generalize it. Earlier in this video I showed you a line. This is a plane. But we can generalize it to any subspace. This is in R3. We can generalize it to Rn, to R100. And that's really the power of what we're doing here. It's easy to visualize it here, but it's not so easy to visualize it once you get to higher dimensions. And actually, one other thing. Let me show that this new definition is pretty much almost identical to exactly what we did with lines. This is identical to saying that the projection onto the subspace x is equal to some unique vector in V such that x minus the projection onto v of x is orthogonal to every member of V. Because this statement, right here, is saying any vector that's orthogonal to any member of v says that it's a member of the orthogonal complement of v. So that statement could be written as x minus the projection onto v of x is a member of v's orthogonal complement. Or we could call some w. So if you call this your v, and if you call this whole thing your w, you get this exact definition right there. You would have w is equal to x minus v. And then if you add v to both sides, you get w plus v is equal to x. We defined v to be, the orthogonal-- the projection of x onto v. w is a member of our orthogonal complement. And I don't want you to get confused. The vector v is the orthogonal projection of our vector x onto the subspace capital V. I probably should use different letters instead of using a lowercase and a uppercase v. It makes the language a little difficult. But I just wanted to give you another video to give you a visualization of projections onto subspaces other than lines. And to show you that our old definition, with just a projection onto a line which was a linear transformation, is essentially equivalent to this new definition. On the next video, I'll show you that this, for any subspace is, indeed, a linear transformation.