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Projection of space curves onto general planes

  1. Nov 6, 2013 #1
    So I've encountered many "what is the projection of the space curve [itex]C[/itex] onto the [itex]xy[/itex]-plane?" type of problems, but I recently came across a "what is the project of the space curve [itex]C[/itex] onto this specific plane [itex]P[/itex]?" type of question and wasn't sure how to proceed. The internet didn't yield me answers so I haven't made much headway. The problem and my attempt at a solution is outlined below:

    1. The problem statement, all variables and given/known data

    Compute the projection of the curve [itex]\vec{\mathbf{r}}(t) = \left\langle \mathrm{cos\:}t, \mathrm{sin\:}t, t \right\rangle[/itex] onto the plane [itex]x + y + z = 0[/itex].

    2. Relevant equations

    I'm having trouble come up with an equation. I've tried drawing the relevant [itex]xy[/itex]-, [itex]yz[/itex]-, and [itex]xz[/itex]-plane projections and seeing where the curves intersect, but I know that these intersection points do NOT necessarily correspond to the projection of the given curve onto the given plane.

    3. The attempt at a solution

    See reasoning above. I really don't know how to do this for a non-standard plane and so I'm completely lost as to how to make headway. I haven't been able to find relevant information on the internet either through a similar problem for some reason.

    Can anybody help me out? If so, is there a way to do this for ANY plane [itex]P[/itex] and ANY space curve [itex]\vec{\mathbf{r}}(t)[/itex]? I feel like there should be yet Stewart's Multivariable Calculus yields nothing (at least the 5th edition doesn't) in this area. Thank you in advance.
     
  2. jcsd
  3. Nov 6, 2013 #2

    tiny-tim

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    Hi Antiderivative! :smile:

    Choose a basis for the plane.

    Then use that basis, and the normal, as a new set of coordinates.
     
  4. Nov 6, 2013 #3
    Hi tiny-tim! Hmm... okay so the normal vector to the plane is N(1,1,1), and the plane goes through A(1/√2,0,-1/√2). Using that, I guess I can try to find an orthonormal basis. Clearly A•N = 0, so A and N are normal. The other vector can be found by finding N x A presumably.

    I'm getting (-1/√2, √2, -1/√2). Since it's an orthonormal basis I'd have to normalize this to get (1/√6, √2/√3, -1/√6).

    Okay so my basis vectors are (1/√2, 0, -1/√2) and (1/√6, √2/√3, -1/√6).

    There's a way to convert the coordinate systems but I'm not sure I know how to do that. It involves a matrix of some kind presumably?
     
  5. Nov 6, 2013 #4
    Okay so I figured out that the three basis vectors in my new coordinate system would be

    [itex]\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)[/itex], [itex]\left( \frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}} \right)[/itex], and [itex]\left( -\frac{1}{\sqrt{6}},\frac{\sqrt{2}}{\sqrt{3}},-\frac{1}{\sqrt{6}} \right)[/itex].

    Graphing these on Wolfram Alpha helped me see that they help to create the coordinate system where the new [itex]xy[/itex]-plane is created by where the given plane exists.

    Should I find the new [itex]\hat{x},\hat{y},\hat{z}[/itex] in terms of the old [itex]x,y,z[/itex]? I feel like that's so much work for a deceptively simple problem. Or is it actually necessary?
     
  6. Nov 6, 2013 #5
    Okay just for completion I want to post that I officially figured it out.

    Using the change of basis, I ended up rewriting the old [itex]x,y,z[/itex] in terms of the new ones, and deriving an equation for this "flattened" helix on the plane:

    [itex]\left( \begin{array}{c}
    \hat{x} \\
    \hat{y} \\
    \hat{z} \end{array} \right) =
    \left( \begin{array}{c}
    \frac{2\mathrm{cos\:}t - \mathrm{sin\:}t - t}{3} \\
    \frac{2\mathrm{sin\:}t - \mathrm{cos\:}t - t}{3} \\
    \frac{2t - \mathrm{sin\:}t - \mathrm{cos\:}t}{3} \end{array} \right)[/itex]

    Graphing this in Mac's Grapher program or an equivalent software produces the attached diagram, which is what we're going for. Thanks tiny-tim for helping me visualize/understand the process!
     

    Attached Files:

    Last edited: Nov 6, 2013
  7. Nov 6, 2013 #6

    LCKurtz

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    You can do it without changing basis vectors. Let's say you have a point ##P_0## on the plane and a unit normal ##\hat n## pointing to the side that ##P_0## is on. Let your curve be ##\vec r(t)##. Then the projection curve is ##\vec p(t)=\vec r(t) - (\vec r(t)-P_0\cdot \hat n)\hat n##.

    I can give you more details later but have to run now.
     
  8. Nov 6, 2013 #7
    Yep I just used that method after the orthonormal one. In the end it's kind of the same thing because you'll end up getting v – (n•v/n•n) n, which is like the Gram-Schmidt process for creating an orthonormal basis. Analogous procedures! Geometric versus algebraic arguments I suppose.
     
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