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Orthogonal projection of 2 points onto a plane

  1. Dec 14, 2009 #1
    edit: This thread might need moved, sorry about that.

    Hi, I have ended up on this site a few times after searching various maths issues; it seems to have a good community so I am asking you good people for a little help understanding this.

    Tomorrow I have a semi-important maths exam, if I fail it I can resit in 2010 but I would of course rather not have to do that. I am studying game design, so I shall be needing maths a lot more soon.

    I missed the part that involved the mathematics of projection, so I need to figure this out fast.
    Reading the materials I understood fairly well until it bluntly proposed a problem without before giving a solution to a similar one, so I need a bit of help.

    The problem is as follows:
    Calculate the orthogonal projection of the points A(3,5,1) and B(-4,2,6) onto the plane -2x+4y-z=3 and determine the equation of the line joining the image points on the plane.

    I understand visually what is going on here, but I don't know where to start. The notes talk about the Par i/j/k matrices (none seem to apply here) and the homogeneous coordinate transformation matrix... now I could plug numbers in and see what I come up with, its not like I am going to run out of paper. But I am scared to get lost in this.

    Who can help walk me through this? It may take a few posts ^_^

    OK, I think I need to start with:

    vector n = (-2,4,-1)

    n being the normal
    brackets should be vertical... but I don't know how to format it that way

    Ok, I have got this far:


    correct so far? I filled in the values for p=r+tn
    Last edited: Dec 14, 2009
  2. jcsd
  3. Dec 14, 2009 #2


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    A normal vector to -2x+ 4y- z= 3 is <-2, 4, -1>. A line through A(3,5,1) in that direction (and so normal to the plane) is given by x= 3- 2t, y= 5+ 4t, z= 1- t. Put those into the equation of the plane to find the point where that line crosses the plane. Do the same with B(-4,2,6): x= -4- 2t, y= 2+ 4t, z= 6- 6.

  4. Dec 14, 2009 #3
    OK, i've found the co-ordinates of the projection of point A. But they seem odd to me.
    I have:
    x = 1/3
    y = 10 and 1/3
    z = -1/3

    I solved t to 4/3, does that seem right?
  5. Dec 14, 2009 #4


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    That's not what I get- what I get is even more peculiar. Show how you solved the equation, please.
  6. Dec 14, 2009 #5
    OK here we go.

    Calculate the orthogonal projection of points A(3,5,-1) and B(-4,2,6) onto the plane -2x+4y-z=3

    r = a + tb
    <X,Y,Z> = <3,5,-1> + t<-2,4,-1>

    X = 3 - 2t
    Y = 5 + 4t
    Z = -1 -t

    aX +bY +cZ = d
    -2(3 - 2t) + 4(5 + 4t) - ((-1) - t) = 3

    t ends up as -18/21 this time

    I calculated X of the projection of point A to be 9/7 - is it good this time?

    Now, can I use any of those results to calculate the projected point B? because there is a problem later on confusing me that asks "A camera is aligned such that it projects the point (1,1,2) to the origin by an orthographic projection. Determine the image of the point (-1,4,-3) under the same projection". Any hints for where to start this one?
  7. Dec 15, 2009 #6


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    So -6+ 4t+ 20+ 16t+ 1+ t= 3
    (4+16+1)t= 3+ 6- 20- 1
    21t= -12
    t= -12/21= -4/7

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