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Orthogonal projection over an orthogonal subspace

  1. Jan 2, 2018 #1
    1. The problem statement, all variables and given/known data
    Being F = (1,1,-1), the orthogonal projection of (2,4,1) over the orthogonal subspace of F is:

    a) (1,2,3)
    b) (1/3, 7/3, 8/3)
    c) (1/3, 2/3, 8/3)
    d) (0,0,0)
    e) (1,1,1)

    The correct answer is B

    2. Relevant equations

    3. The attempt at a solution

    Using the orthogonal projection formula:

    (1,1,-1).(2,4,-1) * (1,1,-1) = 7/3 * (1,1,-1) = (7/3,7/3,-7/3)
    (1,1,-1).(1,1-1)

    I am obviously misunderstanding something, I thought that maybe orthogonal subspace ain't (1,1-1), but the answer makes me believe that I should have (1,1-1) in the formula.
     
    Last edited by a moderator: Jan 2, 2018
  2. jcsd
  3. Jan 2, 2018 #2
    Shouldn't that be (2,4,1).
     
  4. Jan 2, 2018 #3

    Mark44

    Staff: Mentor

    The problem statement is confusing to me. I think it's asking for the projection of <2, 4, 1> onto the the orthogonal subspace of F (i.e., the plane perpendicular to F). It might be helpful to think about the geometry here.
     
  5. Jan 7, 2018 #4
    In that case, the dot product between F and a random vector contained in the plane would be equal to zero? That means that F is the normal of the plane?

    But in the end, what he probably wants? The projection of F over a vector contained in this plane that is perpendicular to F?
     
  6. Jan 7, 2018 #5

    Mark44

    Staff: Mentor

    Yes, that's what it means to be the orthogonal subspace to F.

    It means to find the projection of <2, 4, 1> onto the plane that is orthogonal to (perpendicular to) F = <1, 1, -1>.

    Again, draw a picture.
     
  7. Jan 7, 2018 #6
    Well, I don't know how to draw it, so drawing doesn't help. I can imagine what I would like to draw if I had the skill, and all I see is that the projection is a vector contained in the plane, which means it should satisfy the plane equation.
     
  8. Jan 7, 2018 #7

    Mark44

    Staff: Mentor

    You can't do something like this?
    Projection.png
    Not labelled, but the projection of the vector onto the plane is the one heading off to the right and down.
     
  9. Jan 9, 2018 #8
    I do not understand. There is three vectors and three labels, how can it not be labelled? As far as I understand, it is the one with the label "Ortho F". I also do not understand what do you mean by "heading down". Doesn't it has to be contained in the plane? How can it be heading down, since it implies that it would cross the plane?
     
  10. Jan 9, 2018 #9

    Mark44

    Staff: Mentor

    What I've labelled "Orth F" is the orthogonal subspace of F; i.e. the plane.
    Here's another picture that explains things better.
    Projection2.png
    Anyway, what I asked before was why can't you draw something like this? It took me about 2 minutes to create the first drawing, and another minute to update it to the above drawing. Using pencil and paper, you could probably create a simple drawing like this in about the same amount of time.
    What I'm calling "Proj <2, 4, 1>" is the heavy line segment that is the projection of <2, 4, 1> onto the plane. This projection is perpendicular to F (= <1, 1, -1>) and lies in the plane. Using simple trig (including the dot product), you should be able to find the vector I'm showing with the heavy line.

    Clear?
     
  11. Jan 10, 2018 #10
    Here is. A simple typo. Should have realized that before.

    Because I had not thought about drawing a "rotated" square as my plane representation. And about the second picture, I also did not think about the dotted line.
     
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