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Linear Algebra Plane Projection

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Let 2 be the plane containing the line, l:(x,y,z)= t(6,4,2)+ (3,-4,2) and the point Q(5, -7, 7). Let P be the point (-6, -12,5)
    a) Find the projection of QP onto 2.


    2. Relevant equations
    I know the projection eqn would be (( plane 2 dot QP)/ magnitude on QP) * components of QP


    3. The attempt at a solution

    I got the vector QP to be (-11, -5,-2). I'm just confused on have to find the eqn of plane 2
     
  2. jcsd
  3. Oct 24, 2011 #2

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    Hi lina29! :smile:


    I don't understand your relevant equation.
    What do you mean by "plane 2"?
    And what would the "magnitude on QP" be?



    This is what I would do.

    First find a normal vector n to plane 2, and then use the projection formula:
    [tex]\text{projection of QP on plane 2} = QP - {(QP · n)n \over ||n||^2}[/tex]

    To find the normal vector n, first you need 2 independent vectors in plane 2.
    You can calculate n for instance by taking their cross product.
     
  4. Oct 24, 2011 #3
    I meant plane 2 as the plane containing the line, l:(x,y,z)= t(6,4,2)+ (3,-4,2). The magnitude of QP would be the sqrt(150). How would I find 2 indpt vectors in plane 2?
     
  5. Oct 25, 2011 #4

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    Well, you used the text "plane 2" in your formula in a place where it should be a vector.
    But how is a plane a vector?
    Is it possible that your relevant equation is for a line instead of a plane?


    The first vector is the variable part in your line expression: (6,4,2).
    The second vector is the difference between the 2 points in the plane: Q and (3,-4,2).
     
  6. Oct 25, 2011 #5
    I used plane 2 since that's what I initally thought it would be. Obviously, a plane isn't a vector I just wasn't thinking of that when I wrote the equation :). So I should do the dot product of QP and the first vector for projection?
     
  7. Oct 25, 2011 #6

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    No, not the first vector, but the normal vector (n) which you have yet to calculate.

    TBH, I do not know which equations and concepts you know, and which you don't.
    Are you familiar with the cross product (as opposed to the dot product)?
    Or do you know perhaps how to define a plane by a Cartesian equation?
     
  8. Oct 25, 2011 #7
    So how would I go about calculating the normal vector?
     
  9. Oct 25, 2011 #8

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    Let's start with the 2 vectors that are in the plane:

    The first vector is (6,4,2).
    The second vector is Q(5, -7, 7) - (3,-4,2) = (2, -3, 5).

    You need a vector that is perpendicular to both these vectors.
    That is the vector that we will call "n".
    Being perpendicular means that the dot product is zero.

    So let's see if we can find a vector perpendicular to (6,4,2) shall we?
    Would (1,-1,-1) do the job?
     
  10. Oct 25, 2011 #9
    yes.
     
  11. Oct 25, 2011 #10

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    Soooooo....? :rolleyes:
     
  12. Oct 25, 2011 #11
    lol so I would use that vector for the projection?
     
  13. Oct 25, 2011 #12

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    Well, is it perpendicular to (2, -3, 5) as well?
     
  14. Oct 25, 2011 #13
    I got (-29/3, -19/3, -10/3) which was counted right. Thank you!
    I'm also asked to find the distance between P and l. Using the formula
    |ax0+by0+cz0+d|/ sqrt(a2+b2+c2).Where a,b,c,d are the variables of the line, and x,y,z are variables of the point P. Would the line I used be 6a+4b+2c=0
     
  15. Oct 25, 2011 #14

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    Oh, okay.

    Ah well, the line to be used has (a,b,c)=(6,4,2).
    What you wrote is actually a Cartesian plane equation with normal vector (6,4,2) that intersects the origin.
     
  16. Oct 25, 2011 #15
    so how would I find the formula?
     
  17. Oct 25, 2011 #16

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    Pick a point (x,y,z) on the line and fill it in, in ax+by+cz+d=0.
    From this you can deduce d.
    (Actually you get a plane that is perpendicular to the line and intersects it at the point (x,y,z).)

    Then you need to fill in (x0,y0,z0) which is P - (x,y,z).
     
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