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Finding plane normal vector

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Let us have lines
    (x+3)/2 = (y+2)/3 = (z-6)/-4
    and
    (x-5)/1 = (y+1)/-4 = (z+4)/1

    Find a normal vector to the plane these two lines are on.

    3. The attempt at a solution

    I already know you can solve it by
    s1 x s2 || n ; where s1 = (2, 3, -4) and s2 = (1, -4, 1)
    but Im not interested in that atm.
    I thought of another solution that I thought would be correct, but as it turns out, it doesnt seem to be.

    What I did was use the dot product.
    Let n = (A+3; B+2; C-6), because we know (-3, -2, 6) to be on the plane the lines are on.
    Let us also find s3=s1+s2=(3, -1, -3), because we have 3 variables in the system, so we need 3 equations and s3 is also on the same plane as s1 and s2.

    As we know, for a vector on the plane and a normal vector
    n.s=0 (dot product)

    So
    n.s1=0
    n.s2=0
    n.s3=0

    Yet when solving the system I end up with
    11B-6C=-34
    11B-6C=14
    Which means I get no solutions out of it.
    Where did I astray from logic?

    Thanks in advance
     
  2. jcsd
  3. Apr 21, 2013 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    I'm not sure why you got no solutions, possibly an arithmetic error. You should get infinite solutions. You are correct that you have 3 unknowns but only two equations. But using the sum of the two vectors will not give you a linearly independent equation. Any solution to the first two will give you a solution to the third.

    Remember that saying a vector is orthogonal to the plane does not define that vector uniquely. It only defines its direction. Given n is normal so is 5n and -3n. So you can either impose a third condition in the form of choosing the magnitude (which should give you a unique solution up to sign) or pick one of the components (judiciously since you may find one is necessarily zero) say the x component to be 1. I typically begin solving and choose, if I can, a value to avoid fractions in the vector components. Then I can if I like normalize the result.

    [edit: P.S. It's great that you are exploring alternative methods and -so to speak- playing with the problem. That is how you really learn to understand the subject. Keep it up!]
     
  4. Apr 21, 2013 #3
    Ofcourse! And plus I had made an arithmetic error too, so I should have really thought this through. Thank you for your help!
     
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