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Find a nonzero vector normal to a plane

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a nonzero vector normal to the plane -5x -y +z +9 = 0



    2. Relevant equations



    3. The attempt at a solution
    so the direction of the vector would be (a,b,c) = (-5,-1,1)
    I'm not exactly sure what the final form should look like..
    is r = ri + tv on the right track? where i plug (-5,-1,1) in for v and for ri i can plug in any numbers that make -5x - 6 + z +9 = 0?
    If so I could theoretically plug in (1,4,0)
    so
    r= (1i + 4j + 0k) + t(-5i -1j +1k)
    r = (1-5t)i + (4-t)j + (t)k

    The internet program says i'm wrong.. halp?
     
  2. jcsd
  3. Sep 22, 2013 #2

    Mark44

    Staff: Mentor

    Your normal looks fine, but there are an infinite number of vectors that are perpendicular to (normal to) a given plane. However, all of the vectors are scalar multiples of one another. I suspect that the answer the program was looking for was <5, 1, -1>. Or maybe <1, .2, -.2>.

    Did it mention anything about wanting a unit vector?
     
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