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Determine if plane perpendicular to line

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Find cartesian equations of the line L containing P(2, 0, -3) and Q(1, -1, 6) and determine if plane (8x + y - z = -1) is perpendicular to L


    2. Relevant equations



    3. The attempt at a solution

    PQ = (1-2)i + (-1-0)j + (6+3)k = -i -j +9k

    (x-2)i + (y-0)j + (z+3)k = -ti - tj +9tk

    so, cartesian equation of L is:
    [tex]\frac{x-2}{-1} = \frac{y}{-1} = \frac{z+3}{9}[/tex]

    The plane normal n = (8, 1, -1). If dot product n and L = 0 then they are perpendicular.

    How do I take the dot product of L and n in that form?

    I don't understand the general equation of the line, I just followed an example and plugged in some numbers. This question should take about 5 minutes but has taken me about 2 hours and I still have no clue.
     
  2. jcsd
  3. May 3, 2009 #2
    L is already perpendicular to the plane.

    And what is the characterisitc vector of the line. it is u=<-1,-1,9> right

    and of the plane n=<8,1,-9>.

    If the plane and the line were perpendicular it means that the characteristic vector of the line, namely u, and that of the plance, namely n, have to be parallel, dont they?

    so one can be written as a linear combination of the other.
     
  4. May 3, 2009 #3
    how?

    So a characteristic vector is the denominators in the cartesian equation? And a characteristic vector is perpendicular to the line?

    Is linear combination the same as scalar multiple?
     
  5. May 3, 2009 #4

    I meant the vector n. not L. my bad.
    yes. and no.

    The line is in the same direction as the vector that i wrote in my previous post, it is not perpendicular to it.
    the characterisitc vector of a plane is perpendicular to it, while that of a line is parallel to the line itself.
     
  6. May 3, 2009 #5
    in the case of two vectors, yes.
     
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