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Cartesian equation of a line

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the vector equation of the line that passes through the point Q(2,0,-5) and
    is perpendicular to both the vectors m=(0,1,4) and n=(-2,-1,3).

    2. Relevant equations
    vector equation of a line: (x, y, z)=(x0,y0,z0) + t(a,b,c)
    cartesian equation of a line: (x-x0)/a=(y-y0)/b=(z-z0)/c

    3. The attempt at a solution
    (x0,y0,z0)=(2,0,5)
    To find (a,b,c)I know I can get two equations because the dot product of (a,b,c) with the two perpendicular lines equals zero:
    b+4c=0
    -2a-b+3c=0
    But two equations isn't enough to solve for three variables. Also, shouldn't the point (2,0,-5) also dot product with u or v to equal zero, since it's on the same line?
    Is it correct to assume that the points (0,1,4) and (-2,-1,3) are also points on the line? In which case I can easily find the direction of the line by subtracting one from the other.
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2

    Mark44

    Staff: Mentor

    The cross product of <0, 1, 4> and <-2, -1, 3> will give you a vector that is perpendicular to both. That's the vector you need to write your line in either its vector form or in Cartesian form.
     
  4. Apr 5, 2010 #3
    Man, I can't believe I didn't think of that myself. thanks for your help!
     
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