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Finding line perpendicular to 2 other lines in 3D space

  1. Feb 8, 2009 #1
    Find the equations for a line that is perpendicular to both of the following lines

    x/3 = y/2 = z/2
    x/5 = y/3 = (z-4)/2

    ---

    cross product:
    <3,2,2> x <5,3,2> = <-2,4,-1>

    then the line is
    (x-a)/2 = (b-y)/4 = (z-c)/1

    but the answer is supposed to be
    (1/2)x - (52/7) = (-1/4)y + (52/21) = z - (208/21)

    How on earth do I get that?
     
  2. jcsd
  3. Feb 9, 2009 #2

    HallsofIvy

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    You answer is incorrect because it is not the equation of a single line but rather the general form for a line perpendicular to the two given lines through the point (a, b, c). If the two given lines were either parallel or intersecting, there would exist an infinite number of lines perpendicular to the two given lines and you could pick whatever point (a, b, c) was convenient. However, the two given lines are skew, neither parallel (because the direction vectors <3, 2, 2> and <5, 3, 2> are not parallel) nor intersecting (because the equations x/3= y/2 and x/5= y/3 have x=y= 0 as solution and there is no z satisfying both z/2= 0 and (z-4)/2= 0) so they have a unique perpendicular. You must also determine where that perpendicular crosses one of the lines.

    The given answer is the line perpendicular to the two given lines through the point (104/7, 208/21, 208/21)
     
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