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Homework Help: Plane containing 2 lines

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane that contains the lines,

    [tex]x = 2s + 2[/tex]

    [tex]y = 3s - 4[/tex]

    [tex]z = -5s + 2[/tex]

    and

    [tex]x = 4t + 3[/tex]

    [tex]y = 6t + -4[/tex]

    [tex]z = -10t + 5[/tex]

    2. Relevant equations



    3. The attempt at a solution

    One can quickly note that the two lines are parallel to each other, because the direction vector of the second line is simply 2 times the direction vector of the first line.

    We can find a point on the first line, call it P(2,-4,2).

    We can also find a point on the second line, call Q(3,-4,5)

    Now if we draw a vector from P to Q then,

    [tex]\vec{PQ} = <1,0,3>[/tex]

    Now all we need to do is find another vector call it,

    [tex]\vec{n} = <x,y,z>[/tex]

    That is perpendicular to,

    [tex]\vec{PQ}[/tex].

    Two vectors are perpendicular when their dot product is 0.

    So,

    [tex]\vec{PQ} \cdot \vec{n} = 0[/tex]

    This will give us the following equation,

    [tex]1x + 0y + 3z = 0[/tex]

    Now I could easily pick values for x,y, and z that would satisfy this. For example,

    [tex]x=3[/tex]

    [tex]y=0[/tex]

    [tex]z=-1[/tex]

    So then the vector n would be defined as follows,

    [tex]\vec{n} = <3,0,-1>[/tex]

    Then I could define my plane as follows,
    [tex]
    3(x-3) -(z-5) = 0
    [/tex]

    What's wrong with this approach?

    EDIT: Whoops I think I found my mistake. I want the vector, [tex]\vec{n}[/tex] to be normal to, [tex]\vec{PQ}[/tex] and [tex]\vec{v}[/tex] where v is the direction vector of one the lines.

    So,

    [tex]\vec{n} = <-9,11,3>[/tex]

    So the equation of the plane is,

    [tex]-9x + 11y +3z = -56[/tex]
     
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 29, 2010 #2

    ehild

    User Avatar
    Homework Helper

    It is correct.

    ehild
     
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