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Intersection of a line and a plane, for what value(s) of k ?

  1. Aug 16, 2005 #1
    Well,

    From what I understand, to determine the intersection of a line and a plane, we use parametric form of the line and substitute the values of x, y and z into the Cartesian equation of the plane, correct?

    so, given the line
    x = 2 + 4t
    y = -1 + kt <=== note the 'k' variable
    z = 5 - 3t

    and the plane 7x + 6y - z - 3 = 0

    What must be the value of k for no intersection point, one intersection point, an infinite number of intersection points?


    *** My calculations this far are written below. So far I think I'm headed completely in the wrong direction but I've exhausted the only way I thought possible.

    If I plug the parametric form into the plane equation, I end up with
    31t + 6kt = 0

    this is where I'm lost. The question has me lost, because I have not tried this form (usually I have been given a non-variable value for the parametric equations, plugging them in is easy and I go from there)

    At this point, I'm -guessing- that to have no intersection point, I must have a constant on the other side of the equation and be inconsistent (of the sort, 0t = 123). This is not possible as I can't just invent one? So, there cannot be NO intersection (it must pass through the plane)

    To have one intersection point, we can suppose any value of k, then t = 0?

    And for infinite number of intersection points, basically I need 0t = 0 (dependent system)

    SO, by letting k = - 31/6
    31t + 6kt = 0
    0t = 0
    and thus an infinite number of intersection points (the line is contained in the plane)


    *** The above is probably incorrect ( I feel) but I do not know another way. Please suggest the correct way of going about this. Do not give me the answer, I would rather just some advice on how to work my way back on this question! :smile:
     
    Last edited: Aug 16, 2005
  2. jcsd
  3. Aug 16, 2005 #2

    TD

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    Homework Helper

    Your line is
    [tex]\left( {\begin{array}{*{20}c}
    x \\
    y \\
    z \\

    \end{array}} \right) = \left( {\begin{array}{*{20}c}
    2 \\
    { - 1} \\
    5 \\
    \end{array}} \right) + t\left( {\begin{array}{*{20}c}
    4 \\
    k \\
    { - 3} \\
    \end{array}} \right)[/tex]

    Note that the point (2,-1,5) lies in the plane, which is why you can't find a value of k for which there is no intersection.

    For infinite intersections, the line has to be in the plane and k = -31/6 is correct there.
     
  4. Aug 16, 2005 #3
    aha!

    sorry, I'm slow on the uptake :(

    thanks for the help :D
     
  5. Aug 16, 2005 #4

    TD

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    Homework Helper

    You do understand now, I hope? If there's something unclear, don't hesitate to ask!

    Glad that I could help :smile:
     
  6. Aug 16, 2005 #5
    Yeah, unfortunately I didn't think to look at the initial vector of the line being on the plane. I should have noticed that when I expanded the substitution and it came out to zero with the t variable left over... LOL

    thanks again
     
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