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Find two planes that intersect at a given line.

  1. Sep 30, 2011 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data
    Find two different planes whose intersection is the line [itex]x=-7+t, y=9-t, z=6+2t[/itex]. Write equations for each plane in the form [itex]Ax+By+Cz=D[/itex].


    2. Relevant equations
    The solutions are
    x+y=2;
    2y+z=24;
    but I do not know how to obtain these answers.
    (This is not for class work, I am just trying to teach myself.)


    3. The attempt at a solution

    When t = 0, the point on the line (and also a point on both planes) is P=(-7,9,6).

    I understand that any point on the line must also be a point on both planes. What I really need is to find the 'tilt' or normal vector of each plane, though when I visualize it, I can see an infinite number of intersecting planes that could share this line. Am I visualizing it wrong?

    Understanding that a vector parallel to the line is [itex]L=<i-j+2k>[/itex] and is also the result of the cross product between the normal vectors of each plane, I tried to work the cross product backwards to obtain the two normal vectors. This didn't seem to get me anywhere.

    [itex]n_{1}=<A_{1}i+B_{1}j+C_{1}k>[/itex]
    [itex]n_{2}=<A_{2}i+B_{2}j+C_{2}k>[/itex]
    [itex]n_{1} \times n_{2} = <i-j+2k>[/itex]

    I wound up with three equations and 6 variables.

    [itex]B_{1}C_{2}-B_{2}C_{1}=1[/itex]
    [itex]A_{2}C_{1}-A_{1}C_{2}=-1[/itex]
    [itex]A_{1}B_{2}-A_{2}B_{1}=2[/itex]

    Looking back on the process of finding the line when given the two planes, I notice that I set z to 0 so solve for x in terms of y, and did some substitution to eventually find a point on the line, but I am unsure of what the analogous 'backwards' process of this would be.

    Any insight would be appreciated.
     
    Last edited: Sep 30, 2011
  2. jcsd
  3. Sep 30, 2011 #2

    SammyS

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    I think you mean:

    x+y=2;

    2y+z=24;
     
  4. Sep 30, 2011 #3

    ElijahRockers

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    Ooops. Thanks. It's quite possible I copied the solutions wrong. That really is besides the point though, whatever the solutions may be, I don't know how to get them.
     
  5. Sep 30, 2011 #4

    HallsofIvy

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    Do you understant that there are an infinite number of pairs of planes that intersect at that line? There is no "the" solution- there are many solutions of which the given solution is one.

    The line is given by x=−7+t,y=9−t,z=6+2t and has "direction vector" <1,-1,2>. Also (-7, 9, 6) is a point on that line. Choose any other vector, u, and take the cross product to get a vector perpendicular to both. Find the equation of the plane having that vector as normal vector and containing point (-7, 9, 6). This plane will contain the given line.

    Choose any vector, v, other than u and do the same to get a second plane also containing that line. Now you have two planes whose intersection is that plane.
     
  6. Sep 30, 2011 #5

    ElijahRockers

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    Yes actually, that's what was throwing me off I guess. The computer generated question was multiple choice, so it kind of led me to believe that there was only one solution, when I knew from visualizing the problem (as I said in my original post) that there should be infinitely many solutions. I have a hard time understanding how they would expect me to arrive at one of their 4 multiple choice answers if I need to just choose any vector u at random. Perhaps they expected me to just find the intersecting lines of each possible answer until I found one that matched the given line.

    However, I think I do understand the process now, I will take another look at a similar question to see if I can arrive at one of their solutions.
     
    Last edited: Sep 30, 2011
  7. Sep 30, 2011 #6

    SammyS

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    FYI

    The plane given by x+y=2 is parallel to the z-axis.

    The plane given by 2y+z=24 is parallel to the x-axis.
     
  8. Sep 30, 2011 #7

    ElijahRockers

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    Yea, though I'm uncertain if that information is supposed to be able to help me choose the correct pair of planes just by looking at the parametrization of the line. I am so used to doing calculus in 2 dimensions, there is a much bigger playing field that I'm not used to now.
     
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