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Math Help - Line of intersection of planes

  1. #1
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    Line of intersection of planes

    Prove that the line of intersection of the two planes
    r=\lambda_1i+2\mu_1j+3\nu_1k \lambda_1+\mu_1+\nu_1=1
    r=2\lambda_2i+\mu_1j+2\nu_1k \lambda_2+\mu_2+\nu_2=1
    can be written in terms of a single parameter t as
    6r=(3+t)i+4tj+9(1-t)k

    I managed to get the vector equations into this form:
    r.(6i+3j+2k)=6
    and r.(i+2j+k)=4
    I calculated the direction vector of the line to be (-i-4j+9k)
    I don't know how else to continue.
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Prove that the line of intersection of the two planes
    r=\lambda_1i+2\mu_1j+3\nu_1k \lambda_1+\mu_1+\nu_1=1
    r=2\lambda_2i+\mu_1j+2\nu_1k \lambda_2+\mu_2+\nu_2=1


    Too many equality signs...

    can be written in terms of a single parameter t as
    6r=(3+t)i+4tj+9(1-t)k

    I managed to get the vector equations into this form:
    r.(6i+3j+2k)=6
    and r.(i+2j+k)=4
    I calculated the direction vector of the line to be (-i-4j+9k)
    I don't know how else to continue.
    Thanks!
    How did you get the intersection line's director vector to be such a nice one if in the definition of the two planes you have 6 (!!) parameters??
    The director vector of the line is a normal to both plane's normals, and then you only need one single point belonging to both planes to form the parametric euqation of the intersection line. I can't understand what you did above...in fact, I hardly understand your planes' equations.

    Tonio
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  3. #3
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    I eliminated the \lambda in each, then since the two vectors remaining were parallel to the plane, their cross product I thought would be the direction vector of the plane. And since I had a position vector of a point on each line, found the equation of the plane.
    for the first equation
    r=i+\mu_1(-i+2j)+\nu_1(-i+3k)
    I think I did something wrong.
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