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Math Help - 3D geometry

  1. #1
    MHF Contributor alexmahone's Avatar
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    3D geometry

    Find the value of a such that the planes ax+y+z=0, x+3z=0, and 5y+6z=0 have a line in common.
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    Find the value of a such that the planes ax+y+z=0, x+3z=0, and 5y+6z=0 have a line in common.
    Let
    p_1:x+3z=0

    p_2:5y+6z=0 and

    p_3:ax+y+z=0

    denote the three planes.

    1. Calculate

    p_1 \cap p_2 = l_{p_1,p_2} : \left\{\begin{array}{l}x=-3t \\y=-\frac65t \\z = t\end{array}\right.

    2. Calculate

    p_2 \cap p_3 = l_{p_2,p_3} : \left\{\begin{array}{l}x=\frac1{5a} t \\y=-\frac65t \\z = t\end{array}\right.

    3. Compare the coefficients:

    l_{p_1,p_2} = l_{p_2,p_3}\ if\ -3=\frac1{5a}~\implies~ a=\dfrac1{15}

    4. Thus the third plane has the equation:

    p_3:\dfrac1{15} x+y+z=0

    5. I've attached a sketch of the three planes.
    Attached Thumbnails Attached Thumbnails 3D geometry-gemeins_schnittgerade.png  
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by earboth View Post
    -3=\frac1{5a}~\implies~ a=\dfrac1{15}

    Thus the third plane has the equation:

    p_3:\dfrac1{15} x+y+z=0
    Thanks for your solution, but shouldn't a=-\frac{1}{15}?
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  4. #4
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    Quote Originally Posted by alexmahone View Post
    Thanks for your solution, but shouldn't a=-\frac{1}{15}?
    Yes. Earboth made a simple and obvious typo.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Yes. Earboth made a simple and obvious typo.
    Thanks!

    Quote Originally Posted by alexmahone View Post
    Thanks for your solution, but shouldn't a=-\frac{1}{15}?
    Of course, you are right. The drawing is correct. The negative sign somehow slipped away on the way from my notes to the MHF editor. Sorry.
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