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Math Help - Finding planes which intersect at a given line

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    Finding planes which intersect at a given line

    If you are given a line (r=[4,-5,6] + t[2,0,-1], how can you, from that info, figure out 3 planes which intersect at that point?
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    Hello, theowne!

    If you are given a line: . r\:=\:[4,-5,6] + t[2,0,-1]
    how can you, from that info, figure out 3 planes which intersect at that point?
    What point ??


    If you're referring to the point (4,-5,6),
    . . write three equations of the form: ax + by + cz \:=\:d
    . . which have (4,-5,6) as their solution.


    Here's one system: . \begin{array}{ccc}x + y + z &=& 5 \\ x - y + z &=& 15 \\ x + y - z &=& \text{-}7 \end{array}

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    Sorry, I meant to say, "intersect at that line".
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    Quote Originally Posted by theowne View Post
    If you are given a line (r=[4,-5,6] + t[2,0,-1], how can you, from that info, figure out 3 planes which intersect at that point?
    You only have to add a second direction to this line:

    p: \vec r = \langle 4,-5,6\rangle +  t \langle 2,0,-1\rangle + s \langle a,b,c \rangle

    with \langle a,b,c \rangle \ne k \cdot \langle 2,0,-1\rangle ~,~ k \in \mathbb{R} \setminus \{0 \}
    Last edited by earboth; June 7th 2008 at 10:39 PM.
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    Sorry, I don't understand..can you please clarify? How does this give us 3 planes?
    Last edited by theowne; June 8th 2008 at 12:56 PM.
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    Quote Originally Posted by theowne View Post
    Sorry, I don't understand..can you please clarify? How does this give us 3 planes?
    I've attached a sketch of three planes with a common line (in red). As you can see you only need a second direction vector (colour of the 2nd vector correspond to the colour of the plane's name) to span a plane.

    So choose any numbers for a, b and c to get a new plane which contains the given line:

    p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle a,b,c \rangle

    a = 1, b = 2, c = 3 will yield:

    p: \vec r = \langle 4,-5,6\rangle + t \langle 2,0,-1\rangle + s \langle 1,2,3 \rangle

    General solution:
    The normal vector of the new plane is \vec n = \langle 2,0,-1\rangle \times \langle a,b,c \rangle = \langle b~,~ -2c-a~,~ 2b \rangle

    Considering that the point A(4, -5, 6) must be in the new plane you'll get the general equation:

    p: bx-(2c+a)y + 2bz=5a+16b+10c
    Attached Thumbnails Attached Thumbnails Finding planes which intersect at a given line-dreiebenen_parameterform.gif  
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