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Math Help - Geometry

  1. #1
    Junior Member
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    Geometry

    Can anyone help me with the following question on geometry and vectors:

    The line L through the origin is normal to the plane 2x-y-z=4. Find the point in which L meets the plane x+y-2z=2.

    Thanks to those who can help.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, smiler!

    I'm sure you know enough of the basic to solve this.
    You just have to put it all together . . .


    The line L through the origin is normal to the plane 2x-y-z\:=\:4
    Find the point in which L meets the plane x+y-2z\:=\:2

    The plane has the normal vector: . \vec{n} \:=\:\langle2,-1,-1\rangle

    This is the direction vector of the line through the origin.

    Its equations are: . \begin{array}{ccccc}x & = & 0 + 2t & \Rightarrow & x \:=\:2t \\ y & = & 0 + (\text{-}1)t & \Rightarrow & y \:=\:-t \\ z & = & 0 + (\text{-}1)t & \Rightarrow & z \:=\:-t \end{array} . [1]


    To find its intersection with the plane: . x +y -2z \:=\:2 . [2]
    . . substitute [1] into [2]: . (2t) + (-t) - 2(-t) \:=\:2\quad\Rightarrow\quad 3t \:=\:2\quad\Rightarrow\quad t \:=\:\frac{2}{3}

    Substitute into [1] and we have: . \boxed{\:x \:=\:\frac{4}{3},\;\;y \:=\:-\frac{2}{3},\;\;z \:=\:-\frac{2}{3}\:}

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  3. #3
    Junior Member
    Joined
    Nov 2007
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    direction vector

    i understand the majority of the answer but how do you work out the direction vectors x=0+2t, y=0+(-t), z=0+(-t)?

    thanks
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