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Math Help - Plane with a line

  1. #1
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    Plane with a line

    A plane is given by the equation
    3x-4y+2z+4=0

    Find the equation of a line l which lies in the given plane and has distance 6 away from the xy-plane.

    The line should be given in the form
    x=t \wedge y=t \wedge z=t
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  2. #2
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    Quote Originally Posted by MatteNoob View Post
    A plane is given by the equation
    P: 3x-4y+2z+4=0

    Find the equation of a line l which lies in the given plane and has distance 6 away from the xy-plane.

    The line should be given in the form
    x=t \wedge y=t \wedge z=t
    Construct a plane E parallel to the xy-plane and in a distance of 6 units form the xy-plane:

    E: z = 6

    The line you are looking for is l = P \cap E :

    Substitute z = 6 into the given equation:

    3x-4y+2\cdot 6 + 4 = 0~\implies~ y = \frac34x + 4

    Therefore

    l:\left\{ \begin {array}{l}x = t \\ y=\frac34t+4 \\ z=6 \end{array}\right.


    Remark: This is my last post for the next 4 weeks. Bye and much success!
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  3. #3
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    Thanks a lot!
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  4. #4
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    Quote Originally Posted by MatteNoob View Post
    Thanks a lot!
    Thanks for the thanks - but I've forgotten the second solution.

    The plane E_2: z=-6 is parallel to the xy-plane too and has also the distance of 6 units to the xy-plane.

    Therefore you'll get a second line which satisfies the given conditions:

    <br />
l_2:\left\{ \begin {array}{l}x = t \\ y=\frac34t-2\\ z=-6 \end{array}\right.<br />
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