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Thread: Intersection of planes in three-space?

  1. #1
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    Intersection of planes in three-space?

    Describe the shape of the intersection of the plane z= -3 and the plane y=z in three-space.

    Well, z= -3 would be a horizontal plane 3 units down the z-axis.
    Y=z is an infinite line with x being any value in the y-z plane.
    I'm not sure at what line they intersect. I know z must be -3.
    Would y be any value, since both z=-3 and y=z extend indefinitely along the y-axis? And would x have to be 0?
    Any feedback would be appreciated.
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  2. #2
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    Re: Intersection of planes in three-space?

    y = z is a PLANE in 3D space, not a line.

    Surely you can tell me what graph the intersection of two planes gives...
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    Re: Intersection of planes in three-space?

    Quote Originally Posted by AFireInside View Post
    Describe the shape of the intersection of the plane z= -3 and the plane y=z in three-space.

    Well, z= -3 would be a horizontal plane 3 units down the z-axis.
    Y=z is an infinite line with x being any value in the y-z plane.
    I'm not sure at what line they intersect. I know z must be -3.
    Would y be any value, since both z=-3 and y=z extend indefinitely along the y-axis? And would x have to be 0?
    Any feedback would be appreciated.
    Nope. The intersection line of the two planes is characterized by z=-3 AND y=z (since you're on both planes simultaneously).

    So the intersection line must have y=z=-3. The x-value of the intersection line is not constrained.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Intersection of planes in three-space?

    Being the line $\displaystyle r \equiv \begin{Bmatrix}x=t \\y=-3\\z=-3\end{matrix}\quad (t\in \mathbb{R} )$ or equivalently

    $\displaystyle r \equiv(x,y,z)=(0,-3,-3)+t(1,0,0)\quad (t\in\mathbb{R})$

    we can describe the "shape" of $\displaystyle r$ as the line passing trough $\displaystyle (0,-3,-3)$ and parallel to the $\displaystyle x$ axis.
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