# Thread: Intersection of planes in three-space?

1. ## 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.

2. ## 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...

3. ## Re: Intersection of planes in three-space?

Originally Posted by AFireInside
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.

4. ## Re: Intersection of planes in three-space?

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

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

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