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let there be two planes $$2x-y-5z+11=0$$ and$$2x+2y+z-1=0 $$ show that they intersect

attempt at a solution:

If planes do not intersect they are parralel hence there is a $t\in R$ such that $t(2,-1,-5)=(2,2,1)$ or in a SOE : $\begin{cases} 2t = 2\\ -t=2 \\ -5t=1 \end{cases} $, from $ 2t = 2, t=1$ and applying that to the other two equations we get false statements, hence no such $t$ exists, and hence the planes indeed intersect.


Is this ok, or is there other ways to show that planes intersect?

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Their normal vectors are not parallel, which implies that the planes intersect. – SuperAbound Jul 28 '14 at 14:20

1 Answer 1

This is ok: if the norm vectors $(2,-1,5)$ and $(2,2,1)$ are not parallel, then planes will intersect. Alternatively, you could find a point satisfying both equations - that is, a point which belongs the intersection. For example, eliminating $x$ yields $3y+6z = 12$ so let's say $y = 2$ and $z = 1$, hence $x = -2$. You can easily check that this point belongs to each plane.

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do I somehow need to retouch this proof or is it ok? – Bak1139 Jul 28 '14 at 14:46
@Bak1139: as they very first part of the very first sentence of my answer say, this is ok :) – Ilya Jul 28 '14 at 14:47

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