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How to find disjoint set of two planes in form $ax+by+cz=d$?

How do I need to alter the equations?

mavavilj
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  • You've been around Math.SE awhile, but you've posted two Questions, both about the intersection (or lack of intersection) of two planes, without giving any context. By omitting this context you leave your Readers guessing about what you already understand and where you need help. – hardmath Sep 20 '17 at 14:15

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Keeping $a,b,c$ fixed and taking different values of $d$ gives parallel planes... just because all these planes share the same normal vector $(a,b,c)$.

For example $x+y+z=1$ and $x+y+z=2$ are equations of parallel planes. A proof of this is that if there was a common point $(x,y,z)$, we would have a proof that $1=2$...

Jean Marie
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  • But does this answer my question? What's the disjoint set? – mavavilj Sep 20 '17 at 13:11
  • Let us take an example: the sets of points ${(x,y,z) s.t. x+y+z=1}$ and ${(x,y,z) s.t. x+y+z=2}$ are disjoint. This is the only meaning I see for your question. – Jean Marie Sep 20 '17 at 13:15
  • @JeanMarien I see, I actually asked the wrong question. I didn't mean to ask for disjointness, but intersection :/ – mavavilj Sep 20 '17 at 13:17
  • ... indeed, it is not the same question. In this case, you should ask a new separate question. – Jean Marie Sep 20 '17 at 13:19