0

Of course the answer is no, but I need to prove it. I tried contradiction, i.e. two different points that satisfy both equations, but I can't make much of that.

3 Answers3

1

No, the intersection of two convex sets must be convex.

Chris Culter
  • 26,806
  • This sounds very interesting and I happen to have seen the term convex while reading on my own, however I haven't yet been properly taught it, so I wouldn't know how to use it in a proof. –  Feb 13 '19 at 17:21
0

I'll outline the proof
Get one point of intersection. Call it $A$
Calculate the cross product of the two normals. Call it $\vec v$
Show $A+\lambda\vec v$ satistifies both equation of the planes.

If you don't know how to calculate normal/cross product, get a second point of intersection $B$. Define $\vec v = B-A$

Anvit
  • 3,379
  • So basically, the proof is that if the intersection is even 1 point (or at least 1 point, however you want to phrase it), then it has to be a whole line? –  Feb 13 '19 at 17:20
  • yes that is correct – Anvit Feb 13 '19 at 17:20
  • Thanks, that makes sense, although the tutor proposed that we began the proof with exactly two points (which i suspect is the alternative proof you suggest, using B - A) –  Feb 13 '19 at 17:23
  • do you know about vectors? If not, I can add another elementary proof – Anvit Feb 13 '19 at 17:24
  • yes, I know simple stuff like inner/cross product, vector equations of a line and a plane etc –  Feb 13 '19 at 17:36
0

Given any plane in three space, and given any two distinct points $A$ and $B$ in the plane, the line through $A$ and $B$ must lie in the plane, as well. Thus, if $A$ and $B$ are in the intersection of two planes--meaning that $A$ and $B$ lie in both planes--then the line through $A$ and $B$ lies in both planes, so every point on that line is in the intersection of the two planes.

Cameron Buie
  • 102,994