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Given five planes:

$\pi_1=2x+5y+z-2=0$

$\pi_2=x+y-z-1=0$

$\pi_3=x+4y+2z-4=0$

$\pi_4=3x-y+4z-3=0$

$\pi_5=-6x+2y-8z+k=0$.

How can i find the solid shape that is formed by those planes?

I tried to draw but it's too complex.

I can see that for a solid to be exists $k\neq6$.

Any help?

Thanks.

arony
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  • Maybe use software... – user2345215 Jan 16 '15 at 23:56
  • so drawing is the only "easy" way? – arony Jan 16 '15 at 23:58
  • Five planes in general position can determine 26 regions of space, of which several will be bounded. Which of these is your solid? It's not an easy problem. I suggest replacing each equation with an inequality. There are $2^{5}=32$ choices, each of which will lead to one of the regions. Plot them all, and see which solid(s) are the ones you want. – vadim123 Jan 17 '15 at 00:04
  • i think i got a prism with triangle as a base, by finding the lines from $\pi_1,\pi_3$ and from $\pi_2,\pi_3$. – arony Jan 17 '15 at 00:10

1 Answers1

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How $\pi_2$ , $\pi_3$ are independents planes and $\pi_1$ in linear combination of $\pi_2$ and $\pi_3$ but the except to independent term so is a prisma triagular infinity.

How the plane $\pi_4$ , $\pi_5$ and $k\neq{6}$ are parallel planes then is a triangular prism bounded for this planes.

jimbo
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