Polyhedron

[jasoncurious]

I have this assignment asking us to find the polyhedron described by the equation |x|+|y|+|z|=1. The question also asked us to justify the answer. After some calculation, I got the result that the polyhedron is a cube. I have the feeling that I am wrong with the answer as the top student in our class has a different answer.

 

Below are my attempt to solve the question:

 

From http://en.wikipedia.org/wiki/Octant_(solid_geometry), I got 8 different plane equations:

1. x+y+z=1
2. -x+y+z=1
3. -x-y+z=1
4. -x-y-z=1
5. x-y+z=1
6. x+y-z=1
7. x-y-z=1
8. -x+y-z=1

So, with the knowledge of plane algebra, the normal for each plane is:

1. (1,1,1)
2. (-1,1,1)
3. (-1,-1,1)
4. (-1,-1,-1)
5. (1,-1,1)
6. (1,1,-1)
7. (1,-1,-1)
8. (-1,1,-1)

From this 8 coordinates, I reached my answer indicating that the polyhedron is a cube, but plugging each of this coordinate back into its respective plane equation does not produce 1.

 

I am stuck here. Help from you guys are greatly appreciated. God bless you all.

[imatfaal]

the octahedron is a pair of the cube in many ways - if you truncate a cube (ie cut off the corner) and carry on you get an octadehron. I havent looked at the question really - but just an odd thought that you are looking at the wrong one of a pair of duals.

 

I did a little sketching during a long telephone call - I assumed z = 0 and worked out the line of y=f(x) that crosses that plane for all eight lines. I then assumed z=1/2 and worked out line of y=f(x) for all the lines that cross that plane. the black lines are z=0 and the red are the first five of z=1/2. To me that looks like four planes meeting at a vertex (no way thats a cube) - which for regular shapes has to be an octahedron. If you can see I have sketched an octahedron at the top of the page with the lines in black and red to show the intersection with the planes z=0 and z=1/2 (in this case z is the vertical axis)

 

[jasoncurious]

What my friend did is, he determined where each of the 8 planes intersect the x,y and z-axis. Then he used the coordinates as the vertices for the polyhedron, which is a solid with 8 triangular surfaces.

 

For example, the plane x+y+z=1 intersects x-axis at (1,0,0), y-axis at (0,1,0), z-axis at (0,0,1)

I was wondering what is the reason of doing so. Does this mean that (1,0,0),(0,1,0) and (0,0,1) is the furthest that the plane x+y+z=1 can reach? If so, what about (3,-3,1)? If I plug in (3,-3,1), I will be getting 1 as well.

[imatfaal]

8 regular triangles with four meeting at each vertex = octahedron. Your friend's method will work in this case as the vertices do lie on the axes - but that is not necessarily the case

 

You seem to be forgetting that this shape bounded by the small fraction of these planes where they intercept - not the totallity of these planes, they are infinite. The plane reaches off to infinity - but the shape (singular shape) is the only part of the xyz volume that is bounded by the planes.

 

If in trouble - drop a dimension. Think of the flat xy plane - what polygon is bounded by the lines x=-1 x=+1 y=-1 and y=1. The lines are infinite - but the shape bounded by them is finite, simple, and clear.

[jasoncurious]

In order for me to comprehend this question, first I look at the polyhedron 2-dimensionally (x-y plane, z-y plane, x-z plane), then I combine the three ways of viewing the polyhedron and get an octahedron. Only then I realize why my friend chose to find the vertices at the three axes.

 

Is this ok?

[imatfaal]

Looking at each of xy xz and zy works on this model - but might fail if you cannot recognize the shape from three silhouettes. But also your friends only works on this occasion, however it would not work for occasions where the vertices are not on the axes

 

to show an example where your friends idea would not work: if you took the simple example of x=1 x=-1 y=1 y=-1 z=1 z=-1 - these intersect the axes at the same points as the octahedron does. In the octahedrons case the intersections with the axes are the vertices. but in this example the intersection with the axes is in the centre of the face - and the shape bounded is actually, as you will have guessed, a cube.

 

For a non-simple example I think i would have to sketch slices, sketch the lines where each plane meets, work out vertices and sketch them in - all possibly from more than one view. However - any more complicated shape is gonna be hell to work out and probably wont come up

[jasoncurious]

http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06961.jpg.html?sort=3&o=3

http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06982.jpg.html?sort=3&o=2

http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06991.jpg.html?sort=3&o=1

http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_07002.jpg.html?sort=3&o=0

Sorry for the poor upload.

Edited December 8, 2012 by jasoncurious