Suppose that a plane pi cuts the Unit cube in the first octant, Use Vectors to show that there is a pi which intersects the cube in a regular hexagon.

Question

amistre64 · Answer

a plane "pi" ?  there is a "pi"? which intersects ?

amistre64 · Answer

i cant quite picture the question

anonymous · Answer

it deals with Vectors and calculus 3 stuff

anonymous · Answer

and it says a Plane "pi"  and that "pi" intersects

amistre64 · Answer

i can prolly do calc2 vector stuff, and a few calc 3 ..

amistre64 · Answer

"pi" sounds like its just a name they give it; and not a special attribute, am i right?

anonymous · Answer

"pi" = 3.14

anonymous · Answer

a Hexagon has 6 sides... and so i assumed that there were six vectors to go with it

amistre64 · Answer

ok, so we have a cube that has a corner sliced off of it by a plane

amistre64 · Answer

i some fashion maybe :)

anonymous · Answer

yes! keep going

amistre64 · Answer

unless pi is a 'special' plane, I can only see slice of the cube as triangles and quadreilaterals; but i havent tried all the ways yet ...

amistre64 · Answer

how to fit a regular hex all of its sides touch the box evenly .....

anonymous · Answer

Yes "pi" is a special plane

amistre64 · Answer

how to fit a regular hex, inside of a box so that all of its sides touch the box evenly .....

amistre64 · Answer

http://www.wolframalpha.com/input/?i=plane&a=*C.plane-_*Geometry-

anonymous · Answer

Yes... it looks right

anonymous · Answer

the boxed hexagon sounds right in wolframalpha

amistre64 · Answer

attachment

amistre64 · Answer

now how to show vectors to prove it, i aint got a clue ... yet

amistre64 · Answer

all of it can be seen at http://mathworld.wolfram.com/Cube.html

amistre64 · Answer

As shown above, a plane passing through the midpoints of opposite edges (perpendicular to a  axis) cuts the cube in a regular hexagonal cross section 

             (Gardner 1960; Steinhaus 1999, p. 170; Kasahara 1988,
               p. 118; Cundy and Rollett 1989, p. 157; Holden 1991, 
                  pp. 22-23). 

Since there are four such axes, there are four possible hexagonal cross sections. 

If the vertices of the cube are (+-1, +-1 ,+-1), then the vertices of the inscribed hexagon are:

1. (0,-1,-1)
2. (1,0,-1)
3. (1,1,0)
4. (0,1,1)
5. (-1,0,1)
6. (-1,-1,0)

A hexagon is also obtained when the cube is viewed from above a corner along the extension of a space diagonal (Steinhaus 1999, p. 170).

anonymous · Answer

it also says that the Unit cube is in the first octant!

anonymous · Answer

just another additional info!

amistre64 · Answer

yeah, if anything take and shift these points into the first octant

anonymous · Answer

okay... is that it? as far as the vectors and intersection goes

amistre64 · Answer

wish I could say, but thats as close as I can get for an answer.

the midpoints adjust to:

(1, 1/2, 0)  ->  (1, 0, 1/2)   ->  (1/2, 0, 1) 

    ->  (0, 1/2, 1)  ->  (0, 1, 1/2)  ->  (1/2, 1, 0)

       -> (back to the start)

amistre64 · Answer

there are a few set ups for that, but i drew those and they work ...

anonymous · Answer

okay.. thanks... if you can think of anything else.. let me know...

amistre64 · Answer

the "show" part is tricky.

define the angle between vectors, and the length of each vector to be the same

anonymous · Answer

but so far. so good

anonymous · Answer

ok

amistre64 · Answer

cos(a) = a.b/(|a||b|) right?  

so angle = cos-1(a.b/(|a||b|))

and length is $\sqrt{(\triangle x )^2+(\triangle y)^2+(\triangle z)^2}$

anonymous · Answer

Yes... absolutely!

amistre64 · Answer

angles between them to be a hexagon have to be 120 degrees ...

amistre64 · Answer

and as all of them are the same length, i fell that that is "show" enough :)

anonymous · Answer

yes sir.. thats good!

amistre64 · Answer

good luck with it ;)

anonymous · Answer

Thank you... Now are you a professor ?

amistre64 · Answer

and to make it just that much easier; cos(120) = -1/2.

so the $\frac{a.b}{|a||b|} = -\frac{1}{2}$

amistre64 · Answer

nah, just a student; looking to be a college math teacher in the future tho

anonymous · Answer

Where at? and you are awesome...

amistre64 · Answer

going to a place called StLeo in the fall ...

anonymous · Answer

Awesome!

anonymous · Answer

Thank You!

anonymous · Answer

have another question if you can help! http://openstudy.com/groups/mathematics#/users/benfraser1012

anonymous · Answer

Hey, Good to see you.. I am stuck on this question.. and was wondering if you could help..

anonymous · Answer

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e0cc35d0b8bd74af49df6fe