Question

What are the equations of the other planes of a 4X4X4 cube, given that the equation of one of the planes is 3x + 2y - z =4?
I haven't worked with vectors in 5 years and my sister needs help with this. I have an idea how to do it, but I'm not sure my equations are right.

Answer 1

cubes have parallel sides and perpendicular sides right?

Answer 2

one of the sides will have to be a distance of 4 from this plane, you know what a normal is?

Answer 3

Yes, but can I just change the 4 to 8 or 0 to get a parallel line with distance 4 away?

Answer 4

i do not believe so, you have to measure along the normal vector; not just randomly add stuff. it might work but i cant guarentee it :)

Answer 5

It doesn't work by changing the 4, I wolfram-ed it.

Answer 6

How would you do find the parallel plane?

Answer 7

take a point on the plane, and add 4 unit normals to it to find a point in t parallel plane that is 4 away

Answer 8

3x + 2y - z =4, a point on the plane; let x=0, y=0, z has to be -4

Answer 9

or, let x=0, z=0, y has to 2

Answer 10

how do we know the normal, and how do we unit it?

Answer 11

normal is [3,2,-1] unit is [3,2,-1]/sqrt(14)

Answer 12

good

now we just add vectors

(0,2,0) + (3,2,-1)/sqrt(14)

Answer 13

forgot to times 4 lol

Answer 14

(0,2,0) + 4(3,2,-1)/sqrt(14)

(0,2,0) + (12,8,-4)/sqrt(14)

Answer 15

1/sqrt(14) = sqrt(14)/14 ifn you want to simplify things ... dunno how simple they get tho

Answer 16

That makes sense

Answer 17

(0,2,0) + sqrt(14)(12,8,-4)/14

(0,2,0) + sqrt(14)(6,4,-2)/7

add like parts

point on parallel plane is:

6sqrt(14)/7 , 2 + 4sqrt(14)/7 , -2sqrt(14)/7
a b c


now, the equation of the new plane just differs by a constant,

3x +2y -z = K, such that K=3a +2b -c

Answer 18

What about adding 4 unit vectors to the original equation 3(x-12/sqrt(14)) + 2(y-8/sqrt(14)) - (z+4/sqrt(14))=4

Answer 19

im not sure that would work, we can test it out but id work the confident way first to make sure :)

Answer 20

Ok I'll do that

Answer 21

3(6sqrt(14)/7)+2(2 + 4sqrt(14)/7)-(-2sqrt(14)/7) = 4(1+sqrt(14))

which is about 19 if we round but we dont want to round do we

Answer 22

now comes a question ive been wondering since we started .... does the cube give us any placement of it? if we only know 1 side of it; but not anything else, then were are the corners sopse to be? do we just randomly make one thats 4x4x4?

Answer 23

good news is, my approach and your shifting approach work the same results :)

Answer 24

I think we just randomly work out any cube with 4X4X4

Answer 25

then we should decide on a vertex point to establish the other planes with

Answer 26

3x +2y -z = 4

we have 3 to play with that are the intercepts

0,0,-4
0,2,0
4/3,0,0

Answer 27

0,2,0 seems good

Answer 28

hmm, then it must be a trap lol

Answer 29

lets ride along the vertex from 0,2,0 towards 0,0,-4 since we know they are in the plane and we can make a direction vector from them

Answer 30

It's not. It's actually weird. My sis came to me and showed me a picture that was on her class's whiteboard - the teacher wrote "on tuesday, you'll be given the equation of a plane and asked to create other planes to build a geometric shape in 3d. you will have to show your planes work. you may have dimensions specified"

Answer 31

So my sister and I made up a very general question and I'm trying to work from that to give her an idea of what to do

Answer 32

Ok, makes sense

Answer 33

your doing fine. well cover what we did in the end in a general after we have established a good path to take

Answer 34

0,0,-4 - 0,2,0 = 0,-2,-4, right? this should be our direction vector from 0,2,0

Answer 35

yes

Answer 36

unit it and times it by 4

Answer 37

and cross it with the normal to get the vector in the other direction ...

Answer 38

yes

Answer 39

n x v = p such that p is perp to n and v

these vectors will be the 'normals' to their respective sides

Answer 40

and since they are 4 from 0,2,0 the will also provide points for anchoring