# Is this plane correctly drawn?

• B
Hi,
I was trying to draw a plan 2,0,0 in X,Y,Z co-ordinate system

Is it correctly drawn ? I think ...I'm wrong in drawing this plane

I think I'm wrong because each point in this plane has non zero y & z value....they are not having y=0& z=0 everywhere all the time although x=2 everywhere on this plane....so my drawing to plane (2,0,0) is wrong.

Could you please correct my analysis ?
If I'm wrong then what is correct way to draw the plane (2,0,0) ?

Arman777
Gold Member
Hi,
I was trying to draw a plan 2,0,0 in X,Y,Z co-ordinate system

Is it correctly drawn ?

View attachment 199625

I think ...I'm wrong in drawing this plane

I think I'm wrong because each point in this plane has non zero y & z value....they are not having y=0& z=0 everywhere all the time although x=2 everywhere on this plane....so my drawing to plane (2,0,0) is wrong.

Could you please correct my analysis ?
If I'm wrong then what is correct way to draw the plane (2,0,0) ?
(2,0,0) is a point on 3D space.You draw x=2 plane

Ssnow
Gold Member
Hi, you must specify better your plane because there are infinity planes by a point ##(2,0,0)##. Do you want the plane passed by ##(2,0,0)## and parallel to the plane ##yz##?
Ssnow

Ibix
2020 Award
I agree we're missing some information here. Can you give us some more information? The problem specification in full, or some context around why you're trying to do this?

I'm trying to draw a crystal structure plane (200)

I'm not sure ..how to draw that

Ibix
2020 Award
I did wonder if that was it.

(2,0,0) is not a single plane; rather it's an infinite set of parallel planes. The planes ##(n_x,n_y,n_z)## are the set of planes whose ith member crosses the x axis at ##i/n_x##, working in units where the unit cell's edges are of length 1. Similarly for y and z.

Does that help?

Last edited:
there are infinity planes by a point (2,0,0).
what do you mean by this ? can you show me how they look like ?
I just want to visualize it.

I did wonder if that was it.

(2,0,0) is not a single plane; rather it's an infinite set of parallel planes. The planes ##(n_x,n_y,n_z)## are the set of planes whose ith member crosses the x axis at ##i/n_x##, working in units where the unit cell's edges are of length 1. Similarly for y and z.

Does that help?
very much confusing....would it be possible to post an example with related to this ?

Ibix
2020 Award
very much confusing....would it be possible to post an example with related to this ?
Pick an ##i##, say ##i=1##. What are ##n_x##, ##n_y## and ##n_z## in your case? Therefore, where does the ##i##th plane cross each of the axes? What does "at infinity" mean in practice?

Edit: I'm presuming you know what a unit cell is?

Pick an ##i##, say ##i=1##. What are ##n_x##, ##n_y## and ##n_z## in your case? Therefore, where does the ##i##th plane cross each of the axes? What does "at infinity" mean in practice?
okay..let me try that

i=1 i.e 1st plane
What are nxnxn_x, nynyn_y and nznzn_z in your case
its 2,0,0

Therefore, where does the iiith plane cross each of the axes?
1/2 , 1/0 , 1/0
=>
.5 , infinite , infinite

so I conclude its one plane which passes through at x=.5 and crosses y & z at infinite i.e parallel to YZ

But you said
(2,0,0) is not a single plane; rather it's an infinite set of parallel planes.
I'm getting confused more.

Ibix
2020 Award
Yes - you've worked out one of the planes correctly. Now repeat for all ##i=\ldots, -2, -1, 0 , 1, 2,\ldots## and you have the complete set of planes. Of course, since this is crystallography and everything repeats periodically you'd only be interested in the ##i## that gives a plane that passes through the unit cell. Outside that you are just repeating yourself.

Do remember that unit cells may not have all sides the same length.

okay ...that looks fine.

one more question :

The planes (nx,ny,nz)(nx,ny,nz)(n_x,n_y,n_z) are the set of planes whose ith member crosses the x axis at i/nxi/nxi/n_x, working in units where the unit cell's edges are of length 1.
Is it a theorem ..if so what it is called ?
It would be interesting to know how it is derived where the plane crosses the axis.

one more question ..

Yes - you've worked out one of the planes correctly. Now repeat for all ##i=\ldots, -2, -1, 0 , 1, 2,\ldots## and you have the complete set of planes.
all this happening in a specific unit cell ? i.e are we considering in this example i=....-2, -1, 0 , 1, 2.... for a given unit cell ? is it correct ?

Ibix
2020 Award
In your example there are only two planes in the unit cell, corresponding to ##i=0,1##. The ##i=2## plane is in the same place in the next unit cell along as the ##i=0## plane is in this one. I don't think there's any kind of theorem here. It's just a shorthand notation for a set of planes that uses the natural periodicity of a crystal to save on the writing. Usually the interest is that the crystal can be cut along planes (cleavage planes) and you can just say "this crystal cleaves along the (1,1,1) plane", or whatever.

okay.....thats nice...I understand .

But ...
I don't think there's any kind of theorem here
I don't understand how do you arrive at this part
whose ith member crosses the x axis at i/nxi/nxi/n_x
that division ....thats not clear.

Ibix