# B Is this plane correctly drawn?

1. May 7, 2017

### momentum

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) ?

2. May 7, 2017

### Arman777

(2,0,0) is a point on 3D space.You draw x=2 plane

3. May 7, 2017

### Ssnow

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

4. May 7, 2017

### Ibix

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?

5. May 7, 2017

### momentum

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

I'm not sure ..how to draw that

6. May 7, 2017

### Ibix

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: May 7, 2017
7. May 7, 2017

### momentum

what do you mean by this ? can you show me how they look like ?
I just want to visualize it.

8. May 7, 2017

### momentum

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

9. May 7, 2017

### Ibix

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?

10. May 7, 2017

### momentum

okay..let me try that

i=1 i.e 1st plane
its 2,0,0

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
I'm getting confused more.

11. May 7, 2017

### Ibix

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.

12. May 7, 2017

### momentum

okay ...that looks fine.

one more question :

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.

13. May 7, 2017

### momentum

one more question ..

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 ?

14. May 7, 2017

### Ibix

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.

15. May 7, 2017

### momentum

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

But ...
I don't understand how do you arrive at this part
that division ....thats not clear.

16. May 7, 2017

### Ibix

It's just what crystallographers mean by the (2,0,0) plane. They mean the set of planes parallel to the y-z plane spaced by half the llength of the x side of the unit cell. It's just a convenient notation for a concept that is useful in crystallography. It isn't something you can derive from anywhere - it's just something you need to learn.