Intersecting Planes: (110) & (101) - Direction?

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Have a question from materials science class:

Using appropriate sketches determine the direction of the intersection of the following planes: (110) and (101)

do I need to actually draw out these planes, then find 2 vectors for each. take the cross product, then use the lines perpindicular to the planes to find th edirection? Or can i simply use the planes to get the cross product?


<br /> \begin{array}{*{20}c}<br /> i &amp; j &amp; k \\<br /> 1 &amp; 1 &amp; 0 \\<br /> 1 &amp; 0 &amp; 1 \\<br /> \end{array}<br />
 
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You're going to have to let us know what your notation means. There are many different ways to specify a plane. What exactly do those numbers represent?
 
this is the standard materials science notation for a lattice plane. You take the inverse of the individual coordinates and that gives you the x, y and z intercepts.
 
Well, if I understand your notation correctly then the planes are very simple and just drawing a picture will do it without calculation. You need to find 2 points that are on both planes, and one of them is a given: (1, 0, 0).
 
I'm pretty sure that they want us to use the cross-product.
 
Then why would they ask you to draw a picture? The second point is obvious from the picture.
 
The notation (ABC) refers to the plane with equation Ax+ By+ Cz= 1 so that (1/A, 0, 0), (0, 1/B, 0), (0, 0, 1/C) are the x,y,z intercepts.

Yes, Ai+ Bj+ Ck is perpendicular to that plane. Also Xi+ Yj+ Zk is perpendicular to the plane Xx+ Yy+ Zz= 1 and so their cross product it perpendicular to both and points in the direction of their line of intersection. You still need to find a point on that line.
 
we are using miller indicies, I'm pretty sure that taking the cross product is right
 
Yes, that's what I just said! Have you solved the problem yet?
 
  • #10
I took the crossproduct of the two planes and believe that the vector resulting from the cross product is the direction of the intersection, right?
 
  • #11
JSBeckton said:
I took the crossproduct of the two planes and believe that the vector resulting from the cross product is the direction of the intersection, right?
Yes! You've been told that three times now. Have you solved the problem yet?

By the way- please don't say "the crossproduct of the two planes". You don't take the crossproduct of planes- crossproduct is only defined for vectors. I know you meant "the crossproduct of the normal vectors to the two planes" but I would hope that materials science values precise statement as much as mathematics.
 
  • #12
For the pairs of planes: (110) and (101)

I crossed the normal vectors of the two planes and got <br /> \left[ {1\overline 1 \overline 1 } \right]<br />

I am sorry for the confusion but I did not understand why you said that I still needed to find a point? Thats why I asked if my method was correct again, because I did not find any point.
 
  • #13
Oh those material scientists! Why can't they use the same notation as normal people?

You have determined (I think!) that a vector perpendicular to both planes is i- j- k (those overlines are negatives? WHY??)

You now know a vector pointing in the direction of the line of intersection. But there are an infinite number of (parallel) lines pointing in that direction. You must know a point on the line to choose between them.

Your two planes are x+ y= 1 and x+ z= 1. Any point on the intersection line must satisfy both of those. But there are an infinite number of points on the line and you only need ONE. Let x= 0 (just because it is easy). Then y= 1 and z= 1. The point (0, 1, 1) satisfies both equations so it lies on both planes. It must, then, be on the line of intersection.

Now what are parametric equations for the line through (0, 1, 1) in the direction of the vector i- j- k?
 
  • #14
Well, he only was asking for the direction of the line of intersection so I don't think he actually needs the line.
 
  • #15
Yes, just the direction. By the way I agree with you about the notation, I am Mech Eng major, so hopefully I won't see this a whole lot after this class.
 
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