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Miller Index Question

  1. Oct 23, 2006 #1
    I was asked the write down the Miller indices for a plane.

    The question is further described in


    I took point A as an origin.

    For the plane "opn", the intersections with X axis and Y axis are a/2 and -a respectively.

    Then I really can't figure out the intersection of plane "opn" with the Z axis.

    Is there any one can help me with the question or just simply give me some hints?

    I'd really appreciate your help!!
    Last edited: Oct 23, 2006
  2. jcsd
  3. Oct 23, 2006 #2


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    I'm just going by vague recollection and what I see here


    but I think you might need to work with a plane parallel to "opn". The plane in your diagram intersects all three axes at the origin. In the link I posted, if you click on the 111 example, or the words (not the picture; that is misdirected) for the 101 example or the 102 example for the simple cubic, the planes that are used do not intersect the origin, but surely in each case there is a parallel plane that contains the origin. The plane through the origin is not used.
  4. Oct 23, 2006 #3
    Is there any simplest plane parallel to "opn"?

    I think "opn" intercept with the horizontal X-Y plane by a particular angle.
    And there shouldn't be any simple plane parallel to "opn".

    The 3D vision is so tough to me! :cry:
  5. Oct 23, 2006 #4


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    It is not simple, but I think you can do it this way: Find the coordinates of o, p, and n. Since o is the origin, you can multiply the coordinates of p and n by a scale factor to find new points in the plane with integer coordinates (this is not necessary, but it will make life easier). Translate the plane by adding a constant to any one coordinate of each point (or any two or all three; it does not matter). The new set of coordinates will be for a plane parallel to the original plane but no longer going through the origin. Find the intercepts for the new plane and calculate the indices.

    Finding intercepts is not trivial unless you have the equation of the plane. Getting the equation from three points is a bit tedious, but it is easily programmed. There is a calculator at this site, but it only works for integer coordinates


    For your plane, here is what I got for coordinates

    o: (0, 0, 0)
    p: (2a/3, a/2, a)
    n: (a/2, a, 0)

    Scaling the ccordinates of p and n to integers gives a new set of points in the same plane

    o: (0, 0, 0)
    p: (4a, 3a, 6a)
    n: (a, 2a, 0)

    Adding a constant (a) to the y coordinate of each point gives

    o: (0, a, 0)
    p: (4a, 4a, 6a)
    n: (a, 3a, 0)

    These points lie in a plane parallel to the original plane. Any such plane will do.

    Find the equation of the plane through these points (a is just a scale factor for the whole coordinate system, so you can set it to 1 to calculate the equation).

    12x - 6y - 5z + 6 = 0

    Solve for the intercepts and you are on your way.

    If you don't trust the translation of the plane, play with the calculator to convince yourself that adding a constant to any one coordinate of all three points gives you a plane with the same coefficients; only the constant term changes. All such planes are parallel and their intercepts are proportional.
    Last edited: Oct 23, 2006
  6. Oct 23, 2006 #5
    For 12x - 6y - 5z + 6 = 0, the three intercepts yield -1/2, 1 and 6/5.
    So the plane "opn" should be (-2 1 5/6).
    Am I doing it right?

    There's also a question I want to ask.

    If I reduced the point p to (2, 2, 3), I'd get a plane 2x - 1y - 1z + 1 = 0 which is different from 12x - 6y - 5z + 6 = 0.
    If you could turn p: (2/3, 1/2, 1) to p: (4, 3, 6).
    Why can't I turn p: (4, 4, 6) to p: (2, 2, 3)?
  7. Oct 23, 2006 #6


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    I think you need to clear the fractions to have Miller indices, so you need to multiply through by 6.

    You cannot turn (4, 4, 6) into (2, 2, 3) because (4, 4, 6) is a point in the translated plane that does not contain the origin. When the plane contains the origin, the position vector to each point in the plane lies in the plane. Scaling the vector (all three cordinates) gives you a new position in the plane. When the plane is translated, the position vector to a point in the plane is no longer in the plane. Scaling the vector takes you to a new point that is out of the plane. Moving one of the three points out of the plane gives you a new plane defined by the three points that is not parallel to the original. The new and old planes intersect along the line between the two stationary points.
  8. Oct 23, 2006 #7
    Got it!! I really appreciate your help!
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