1.9 the planes (100), To see why, we can make use of elementary vector algebra relationships discussed in Appendix 1, Section A1.1. Consider the plane (hkl) shown in Figure 1.10. The vector normal to this plane, n, must be parallel to the cross product
(1.7)
and so, after some straightforward mathematical manipulation, making use of the identities
it is apparent that n is parallel to the vector ha* + kb* + lc*. That is:
(1.8)
for a constant of proportionality, ξ. The vectors a*, b* and c* in Eq. (1.8) are termed reciprocal lattice vectors, defined through the equations:
(1.9)
If the normal to the (hkl) set of planes is simply taken to be the vector
(1.10)
the magnitude of n is inversely proportional to the spacing of the hkl planes; that is, it is inversely proportional to the distance OP in Figure 1.10 (Section A1.2), irrespective of the orientations of the x‐, y‐ and z‐axes with respect to one another.
Furthermore, it is evident that the scalar product of a normal to a set of planes, n, with a vector r = [uvw] lying on one of these planes must be zero. That is, r · n = 0. Writing out this dot product explicitly, we obtain the result:
which is the Weiss zone law. This demonstrates that the Weiss zone law is a scalar product between two vectors, one of which lies in one of a set of planes and the other of which is normal to the set of planes.
Given the indices of any two planes, say (h1k1l1) and (h2k2l2), the indices of the zone [uvw] in which they lie are found by solving the simultaneous equations:
(1.11)
(1.12)
Since it is only the ratios u : v : w which are of interest, these equations can be solved to give:
(1.13)
There are other methods of producing the same result. For example, we could write down the planes in the form:
We then cross out the first and the last columns and evaluate the 2 × 2 determinants from (i) the second and third columns, (ii) the third and fourth columns, and (iii) the fourth and fifth columns:
(1.14)
Therefore, we find [uvw] = [k1l2 − k2l1, l1h2 − l2h1, h1k2 − h2k1]. A third method is to evaluate the determinant
(1.15)
to determine [uvw]. The result is [uvw] = (k1l2 − k2l1) a + (l1h2 − l2h1) b + (h1k2 − h2k1) c.
Thus, for example, supposing (h1k1l1) = (112) and (h2k2l2) = (
and so [uvw] = [−5, −5, 5] ≡
(1.16)
(1.17)
Using a similar method to the one used to produce Eq. (1.14), we draw up the three 2 × 2 determinants as follows:
(1.18)
to find that (hkl) = (v1w2 − v2w1, w1u2 − w2u1, u1v2 − u2v1). The method equivalent to Eq. (1.15) is to evaluate the determinant:
(1.19)
It is also evident from Eqs. (1.11) and (1.12),
