Ha 1 þ Ka 2 þ Lc ¼ ha 1 þ ka 2 þ ia 3 + lc
ð4:1Þ
Since, the axes a 1 , a 2 and a 3 are related to each other by a rotation of 120°, the
vector sum
a 1 þ a 2 þ a 3 ¼ 0 or a 3 ¼ À a 1 þ a 2
ð
Þ
Substituting the value of a 3 , Eq. 4.1 becomes
Ha 1 þ Ka 2 þ Lc ¼ ha 1 þ ka 2 - i (a 1 + a 2 ) + lc
ð4:2Þ
Comparing the coefficients of a 1 , a 2 and c in Eq. 4.2, we obtain
H ¼ h À i; K ¼ k À i and L ¼ 1
An inverse transformation gives us
h ¼
1
3
2H À K
ð
Þ ; k =
1
3
2K À H
ð
Þ ; i ¼ À h þ k
ð
޼1
3
H þ K
ð
Þ,l ¼ L
Figure 4.15 shows the Miller–Bravais indices of principal directions in the basal
plane of hexagonal crystal system.
Solved Examples
Example 1 Change the three index system of Miller indices (310), 123
ð
Þ, (011),
(346), and 4 23
ð
Þ into four index system of Miller–Bravais indices.
Solution: We can determine the value of the index i and hence the Miller–Bravais
indices using the formula i = –( h + k). The Miller–Bravais indices are shown in
the right column.
Fig. 4.15 Miller–Bravais
indices of principal directions
in the basal plane of
hexagonal crystal system
4.3 Miller–Bravais Indices
151