Solution: Construct a cubic unit cell; select its origin and the three crystallographic
axes. Follow the above said procedure and draw the given plane by shifting the
origin from O to O′ as shown in Fig. 4.11. The Miller indices of directions are also
shown. They are: 11 1
½
Š; 111
½
Š and their opposite 1 11
½
Šand 1 1 1
½
Š:
Example 7 Draw the planes (110) and (111) in a cubic unit cell. Determine the
Miller indices of the line of their intersection.
Solution: Consider the conventional orthogonal axes and construct a unit cell with
O as the origin. Follow the above said procedure and draw the given planes as
shown in Fig. 4.12. The Miller indices of the common direction of the two planes
are shown as 1 10
½
Š or 110
½ Š.
Example 8 Determine the directions connecting the origin with the nearest lattice
sites whose coordinates are
1
3
1
3
2
3
 Ã
Â
à ;
1
3
2
3
1
3
 Ã
Â
Ã
and
1
2
1
2 0
Â
Ã
Â
à : Show them in a cubic unit
cell.
Solution: Given: The nearest lattice sites whose fractional coordinates are:
1
3
1
3
2
3
 Ã
Â
à ;
1
3
2
3
1
3
 Ã
Â
Ã
and
1
2
1
2 0
Â
Ã
Â
à : Removing fractions by multiplying with 3, 3 and 2
respectively, their position coordinates and the directions become:
1
3
1
3
2
3
! !
! 112
½
Š
½
Š!½112Š
1
3
2
3
1
3
! !
! 121
½
Š
½
Š!½121Š
Fig. 4.11 1 10
ð
Þ plane and
the directions that lie on this
plane
Fig. 4.10 [112], [121] and
[220] directions
148
4 Unit Cell Representations of Miller Indices