Page 96 - Materials Science and Engineering An Introduction

68 • Chapter 3 / The Structure of Crystalline Solids

EXAMPLE PROBLEM 3.7

Determination of Directional Indices z
Determine the indices for the direction shown in
the accompanying figure.
x = 0a
2
Solution y = b
2
z = c/2
It is first necessary to take note of the vector tail O 2
and head coordinates. From the illustration, tail c y
coordinates are as follows: a
x = a
1
x 1 = a y 1 = 0b z 1 = 0c y = 0b b
1
z = 0c
1 x
For the head coordinates,
x 2 = 0a y 2 = b z 2 = c/2
Now taking point coordinate differences,
x 2 - x 1 = 0a - a = -a
y 2 - y 1 = b - 0b = b
z 2 - z 1 = c/2 - 0c = c/2
It is now possible to use Equations 3.10a through 3.10c to compute values of u, v, and w.
However, because the z 2 z 1 difference is a fraction (i.e., c/2), we anticipate that in order to
have integer values for the three indices, it is necessary to assign n a value of 2. Thus,

-a
x 2 - x 1
u = na b = 2a b = -2
a a
y 2 - y 1 b
v = na b = 2a b = 2
b b
c>2
z 2 - z 1
w = na b = 2a b = 1
c c
And, finally enclosure of the 2, 2, and 1 indices in brackets leads to [221] as the direction
designation. 4
This procedure is summarized as follows:

x y z
Head coordinates (x 2 , y 2 , z 2 ,) 0a b c/2
Tail coordinates (x 1 , y 1 , z 1 ,) a 0b 0c
Coordinate differences a b c/2
Calculated values of u, v, and w u 2 v 2 w 1
Enclosure [221]

4 If these u, v, and w values are not integers, it is necessary to choose another value for n.