a 2 ¼ a 1 + c 1 , b 2 ¼ a 1 + b 1 , c 2 ¼ b 1 þ c 1
Now, using these equations, we can write the following transformation equations
H ¼ 1h þ 0k þ 1l
K ¼ 1h þ 1k þ 0l
L ¼ 0h þ 1k þ 1l
ð5:21Þ
The matrix form of these equations is
H
K
L
0
@
1
A =
1 0 1
1 1 0
0 1 1
0
@
1
A
h
k
l
0
@
1
A
ð5:22Þ
The determinant of the matrix in Eq. 5.22, D ¼ 2: Therefore Using Eq. 5.3, we
can obtain the volume relationship between the two unit cells
V BCC ¼ 2V P
The reverse relationships can be found by determining the inverse of the matrix
in Eq. 5.22, that is,
1 0 1
1 1 0
0 1 1
0
@
1
A
À1
=
1
1 À1
À1 1
1
1 À1 1
0
@
1
A
Fig. 5.5 BCC and its
primitive unit cell axes
5.2 Transformation of Indices of Crystal Planes (Unit Cell)
189