Similarly, for the second set of fractional coordinates, we have
L 2 ¼
0:200 À 0:300
ð
Þ
2 Â 6
2 þ 0:150 À 0:050
ð
Þ
2 Â 7
2 þ 0:333 þ 0:123
ð
Þ
2 Â8
2 þ
0 þ 0 þ 2 Â 8 Â 6 0:333 þ 0:123
ð
Þ0:200 À 0:300
ð
Þ cos 115
"
# 1=2
¼ À0:100
ð
Þ
2 Â36 þ 0:100
ð
Þ
2 Â49 þ 0:456
ð
Þ
2 Â64 þ 96 0:456
ð
Þ À0:100
ð
ÞðÀ0:423Þ
h
i 1=2
¼ 0:100 Â 36 þ 0:010 Â 49 þ 0:208 Â 64 þ 1:852
½
Š
1=2
¼ 0:360 þ 0:490 þ 13:308 þ 1:852
½
Š
1=2
¼ 16:01
½
Š
1=2 ¼ 4:00 ˚
A
Example 5 A unit cell has the following dimensions: a = 5 Å, b = 7 Å, c = 9 Å,
a = 120°, b = 85° and c = 75°. Calculate the length of the vectors with the help of
following components: (i) [0.2, −0.3, 0.6] and (ii) [0.5, 0.7, −0.1].
Solution: Given: a = 5 Å, b = 7 Å, c = 9 Å, a = 120°, b = 85° and c = 75°.
Vectors components: (i) x 2 À x 1 ¼ 0:2; y 2 À y 1 ¼ À0:3; z 2 À z 1 ¼ 0:6;
(ii) x 2 À x 1 ¼ 0:5; y 2 À y 1 ¼ 0:7; z 2 À z 1 ¼ À0:1:
Making use of Eq. 3.8 in slightly modified form, we can obtain the length of the
vectors for the given components as:
U ¼ u 1 a
ð Þ
2 þ u 2 b
ð Þ
2 þ u 3 c
ð Þ
2 þ 2u 1 u 2 ab cos c þ 2u 2 u 3 bc cos a þ 2u 3 u 1 ca cos b
h
i 1=2
¼
5 Â 0:2
ð
Þ
2 þ 7 Â À0:3
ð
Þ
2 þ 9 Â 0:6
ð
Þ
2 þ 2 Â 0:2 Â À0:3 Â 5 Â 7 cos 75
þ
2 Â À0:3 Â 0:6 Â 7 Â 9 Â cos 120
þ 2 Â 0:6 Â 0:2 Â 9 Â 5 Â cos 85
"
# 1=2
¼ 1:0
ð Þ
2 þ À2:1
ð
Þ
2 þ 5:4
ð Þ
2 À4:2 Â 0:259 À 22:68 Â À0:5 þ 10:8 Â 0:087
h
i 1=2
¼ 1 þ 4:41 þ 29:16 À 1:087 þ 11:34 þ 0:94
½
Š
1=2
¼ 45:764
½
Š
1=2 ¼ 6:76 ˚
A
Similarly,
V ¼
5 Â 0:5
ð
Þ
2 þ 7 Â 0:7
ð
Þ
2 þ 9 Â À0:1
ð
Þ
2 þ 2 Â 0:5 Â 0:7 Â 5 Â 7 cos 75
þ
2 Â 0:7 Â À0:1 Â 7 Â 9 Â cos 120
þ 2 Â À0:1 Â 0:5 Â 9 Â 5 Â cos 85
"
# 1=2
¼ 2:5
ð Þ
2 þ 4:9
ð Þ
2 þ À0:9
ð
Þ
2 þ 24:5 Â 0:259 À 8:82 Â À0:5 À 4:5 Â 0:087
h
i 1=2
¼ 6:25 þ 24:01 þ 0:81 þ 6:34 þ 4:41 À 0:39
½
Š
1=2
¼ 41:43
½
Š
1=2 ¼ 6:44 ˚
A
108
3 Unit Cell Calculations