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(3) Let s q ( z ) = z M q , where M q is the 3 ¥ 3 matrix that represents the linear trans-
formation s q . If
q
=+ + +
r
a
i
b
j
c
k
and
q
=
1,
then
2
2
Ê
12 2 2 2 2 2
22 122 22
22
--
b
c
rc
+
ab
ac
-
rb
ˆ
Á
Á
˜
˜
2
2
M
q =
ab
-
rc
-
c
-
a
ra
+
bc
.
2
2
Ë
rb
+
ac
2 2 122
bc
-
ra
-
a
-
b
¯
Proof.
To prove (1), note that
1
-
1
qz q
=
qq qz q
by Proposition 20.2.4(3). Repeated use of (7), (4), and (3) in Proposition 20.2.3 shows
that
(
)
qzq
+
qzq
qz z q
+
(
) =
re q z q
=
=
0,
2
2
and (1) is proved.
The first part of (2) follows from the fact that
() -
() =
(
)
-
1
-
1
s
z
s
z
q z
-
z
q
=
q z
-
z
q
=
z z
.
q
1
q
2
1
2
1
2
12
Next, observe that there is no loss in generality if we assume that q is a unit quater-
nion because the map s q stays the same if we replace q by any nonzero real multiple
of q . Therefore, let
q
=
cos
q
+
sin
q
n
,
where q and n are as described in Proposition 20.3.1.
Claim.
s q fixes every point on the line L .
To prove the Claim, observe that
() =
(
)
(
)
s
q n
cos
q
+
sin
q
n n
cos
q
-
sin
q
n
2
2
2
2
3
=
cos
q
n
-
cos
q
sin
q
n
+
sin
q
cos
q
n
-
sin
q
n
(
)
2
2
2
=
cos
q
-
sin
q
nn
=
n
,
since n 2 =-1. This clearly proves the Claim.
To prove the rest of (2), let
ni
=+ +
nn n
j
k
,
i
2
3
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