Graphics Reference
In-Depth Information
2.2.4.2. Theorem.
Every motion M can be written uniquely in the form M = T
1
M
0
= M
0
T
2
, where T
i
is a translation and M
0
is a motion that fixes the origin, that is,
M
0
(
0
) =
0
.
Proof.
Define the translation T
1
by T
1
(
p
) =
p
+ M(
0
) and let M
1
= T
1
-1
M.
Clearly, M
1
(
0
) =
0
and M = T
1
M
1
. Similarly, if we define the translation T
2
by
T
2
(
p
) =
p
- M
-1
(
0
) and let M
2
= MT
-1
, then M
2
(
0
) =
0
and M = M
2
T
2
. Next, we
show that M
2
= M
1
. But by Theorem 2.2.4.1, the motions M
i
are linear transforma-
tions and so
()
=
()
=
()
+
()
MTM M M
p
p
p
0
11
1
and
(
)
()
=
()
=
()
-
-
1
()
MMT M MM
p
p
p
0
.
22
2
2
Therefore, for all
p
,
(
)
=
()
-
-
1
()
()
+
()
MMM
p
0
MM
p
0
.
2
2
1
The special case where
p
is
0
shows that -M
2
(M
-1
(
0
)) = M(
0
). In other words, we can
cancel those terms to get that M
2
(
p
) = M
1
(
p
). The uniqueness part of the theorem is
proved in a similar way.
2.2.4.3. Lemma.
Let M be a motion and assume that M(
0
) =
0
. Then M(
u
)•M(
v
) =
u
•
v
for all vectors
u
and
v
.
Proof.
The following string of equalities hold because M is a distance preserving
map and, by Theorem 2.2.4.1, also a linear transformation:
(
)
(
)
u•u
+
2
u•v
+
v•v
=
u
+
v u v
uv uv
uu
•
+
(
)
(
)
=
MM
MM MM MM
MM
+
•
+
()
()
+
()
()
+
()
()
=
•
2
uv
•
v
•
v
()
()
+
=
u•u
+
2
u
•
v
v•v
Now cancel the terms
u
•
u
and
v
•
v
from both sides and divide by 2.
2.2.4.4. Theorem.
If M is a motion, then
() ()
() ()
=
MM
AB
•
MM
AC B• C
for all points
A
,
B
, and
C
.
Proof.
By Theorem 2.2.4.2 we can express M in the form M = TM
0
, where T is a
translation and M
0
is a motion with M
0
(
0
) =
0
. It is easy to check that
() ()
=
()
()
MM
AB
M M
A B
0
0
