Graphics Reference
In-Depth Information
z
Figure 2.31.
Example 2.5.2.3.
u
3
y
u
2
2
1
1
X
u
1
x
1
2
Ê
ˆ
10 0
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
u
u
u
Ê
ˆ
1
1
2
1
2
Á
Á
˜
˜
A
-
1
=
=
0
,
2
Ë
¯
1
2
1
2
3
0
-
Ë
¯
and that
e
i
A
-1
=
u
i
.
The approach used in the last two examples generalizes.
2.5.2.3. Example.
To find a rigid motion M that moves the origin to the point
p
=
(0,0,1) and the x-y plane to the plane
X
defined by x + y + 2z = 2. See Figure 2.31.
Solution.
All we have to do is to find an orthonormal basis (
u
1
,
u
2
,
u
3
), so that
u
1
and
u
2
are a basis for
X
. Then the motion defined by the frame F = (
u
1
,
u
2
,
u
3
,
p
) will do the
job. The equation for
X
tells us that
n
= (1,1,2) is a normal vector for
X
. There are many
ways to find a basis for
X
. Clearly, (2,0,-1) is a vector orthogonal to
n
. Therefore, let
n
n
1
6
1
5
1
30
==
(
)
(
)
(
)
u
11 2
,, ,
u
=
20 1
, ,
-
,
and
u
=¥=
u
u
- -
15 2
, ,
.
3
1
2
3
1
2.5.2.4. Example.
To find a rigid motion M which moves the point
p
= (0,0,1) to
the point
q
= (0,-1,0) and the plane
X
defined by x + y + 2z = 2 to the plane
Y
defined
by -x - y + z = 1.
Solution.
Let F be the frame defined in Example 2.5.2.3. The motion F
-1
will map
the plane
X
to the x-y plane and
p
to the origin. We simply need to define a frame
G = (
w
1
,
w
2
,
w
3
,
q
) that will send the x-y plane to the plane
Y
and the origin to
q
and
set M = GF
-1
. This is another problem like the one in Example 2.5.2.3. Let
1
3
1
2
1
6
(
)
(
)
(
)
w
=--
111
,,,
w
=
101
,,,
and
w
=¥= -
w
w
121
,,.
3
1
2
3
1
