Graphics Reference
In-Depth Information
¢x ,
¢y t,
¢z .
But (x¢,y¢,z¢) belongs to
X
if and only if
2
xt
-+=,
yt
zt
6
that is,
6
t
=
.
2
xyz
-+
In other words, S is defined by the equations
x
xyz
6
y
xyz
6
z
xyz
6
x
¢=
,
y
¢=
,
z
¢=
.
2
-+
2
-+
2
-+
A second way to solve the problem is to transform the problem into simpler ones.
The y-, x-, and z intercepts of
X
are the points
A
(0,-6,0),
B
(3,0,0), and
C
(0,0,6), respec-
tively. See Figure 3.16. To find the map S, we shall reduce this problem to one we can
handle by describing S as a composition of three maps for which we already know
how to derive the equations. First of all, let T be the linear transformation of
R
3
, which
sends
A
to
A
¢(0,-6,6),
B
to
B
¢(3,0,6), and
C
to
C
¢=
C
. It is easy to see that T sends the
plane
X
to the plane defined by z = 6. Next, let R be the translation that translates this
plane to the x-y coordinate plane. Then RT sends the origin to (0,0,-6). It follows that
S = T
-1
R
-1
C
1/6
RT. The maps have the following matrices (with respect to homogeneous
coordinates):
10 2 0
01 10
00 1 0
00 0 1
10 20
01 1 0
00 1 0
00 0 1
-
Ê
ˆ
Ê
ˆ
Á
Á
Á
˜
˜
˜
Á
Á
Á
˜
˜
˜
-
-
1
T
´
T
´
Ë
¯
Ë
¯
z
C(0,0,6)
y
x
B(3,0,0)
A(0,-6,0)
Figure 3.16.
The plane in Example 3.5.1.1.
