Graphics Reference
In-Depth Information
x
1
1
z
1
b
=
x
2
1
z
2
x
3
1
z
3
x
1
y
1
1
c
=
x
2
y
2
1
x
3
y
3
1
and
d
=
−
(
ax
1
+
by
1
+
cz
1
)
thus
120
104
112
a
=
=0
010
014
312
b
=
=12
021
001
311
c
=
=6
d
=
−
(0
×
0+12
×
2+6
×
0) =
−
24
therefore, the plane equation is
12
y
+6
z
= 24
(11.11)
If we substitute a point (
x
0
,y
0
,z
0
) in the LHS of (11.11) and obtain a value
of 24, then the point is on the plane.
The following table shows various values of
r
,
s
and
t
, and the correspond-
ing position of
P
0
. The table also confirms that
P
0
is always on the plane
containing the three points.
r
s
t
x
0
y
0
z
0
12
y
0
+6
z
0
1
0
0
0
2
0
24
0
1
0
0
0
4
24
0
0
1
3
1
2
24
1
4
1
4
1
2
1
2
1
2
24
1
2
1
2
1
2
1
2
0
3
24
1
2
1
2
0
0
1
2
24
1
3
1
3
1
3
1
1
2
24
Now we are in a position to test whether a point is inside, on the boundary
or outside a 3D triangle.
