Graphics Reference
In-Depth Information
Y
P
3
v
3
P
v
2
p
P
4
P
2
P
1
X
Z
Fig. 11.24.
A tetrahedron.
To demonstrate this, consider the tetrahedron shown in Figure 11.24. Now
the volume of a tetrahedron is given by
x
1
y
1
z
1
V
=
1
6
x
2
y
2
z
2
x
3
y
3
z
3
where [
x
1
y
1
z
1
]
T
,
[
x
2
y
2
z
2
]
T
,and[
x
3
y
3
z
3
]
T
are the three vectors extending
from the fourth vertex to the other three vertices. However, if we locate the
fourth vertex at the origin, (
x
1
,y
1
,z
1
)
,
(
x
2
,y
2
,z
2
)and(
x
3
,y
3
,z
3
) become
the coordinates of the three vertices.
Let's locate a point
P
(
x
P
,y
P
,z
P
) inside the tetrahedron with the follow-
ing barycentric definition
P
=
r
P
1
+
s
P
2
+
t
P
3
+
u
P
4
(11.15)
where
P
,
P
1
,
P
2
,
P
3
and
P
4
are the position vectors for
P, P
1
,P
2
,P
3
and
P
4
respectively.
The fourth barycentric term
u
P
4
can be omitted as
P
4
has coordinates
(0,0,0).
Therefore, we can state that the volume of the tetrahedron formed by the
three vectors
p
,
v
2
and
v
3
is given by
x
P
y
P
z
P
V
=
1
6
x
2
y
2
z
2
(11.16)
x
3
y
3
z
3
Substituting (11.14) in (11.15) we obtain
rx
1
+
sx
2
+
tx
3
ry
1
+
sy
2
+
ty
3
rz
1
+
sz
2
+
tz
3
V
=
1
6
x
2
y
2
z
2
(11.17)
x
3
y
3
z
3
which expands to
