COMPUTATIONS FOR GEOMETRIC DESIGN 283
C (x 3 ,y 3 ,z 3 ) such that the normal to the plane is given by n= AB×BC. The volume of the tetrahedron
ABCDis computed by calculating the determinant Δ as
Δ =
1
1
1
1
111
222
333
444
xyz
xyz
xyz
xyz
(9.3)
PointD can be placed in three possible ways with respect to the plane ABC.
(a) Point D lies on the same side of the plane as its normal if Δ is negative.
(b) Point D lies on the opposite side of the normal if Δ is positive.
(c) Point D lies in the plane if Δ is zero.
The normal n = AB×BC is given by
n
rjk
= – – –
- – –
212121
323232
xxyyzz
xxyyzz
IfD is placed on the same side of the normal, then
nBD > 0 or =
- – –
- – –
- – –
1 > 0
424242
212121
323232
⋅Δ
xxyyzz
xxyyzz
xxyyzz
Performing a few row operations in Δ in Eq. (9.3) results in
Δ=
1
- – – 0
- – – 0
- – – 0
=
1
- – – 0
- – –
111
212121
323232
424242
111
424242
212121
xyz
xxyyzz
xxyyzz
xxyyzz
xyz
xxyyzz
xxyyzz 00
- – – 0
= –
323232
1
xxyyzz
Δ
Thus, for D on the side of the normal, Δ is negative and vice-versa.
Example 9.4. Three points A (1, 0, 0), B (0, 1, 0) and C (0, 0, 1) define a triangular lamina (Figure
9.10a). Find how the points: (a) D (0, 0, 0), (b) D (1, 1, 1), (c) D (1/3, 1/3, 1/3) and (d) D (1, 1, –1)
are placed with respect to this lamina.
(a)D (0, 0, 0). Find ΔABCD
Δ =
1001
0101
0011
0001
= 1 (> 0)