Graphics Reference
In-Depth Information
As T x
T
y
T
z
T
must satisfy both plane equations,
n
1
·
t
=
d
1
and
n
2
·
t
=
d
2
Furthermore, as
v
and
t
are orthogonal,
v
·
t
=
0
These three vector equations can be combined and written as
⎡
⎤
⎡
⎤
⎡
⎤
d
1
d
2
0
a
1
b
1
c
1
a
2
b
2
c
2
a
v
b
v
c
v
x
T
y
T
z
T
⎣
⎦
=
⎣
⎦
·
⎣
⎦
and solved using Cramer's rule, where
d
1
b
1
c
1
d
2
b
2
c
2
0 b
v
c
v
a
1
d
1
c
1
a
2
d
2
c
2
a
v
0 c
v
a
1
b
1
d
1
a
2
b
2
d
2
a
v
b
v
0
x
T
=
y
T
=
z
T
=
DET
DET
DET
a
1
b
1
c
1
a
2
b
2
c
2
a
v
b
v
c
v
DET
=
and
i j k
a
1
b
1
c
1
a
2
b
2
c
2
v
=
To illustrate this technique, let's consider a scenario whose outcome can be predicted as shown
in Fig. 6.27, where two planes intersect along the line
p
=
t
+
v
, where
t
=
j
and
v
=
i
.
The plane equations are
y
−
z
=
1
and y
+
z
=
1
Therefore,
n
1
=
j
−
k
and
n
2
=
j
+
k
and
i j k
01
v
=
n
1
×
n
2
=
1
01 1
−
=
2
i
01
1
01 1
20 0
−
DET
=
=
4
