Graphics Reference
In-Depth Information
6.5.5 Example.
Find the intersection
I
of the lines
L
and
L
¢ in
R
3
, where
L
con-
tains the points
A
(0,3,1) and
B
(2,3,3) and
L
¢ contains the points
C
(2,1,1) and
D
(0,5,3).
Solution.
A direction vector for
L
¢ is
CD
= (-2,4,2). Two orthogonal vectors normal
to
L
¢ are
N
1
= (2,1,0) and
N
2
= (-1,2,-5). Then
(
)
(
)
AC • N
AB • N
220 210
202
,
, •, ,
,, • ,,
-
1
1
t
1
=
=
=
12
(
)
(
)
210
and
(
)
(
)
AC • N
AB • N
220
,
-
,
•,
--
125
,
2
2
t
2
=
=
=
12
.
(
)
(
)
202
,, •
--
12
,,
5
Since t
1
= t
2
, the intersection
I
exists and
I
=
(
)
+
( (
)
=
(
)
031
,,
12 202
,,
132
,, .
The reader may wonder where
N
1
and
N
2
came from. One can either assume that
they were given or find two such vectors as follows: Take any two vectors orthogonal
to
CD
and then apply the Gram-Schmidt orthogonalization algorithm to these. An
alternate solution to this problem is to observe that
L
and
L
¢ intersect if and only if
they lie in a plane
X
. A normal to this plane is
N
=
AB
¥
CD
. Therefore the lines inter-
sect if
C
and
D
satisfy the plane equation
(
)
= 0
NPA
•
-
.
To actually find the intersection, find a normal
N
1
to
L
in
X
(using, for example, the
Gram-Schmidt algorithm on
N
,
AB
,
CD
). Now the problem is to find the intersection
of a line
L
¢ with the hyperplane
L
in
X
with normal
N
1
. The formula from Problem
6.5.2 applies to this variation of the intersection problem also.
We should point out that Formula 6.6.5 in Section 6.6 provides a more direct
formula for the intersection of two lines in
R
3
.
6.5.6 Problem.
To find the intersection
I
of three planes
X
i
, i = 1,2,3, which are
defined by points
p
i
and normal vectors
N
i
. We assume that the vectors
N
i
are linearly
independent, that is, the planes are pairwise nonparallel.
Solution.
One needs a simultaneous solution to the equations
N
i
•(
p
-
p
i
) = 0, i =
1,2,3. The solution is
(
)
(
)
+
(
)
(
)
+
(
)
(
)
pNN N
•
¥
p NN N
•
¥
pNN N
•
¥
1
1
2
3
2
2
3
1
3
3
1
2
I
=
.
(6.12)
(
)
NNN
•
¥
1
2
3
