Graphics Reference
In-Depth Information
(
)
0 =
OI
•
N
=
P
+
t
PQ
-
O
•
N
.
Solving for t leads to
t =
PO • N
PQ • N
.
(6.9)
Of course, t is only defined if
PQ
and
N
are not orthogonal, that is,
L
is not parallel
to
X
. The parallel case is again a tricky case for a computer program. One needs to
determine whether or not
L
lies in
X
.
6.5.3 Example.
Find the intersection
I
of the line
L
containing points
P
(1,1,1) and
Q
(2,0,3) with the plane
X
which has normal vector
N
= (-1,2,0) and contains
O
(3,1,2).
Solution.
Substituting into (6.9) gives
(
)
(
)
201
,, •
-
120
,,
t =
=
23
.
(
)
(
)
112
,
-
,
• ,
-
120
,
Therefore
=
(
)
+
(
)
(
)
=
(
)
I
=+
PPQ
111
t
,,
2 3 1
,
-
1 2
,
53 13 7 3
,
,
.
Problem 6.5.2 easily generalizes to the case where
X
is a hyperplane in
R
n
. It also
generalizes to
6.5.4 Problem.
Find the intersection of a line
L
with a k-dimensional plane
X
in
R
n
. Assume that
N
1
,
N
2
,..., and
N
n-k
are orthogonal normal vectors for
X
. Assume
also as before that
P
and
Q
are points on
L
and
O
is a point on
X
.
Solution.
Define numbers t
i
by
=
PO • N
PQ • N
i
i
t
i
.
(6.10)
The t
i
will be defined provided that
L
is not parallel to
X
, which is a special case that
must be treated separately. If t
1
= t
2
= ...= t
n-k
, then
L
intersects the plane
X
in the
point
IP PQ
=+t
i
.
(6.11)
