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n
=
0, then the ray
is parallel to the plane; that means it's either contained in the plane (in which
case there are infinitely many intersections) or disjoint from the plane, in which
case there are none. In the first case, both the numerator and denominator are zero;
in the second, only the denominator is.
These two concerns, it turns out, are identical! If
d
·
Inline Exercise 7.12:
What happens if we solve for
t
and find that it's negative?
That can happen if either the numerator or denominator is negative, and the
other is positive. Describe each of these situations geometrically, as in “In the
first case,
P
is on the positive side of the plane (i.e., the half-space into which
n
points), and . . .”
Let's carry out the computation again, in this case with a plane specified by a
point
Q
that lies in the plane, and two vectors,
u
and
v
, that lie in the plane and
are linearly independent. Now, points in the plane can be written in the form
Q
+
α
u
+
β
v
(7.96)
for some real numbers
. This version of the problem seems distinctly
more difficult, in the sense that a solution will give us
t
,
α
and
β
α
β
, and
; that impression
is partially correct, as we'll see.
We want to find a value
t
≥
0 with the property that
P
+
t
d
=
Q
+
α
u
+
β
v
(7.97)
for
some
values
. The first algebraic steps are similar. Move the points
around so that a difference of points appears, and we'll be working only with
vectors:
α
and
β
P
+
t
d
=
Q
+
α
u
+
β
v
(7.98)
P
−
Q
+
t
d
=
α
u
+
β
v
(7.99)
P
−
Q
=
α
u
+
β
v
−
t
d
.
(7.100)
Letting
h
=
P
Q
, we can see that now the problem is “Express the vector
h
as a
linear combination of
u
,
v
, and
d
.” We could let
M
be the matrix whose columns
are these three vectors, and the solution becomes
αβ−
−
t
=
M
−
1
(
P
−
Q
)
.
(7.101)
To implement this, we'd need to invert a 3
3 matrix, which is not difficult, but
masks some of the essential features of the problem. First, it's possible that
M
is
not invertible, but even if this happens, there may be a solution to
M
αβ−
×
t
=(
P
−
Q
)
.
(7.102)
(
M
is noninvertible when the ray is parallel to the plane; the solution exists when
the ray lies
in
the plane, and indeed, in this case there are infinitely many solu-
tions.) Second, the inversion must be redone for each new ray we want to intersect
with the plane; that makes ray-plane intersection computationally intensive, which
is bad.
