Graphics Reference
In-Depth Information
Figure 4.16.
Uniqueness of closest point
x
in plane.
p
x
X
xx¢
x¢
If
X
and
Y
are transverse planes in
R
n
with
4.5.13. Theorem.
dim
XY
+
dim
£ n
,
then there are
unique
points
x
Œ
X
and
y
Œ
Y
, so that
(
)
==
(
)
dist
xy
,
xy
dist
XY
,
.
The points
x
and
y
are defined by the condition that the vector
xy
is orthogonal to
both
X
and
Y
.
Proof.
It is easy to show that proving the theorem reduces to proving the following
two cases:
Case 1:
dim
X
+ dim
Y
= n
Case 2:
dim
X
+ dim
Y
= n - 1
In Case 1 the planes intersect in a single point. We sketch the proof for Case 2
and leave the rest to the reader (Exercise 4.5.4). Case 2 applies to skew lines in
R
3
,
for example. The fact that there are points
x
and
y
at which the distance between
X
and
Y
is minimized and the fact that
xy
is orthogonal to
X
and
Y
is proved just like
in Theorem 4.5.12 by parameterizing points of
X
and
Y
via tuples
s
and
t
, expressing
the distance between points of
X
and
Y
as a function d(
s
,
t
) of the variables
s
and
t
,
and looking for the critical points of that function by setting the partial derivatives of
d(
s
,
t
) to zero. It remains to show that the orthogonality condition defines
x
and
y
uniquely.
Assume that there are other points
x
¢Œ
X
and
y
¢Œ
Y
with the property that the
vector
x
¢
y
¢ is orthogonal to both
X
and
Y
. See Figure 4.17. Our hypothesis about the
dimensions of
X
and
Y
implies that all vectors that are orthogonal to
both X
and
Y
are multiples of each other. Therefore,
xy
and
x
¢
y
¢ are parallel. If (
x
,
y
) π (
x
¢,
y
¢), then
the points
x
,
y
,
x
¢, and
y
¢ lie in a two-dimensional plane
Z
. Assume that
x
π
x
¢ and
y
π
y
¢ and consider the line
L
through
x
and
x
¢ and the line
L
¢ through
y
and
y
¢. (The
