Graphics Reference
In-Depth Information
between
P
and the point
Q
corresponding to the orthogonal projection of
P
onto
L
.
Consequently, the line
PQ
is orthogonal to
L
. Because this reasoning holds for both
L
1
(
s
) with respect to
L
2
and
L
2
(
t
) with respect to
L
1
,
v
must be perpendicular to both
lines. For nonparallel lines,
v
is unique.
The problem is now finding values for
s
and
t
satisfying these two perpendicularity
constraints:
d
1
·
v
(
s
,
t
)
=
0
d
2
·
v
(
s
,
t
)
=
0.
Substituting the parametric equation for
v
(
s
,
t
) gives:
d
1
·
(
L
1
(
s
)
−
L
2
(
t
))
=
d
1
·
((
P
1
−
P
2
)
+
s
d
1
−
t
d
2
)
=
0
d
2
·
(
L
1
(
s
)
−
L
2
(
t
))
=
d
2
·
((
P
1
−
P
2
)
+
s
d
1
−
t
d
2
)
=
0.
This can be expressed as the 2
×
2 system of linear equations
(
d
1
·
d
1
)
s
−
(
d
1
·
d
2
)
t
=−
(
d
1
·
r
)
(
d
2
·
d
1
)
s
−
(
d
2
·
d
2
)
t
=−
(
d
2
·
r
),
where
r
P
2
.
Written symbolically, in matrix notation, this corresponds to
=
P
1
−
a
s
t
−
,
−
b
c
=
b
−
e
−
f
where
a
=
d
1
·
d
1
,
b
=
d
1
·
d
2
,
c
=
d
1
·
r
,
e
=
d
2
·
d
2
, and
f
=
d
2
·
r
. This system of
equations is solved, for example, using Cramer's rule to give
=
−
s
(
bf
ce
)/
d
=
−
t
(
af
bc
)/
d
,
d
1
d
2
(
d
1
d
2
cos(
2
2
b
2
. Note that
d
))
2
where
d
=
ae
−
≥
0, in that
d
=
−
θ
=
(
d
1
d
2
sin(
))
2
. When
d
0, the two lines are parallel, which must be han-
dled separately. In this case, any point
P
can be selected on the one line. On the
other line, the point closest to
P
is selected using the projection method described in
Section 5.1.2.
θ
=
