Graphics Reference
In-Depth Information
r
is parallel to
n
, therefore
r
=
λ
n
(12.2)
where
λ
is some scalar.
Taking the scalar product of (12.2)
.
r
=
λ
n
.
n
n
(12.3)
but as
r
=
q
−
p
(12.4)
.
r
=
n
.
q
−
n
.
p
n
(12.5)
substituting (12.1) and (12.3) in (12.5) we obtain
.
.
λ
n
n
=
−
c
−
n
p
(12.6)
therefore
.
p
+
c
)
n
λ
=
−
(
n
.
n
From (12.4) we get
q
=
p
+
r
(12.7)
substituting (12.2) in (12.7) we obtain the position vector for
Q
q
=
p
+
λ
n
The distance
PQ
must be the magnitude of
r
:
PQ
=
r
=
λ
n
Let's test this result with an example where the answer can be predicted.
Figure 12.9 shows a line whose equation is
x
+
y
1 = 0, and the associated
point
P
(1,1). By inspection, the nearest point is
Q
(
2
,
−
1
2
) and the distance
PQ
=0
.
7071.
From the line equation
a
=1
b
=1
c
=
−
1
therefore
2
−
1
1
2
λ
=
−
=
−
2
therefore
1
2
×
1=
1
x
Q
=
x
P
+
λx
n
=1
−
2
1
2
×
1=
1
y
Q
=
y
P
+
λy
n
=1
−
2
The nearest point is
Q
(
2
,
1
1
2
) and the distance is
PQ
=
λ
n
=
2
i
+
j
=
0
.
7071
