Graphics Reference
In-Depth Information
Figure 3.24 shows the graph of the line equation y
=
1. Thus,
n
ˆ
=
j
and c
=
1. The reference
point is P 12, and it is obvious that the reflected point must be Q10.
Using Eq. (3.50), we obtain
q
=
p
−
2
n
ˆ
·
p
−
c
n
ˆ
=
i
+
j
−
2
j
·
i
+
j
−
1
j
q
=
i
+
2
j
−
2
j
·
i
+
2
j
−
1
j
=
i
Therefore,
Q
=
10
3.9.2 The parametric form of the line equation
Figure 3.25 illustrates a line whose direction is given by
v
. T xyis some point on the line
with position vector
t
y
j
, and P is some arbitrary point with position vector
p
. Q is the
reflection of P with position vector
q
.
=
x
i
+
Y
n
Q
T
λ
v
ε
n
R
v
t
ε
n
q
r
P
p
O
X
Figure 3.25.
The following analysis exploits the fact that
−
PQ is perpendicular to
v
, and the objective is to
find
q
in terms of
t
p
, and
v
. We begin with
=
+
r
t
v
(3.52)
Using
v
, take the dot product of Eq. (3.52):
v
·
r
=
v
·
t
+
v
·
v
(3.53)
As
p
and
r
have a common projection on
v
, we have
v
·
p
=
v
·
r
(3.54)
Substituting Eq. (3.54) in Eq. (3.53) gives
v
·
p
=
v
·
t
+
v
·
v
Therefore,
·
−
·
−
v
p
t
v
p
t
=
=
(3.55)
·
2
v
v
v
