Graphics Reference
In-Depth Information
As
r
is mid-way between
p
and
q
, we have
=
1
2
p
+
1
2
q
r
and
q
=
2
r
−
p
(3.56)
Substituting Eq. (3.52) in Eq. (3.56) gives
q
=
2
t
+
v
−
p
If
v
is a unit vector
v
, then
ˆ
=ˆ
v
·
p
−
t
(3.57)
q
=
2
t
+
ˆ
v
−
p
(3.58)
Let's test Eq. (3.4) with a simple example.
Figure 3.26 illustrates a line whose direction is given by
ˆ
=
1
√
2
−
+
v
i
j
. T 10 is some point
on the line with position vector
t
=
i
, and P 11 is some arbitrary point with position vector
p
=
i
+
j
.
Y
P
(1,1)
p
1
n
v
T
t
Q
1
X
Figure 3.26.
Using Eq. (3.57) gives
1
√
2
1
√
2
=ˆ
v
·
p
−
t
=
−
i
+
j
·
i
+
j
−
i
=
1
√
2
=
Using Eq. (3.58) gives
2
i
j
1
√
2
1
√
2
q
=
2
t
+
v
ˆ
−
p
=
+
−
i
+
−
i
−
j
=
2
i
−
i
+
j
−
i
−
j
=
0
Therefore, the reflected point is Q00.
3.10 A line perpendicular to a line through a point
Here is another geometric problem. Given a line m and a point P, what is the equation of a
straight line that passes through P and is perpendicular to m? The algebraic solution is relatively
easy to solve, so let's consider a vector solution.