Graphics Reference
In-Depth Information
=
√
3
2
+4
2
=5
Therefore if
n
=[34]
T
,
n
The equation of the line is
3
5
x
+
4
5
y
−
1=0
•
Example 2
. Given
y
=2
x
+ 1, what is the Hessian normal form?
Rearranging the equation, we get
2
x
−
y
+1=0
If we want the normal vector to point away from the origin we multiply by
−
1:
−
2
x
+
y
−
1=0
Normalize the normal vector to a unit form, i.e.
i
.
e
.
(
−
2)
2
+1
2
=
√
5
1
√
5
=0
Therefore, the perpendicular distance from the origin to the line and the unit
normal vector are respectively
2
√
5
x
+
1
√
5
y
−
−
and
−
T
1
√
5
2
√
5
1
√
5
The two signs from the square root provide the alternate directions of the
vector, and the sign of
d
.
As the Hessian normal form involves a unit normal vector, we can incor-
porate the vector's direction cosines within the equation:
x
cos(
α
)+
y
sin(
α
)
−
d
= 0
(10.32)
where
α
is the angle between the perpendicular and the
x
-axis.
10.2.3 Space Partitioning
The Hessian normal form provides a very useful way of partitioning space
into two zones: points above the line in the partition that includes the normal
vector, and points in the opposite partition. This is illustrated in Figure 10.20.
Given the equation
ax
+
by
−
d
= 0
(10.33)
a point (
x
,
y
) on the line satisfies the equation. But if we substitute another
point (
x
1
,y
1
) which is in the partition in the direction of the normal vector,
it creates the inequality
ax
1
+
by
1
−
d>
0
(10.34)
