Graphics Reference
In-Depth Information
We are almost there, and something is staring us in the face, which we must learn to recognise:
ˆ
n
·ˆ
n
. As the angle between
n
and
ˆ
n
is zero degrees,
ˆ
n
ˆ
·ˆ
n
= ˆ
n
ˆ
n
=
1
Therefore, we can write Eq. (3.6) as
n
ˆ
·
p
=
c
(3.7)
Expanding Eq. (3.7), we get
a
i
+
b
j
·
x
i
+
y
j
=
c
ax
+
by
=
c
which we recognise as the Cartesian form of the line equation where
xy are the coordinates of a point on the line,
a and b are the components of a unit vector normal to the line, and
c is the perpendicular distance from the origin to the line.
Although the above interpretation is correct for our example, we must be careful not to be too
hasty when interpreting straight-line equations. For example, in the following equation,
3x
+
4y
=
10
(3.8)
10 does not represent the perpendicular distance from the origin to the line. This is because the
3 and 4 are not the components of a unit vector; they are the components of a vector 5 units
long.
Therefore, if we scale Eq. (3.8) by
1
5
, we obtain
06x
+
08y
=
2
(3.9)
where 0.6 and 0.8 are the components of a unit vector, and 2 is the perpendicular distance from
the origin to the line. Figure 3.7 shows the line equation representing Eq. (3.9).
Y
n
=
0.6
i
+
0.8
j
1
2
2
0.6
x
+
0.8
y
=
2
2
X
3
3
Figure 3.7.
