Graphics Reference
In-Depth Information
This is just one way of solving this problem. There is another way, which is much simpler and
exploits the perp operator.
We discovered in Section 2.10 that given
n
=
a
i
+
b
j
, a perpendicular vector
n
⊥
exists such
that
n
⊥
=−
b
i
+
a
j
. Therefore, given a unit vector
n
ˆ
=
x
n
i
+
y
n
j
, a perpendicular unit vector
v
ˆ
is given by
v
ˆ
=−
y
n
i
+
x
n
j
and the line equation is
p
=
t
+
v
ˆ
(3.23)
But Eq. (3.23) does not refer explicitly to the original normal vector
n
, and if this is preferred, we
ˆ
+
−
+
must find a way of transforming x
n
i
y
n
j
into
y
n
i
x
n
j
. Fortunately, we can use a determinant
to do this:
=−
i j
x
n
y
n
−
y
n
i
+
x
n
j
Thus, the line equation is
i j
x
n
y
n
p
=
t
−
(3.24)
Whether we use Eq. (3.23) or Eq. (3.24), we can write the line equation as
x
p
=
x
t
−
y
n
(3.25)
y
p
=
y
t
+
x
n
1
√
2
i
Taking the previous example, where
n
ˆ
=
+
j
and T
=
30, we can substitute these in
Eq. (3.25) and obtain
1
x
p
=
3
−
√
2
1
y
p
=
0
+
√
2
When
=
0,
=
P
30
=
√
2,
When
P
=
21
3.7 The position and distance of a point on a line perpendicular to the origin
In this section we apply vector analysis to a simple geometric problem involving a straight line.
Imagine, then, a straight line that does not intersect the origin. We are going to calculate the
shortest distance from the origin to the line and the coordinates of the point on the line.
We examine two strategies: one for the Cartesian form of the line equation, and the other
using the parametric form.
