Graphics Reference
In-Depth Information
Y
P
p
O
Z
X
Figure 4.1.
Step 2
Construct a line from O to a point C on the plane such that
−
OC is perpendicular to the plane.
It is convenient to make
−
OC a scalar multiple of a unit vector
c
k
, i.e.,
−
OC
n
ˆ
=
a
i
+
b
j
+
=
d
n
,
ˆ
where d is some scalar, as shown in Fig. 4.2.
Y
n
C
C
P
ˆ
d
n
P
p
O
Z
X
Figure 4.2.
We can now write
−
OP
=
−
OC
+
−
CP. Substituting vector names gives
+
−
CP
p
=
d
n
ˆ
(4.1)
Although we know d and
n
, and the fact that
p
points to any point on the plane, we do not
know
−
CP. Somehow,
−
CP has to be eliminated, which can be achieved by the same cunning
subterfuge used in Section 3.3.
Remember that the dot product of two perpendicular vectors is zero. Then d
ˆ
·
−
CP must
n
ˆ
·
−
CP
equal zero. But the d is superfluous; therefore,
0.
Unfortunately, Eq. (4.1) does not contain such a term, but there is nothing to stop us from
introducing one by multiplying Eq. (4.1) by
n
ˆ
=
n
using the dot product:
ˆ
·
−
CP
n
ˆ
·
p
=
d
n
ˆ
·ˆ
n
+ˆ
n
which now reduces to
n
ˆ
·
p
=
d
n
ˆ
·ˆ
n
(4.2)
