Graphics Reference
In-Depth Information
We identify a point P on the plane using a position vector
p
. And as we have seen in previous
chapters, we can state
n
·
p
=
d
(6.15)
which reminds us that if
n
is a unit vector, the projection of
p
on
n
equals the perpendicular
distance d from the origin to the plane.
Y
v
P
n
T
p
t
X
Z
Figure 6.9.
Figure 6.9 shows the scenario described above where the line equation is given by
p
=
t
+
v
The line and plane equations could be satisfied for some value of , which can be found by
substituting
p
into Eq. (6.15):
n
·
t
+
v
=
d
n
·
t
+
n
·
v
=
d
where
d
−
n
·
t
=
n
·
v
However, if the line is perpendicular to the plane's normal,
n
·
v
=
0 and will not have a valid
value. Furthermore, if both
n
and
v
are unit vectors, then
=
d
−
n
·
t
Finally, the position vector for the point of intersection is given by
p
=
t
+
v
To test the technique, consider the scenario shown in Fig. 6.10.
