Graphics Reference
In-Depth Information
substituting these components in (10.80) and (10.81) we have
λ
=
(0)(0)
−
(
−
1)
×
1
=
1
2
2
×
1
−
(0)
=
(0)(
−
1)
−
(0)
×
2
=0
2
×
1
−
(0)
therefore
x
P
=0+
1
2
×
0+0
×
1=0
y
P
=0+
1
0=
1
2
2
×
1+0
×
z
P
=1+
1
2
(
−
1) + 0
×
0=
1
2
The point
0
,
has position vector
p
,where
1
2
,
1
2
=
0
2
+
1
2
2
√
2
2
2
+
1
2
=
1
p
the plane equation is
1
2
1
2
2
√
2=0
1
0
x
+
y
+
z
−
2
√
2
2
√
2
1
1
which simplifies to
2
√
2
y
+
1
2
√
2
z
2
√
2=0
1
1
−
or
y
+
z −
1=0
10.7.5 Plane Equation from Three Points
Very often in computer graphic problems we require to find the plane equation
from three known points. To begin with, the three points must be distinct and
not lie on a line. Figure 10.34 shows three points
R
,
S
and
T
,fromwhichwe
create two vectors
u
=
−
RS
and
v
=
−
RT
. The vector product
u
v
then
provides a vector normal to the plane containing the original points. We now
take another point
P
(
x
,
y
,
z
) and form a vector
w
=
−
RP
. The scalar product
w
×
v
)=0if
P
is in the plane containing the original points. This condition
can be expressed as a determinant and converted into the general equation of
a plane. The three points are assumed to be in a counter-clockwise sequence
viewed from the direction of the surface normal.
·
(
u
×
