Graphics Reference
In-Depth Information
Figure 4.17 shows two points, P
1
x
1
y
1
z
1
and P
2
x
2
y
2
z
2
, joined by a connecting line.
Let the plane equation be of the form
ax
+
by
+
cz
=
d
Let the plane's normal
n
be
n
=
p
2
−
p
1
(4.22)
and
n
=
a
i
+
b
j
+
c
k
Let Qxyz be a point on
−−→
P
1
P
2
such that
q
=
−
t
p
1
+
1
t
p
2
(4.23)
Therefore,
n
·
q
=
d
(4.24)
Substituting Eq. (4.22) and Eq. (4.23) in Eq. (4.24) gives
d
=
p
2
−
p
1
·
1
−
t
p
1
+
t
p
2
Finally, we have
a
=
x
2
−
x
1
b
=
y
2
−
y
1
c
=
z
2
−
z
1
=
−
tx
1
+
+
−
ty
1
+
+
−
tz
1
+
d
a1
tx
2
b1
ty
2
c 1
tz
2
Let's test the above equations with the scenario shown in Fig. 4.18.
Y
P
2
Q
P
1
10
X
Z
Figure 4.18.
1
2
, we have
Given P
1
1000, P
2
0100, and t
=
a
=
x
2
−
x
1
=−
10
b
=
y
2
−
y
1
=
10
c
=
z
2
−
z
1
=
0
d
=−
105
+
105
=
0
Therefore, the plane equation is
−
10x
+
10y
=
0
