Graphics Reference
In-Depth Information
Let's try both, just to prove the point:
x
i
=0+
1
3
3=1
x
i
=0+
1
2
2=1
y
i
=1+
1
3
3=2
y
i
=
1
2
+
1
2
3=2
z
i
=0+
1
3
3=1
z
i
=0+
1
2
2=1
Therefore, the point of intersection point is (1, 2, 1).
Now let's take two lines that don't intersect, and also exhibit some linear
dependency:
⎡
⎤
⎡
⎤
⎡
⎤
⎡
⎤
0
1
0
2
2
0
0
2
0
2
2
1
⎣
⎦
⎣
⎦
⎣
⎦
⎣
⎦
a
1
=
b
1
=
a
2
=
b
2
=
Taking the
x
and
y
components we discover that the determinant ∆ is zero,
which has identified the linear dependency. Taking the
y
and
z
components
the determinant is non-zero, which permits us to computer
r
and
s
using
r
=
z
b
2
(
y
a
2
−
y
a
1
)
−
y
b
2
(
z
a
2
−
z
a
1
)
(10.74)
y
b
1
z
b
2
−
z
b
1
y
b
2
s
=
z
b
1
(
y
a
2
−
y
a
1
)
−
y
b
1
(
z
a
2
−
z
a
1
)
(10.75)
y
b
1
z
b
2
−
z
b
1
y
b
2
r
=
1(2
−
1)
−
2(0
−
0)
=
1
2
2
×
1
−
0
×
2
s
=
0(2
−
1)
−
2(0
−
0)
=
0
2
=0
But these values of
r
and
s
must also apply to the
x
components:
2
×
1
−
0
×
2
rx
b
1
−
sx
b
2
=
x
a
2
−
x
a
1
1
2
×
2
−
0
×
2
=0
−
0
which they clearly do not, therefore the lines do not intersect.
Now let's proceed with the equation of a plane, and then look at how to
compute the intersection of a line with a plane using a similar technique.
10.7 Equation of a Plane
We now consider four ways of representing a plane equation: the Cartesian
form, general form, parametric form and a plane from three points.
