Graphics Reference
In-Depth Information
Step 3:
Equate the two line equations:
t
+
a
=
s
+
b
x
t
i
+
y
t
j
+
z
t
k
+
x
a
i
+
y
a
j
+
z
a
k
=
x
s
i
+
y
s
j
+
z
s
k
+
x
b
i
+
y
b
j
+
z
b
k
and collect like components:
x
t
−
x
s
+
x
a
−
x
b
i
+
y
t
−
y
s
+
y
a
−
y
b
j
+
z
t
−
z
s
+
z
a
−
z
b
k
=
0
For this vector to be null, its components must vanish. Therefore, we have
x
a
−
x
b
=
x
s
−
x
t
y
a
−
y
b
=
y
s
−
y
t
z
a
−
z
b
=
z
s
−
z
t
which provide values for and , which, when substituted in the original line equations, reveal
the intersection point. For example, the lines in Fig. 6.8 have equations:
p
=
t
+
a
and
q
=
s
+
b
(6.14)
where
t
=
j
+
2
k
and
a
=
2
i
+
j
−
2
k
and
s
=
2
i
+
j
and
b
=−
2
i
+
j
+
2
k
The first step is to discover whether the lines are parallel, i.e., if
a
×
b
=
0.
We calculate
a
×
b
:
i
j
k
a
2
1
-2
b
-2
1
2
a
×
b
4
0
4
But as
a
4
k
, the lines are not parallel.
The second step is to discover if the distance d between two skew lines is zero:
×
b
=
4
i
+
=
t
−
s
·
a
×
b
=
−
2
i
+
2
k
·
4
i
+
4
k
d
×
+
a
b
4
i
4
k
=
−
2
i
+
2
k
·
4
i
+
4
k
=
−
8
√
32
=
8
+
d
0
4
i
+
4
k
which it is; therefore, the lines do intersect.
