Example 1: Find the intersection point of two lines
Find the point of intersection of the lines
\[
\mathbf{r}=
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}
+
\lambda
\begin{pmatrix}
2\\
-1\\
4
\end{pmatrix}
\]
\[
\mathbf{r}=
\begin{pmatrix}
5\\
0\\
11
\end{pmatrix}
+
\mu
\begin{pmatrix}
-1\\
1\\
-2
\end{pmatrix}.
\]
Solution
Set the two expressions equal:
\[
\begin{pmatrix}
1+2\lambda\\
2-\lambda\\
3+4\lambda
\end{pmatrix}
=
\begin{pmatrix}
5-\mu\\
\mu\\
11-2\mu
\end{pmatrix}.
\]
Compare coordinates:
\[
1+2\lambda=5-\mu
\]
\[
2-\lambda=\mu
\]
\[
3+4\lambda=11-2\mu.
\]
From \( 2-\lambda=\mu \), substitute into the first equation:
\[
1+2\lambda=5-(2-\lambda)=3+\lambda
\]
\[
\lambda=2.
\]
Then
\[
\mu=2-\lambda=0.
\]
Check in the third equation:
\[
3+4(2)=11-2(0)
\]
\[
11=11.
\]
So the lines intersect. Substitute \( \lambda=2 \) into the first line:
\[
\mathbf{r}=
\begin{pmatrix}
1\\
2\\
3
\end{pmatrix}
+
2
\begin{pmatrix}
2\\
-1\\
4
\end{pmatrix}
=
\begin{pmatrix}
5\\
0\\
11
\end{pmatrix}.
\]
The point of intersection is
\[
\begin{pmatrix}
5\\
0\\
11
\end{pmatrix}.
\]
Example 2: Show that two lines do not intersect
Determine whether the lines
\[
\mathbf{r}=
\begin{pmatrix}
1\\
-2\\
4
\end{pmatrix}
+
\lambda
\begin{pmatrix}
3\\
1\\
-2
\end{pmatrix}
\]
\[
\mathbf{r}=
\begin{pmatrix}
7\\
1\\
0
\end{pmatrix}
+
\mu
\begin{pmatrix}
-1\\
2\\
1
\end{pmatrix}
\]
intersect.
Solution
Set the coordinates equal:
\[
1+3\lambda=7-\mu
\]
\[
-2+\lambda=1+2\mu
\]
\[
4-2\lambda=\mu.
\]
From the third equation,
\[
\mu=4-2\lambda.
\]
Substitute into the first equation:
\[
1+3\lambda=7-(4-2\lambda)=3+2\lambda
\]
\[
\lambda=2.
\]
Then
\[
\mu=4-2(2)=0.
\]
Check the second equation:
\[
-2+2=1+2(0)
\]
\[
0\neq 1.
\]
The equations are inconsistent, so the lines do not intersect.
Example 3: Find the intersection using parametric form
The lines are given by
\[
x=2+t,\qquad y=1-2t,\qquad z=3+t
\]
\[
x=5-s,\qquad y=-5+4s,\qquad z=6-s.
\]
Find whether the lines intersect.
Solution
Compare coordinates:
\[
2+t=5-s
\]
\[
1-2t=-5+4s
\]
\[
3+t=6-s.
\]
From the first and third equations:
\[
t+s=3
\]
\[
t+s=3.
\]
So these are consistent. Use \( s=3-t \) in the second equation:
\[
1-2t=-5+4(3-t)
\]
\[
1-2t=7-4t
\]
\[
2t=6
\]
\[
t=3.
\]
Then
\[
s=3-t=0.
\]
Substitute into either line:
\[
(x,y,z)=(2+3,\;1-2(3),\;3+3)=(5,-5,6).
\]
The lines intersect at \((5,-5,6)\).
Example 4: Parallel lines
Determine the relationship between the lines
\[
\mathbf{r}=
\begin{pmatrix}
1\\
3\\
-2
\end{pmatrix}
+
\lambda
\begin{pmatrix}
2\\
-1\\
4
\end{pmatrix}
\]
\[
\mathbf{r}=
\begin{pmatrix}
5\\
1\\
6
\end{pmatrix}
+
\mu
\begin{pmatrix}
-4\\
2\\
-8
\end{pmatrix}.
\]
Solution
Compare the direction vectors:
\[
\begin{pmatrix}
-4\\
2\\
-8
\end{pmatrix}
=-2
\begin{pmatrix}
2\\
-1\\
4
\end{pmatrix}.
\]
The direction vectors are multiples, so the lines are parallel or identical.
Check whether the point
\(
\begin{pmatrix}
5\\
1\\
6
\end{pmatrix}
\)
lies on the first line:
\[
1+2\lambda=5 \Rightarrow \lambda=2
\]
\[
3-\lambda=1 \Rightarrow \lambda=2
\]
\[
-2+4\lambda=6 \Rightarrow \lambda=2.
\]
All three coordinates give the same value of \( \lambda \), so the point lies on the first line.
The two lines are the same line.