Points of Intersection
Points of Intersection
- If two lines are parallel, then their direction vectors will be multiples of one another.
- If they are the same line, then the position vector of one vector has to be on the same line as the other vector and their direction vectors have to be the same or multiplied by a scalar.
Solving points of intersection
- First, write out two parts of each vector as single vectors.
- Then you solve the equations simultaneously using the x and y components.
- Once you find values for \(\lambda\) and \(\mu\), sub them into the z component. If these values work, then they two lines intersect.
- Finally, sub in either \(\lambda\) or \(\mu\) to find the point of intersection.
The line \(l_{1}\) has equation \(\mathbf{r}=\begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}+\lambda \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}\). The line \(l_{2}\) has equation \(\mathbf{r=\begin{pmatrix}1 \\ 3 \\ 6\end{pmatrix}}+\mu \begin{pmatrix}2 \\ 1 \\ -1\end{pmatrix}\). Show that these lines do not intersect.
- Multiply out:
- \(l_{1}=\begin{pmatrix}1+\lambda \\ \lambda \\ -1\end{pmatrix}\)
- \(l_{2}=\begin{pmatrix}1+2\mu \\3+\mu \\6-\mu \end{pmatrix}\)
- Use simultaneous equations:
- \(1+\lambda=1+2\mu\)
- \(\lambda=3+\mu\)
- \(\lambda=6\)
- \(\mu=3\)
- See if it works for \(z\):
- \(6-3=-1\)
- \(3 \neq-1\)
- Therefore points do not intersect.
Find the point of intersection between: \(l:\mathbf{r}=\begin{pmatrix}-1 \\ 1 \\ -5\end{pmatrix}+\lambda \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}\) and \(\Pi:\mathbf{r}\times \begin{pmatrix}1 \\ 2 \\3\end{pmatrix}=4\)
- Sub in \(\mathbf{r}\)
- \(\begin{pmatrix}-1+\lambda \\1+\lambda \\-5+2\lambda \end{pmatrix}\times \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}=4\)
- Rearrange to find \(\lambda\)
- \(-1+\lambda+2+2\lambda-15+6\lambda=4\)
- \(-14+9\lambda=4\)
- \(9\lambda=18\)
- \(\lambda=2\)
- Point of intersection:
- \(\begin{pmatrix}-1+(2) \\1+(2) \\-5+2(2) \end{pmatrix}=\begin{pmatrix}1 \\3\\-1\end{pmatrix}\)