Equation of a Line in Three Dimensions
\(\(\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}\)\)
- \(\mathbf{a}\) is the position vector to start.
- \(\mathbf{b}\) is the direction vector that determines which direction we are travelling in.
- \(\lambda\) is the variable scalar that allows us to get to any point on our line.
- \(\mathbf{r}\) is a position vector of our final point on the line.
\(l_{1}: \mathbf{r}=\begin{pmatrix}3 \\ 0 \\ -2\end{pmatrix}+\lambda \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}\)
Find the vector equation of a line parallel to \(l_{1}\) which passes through \((2, 5, 1)\)
\(l_{2}: \mathbf{r}=\begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix}+\lambda \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix}\)
- This works because the start position is different, but the direcion vector is the same as \(l_{1}\) which means they will be parallel.
\(l_{1}: \mathbf{r}=\begin{pmatrix}-1 \\ 0 \\ 3\end{pmatrix}+\lambda \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}\)
Find the coordiantes of the points on \(l_{1}\) which are a distance of 3 away from (3, 4, 4).
- Generic point on line: \(\begin{pmatrix}-1+\lambda \\\lambda \\3\end{pmatrix}\)
- Point we are finding about: \((3, 4, 4)\)
- Distance between: \(\sqrt{ (-1+\lambda-3)^3+(\lambda-4)^2+(3-4)^2 }=3\)
- \(\lambda^2-8\lambda+12=0\)
- Solve for \(\lambda\) gives us: \(\lambda=2\) or \(\lambda=6\)
- Sub back into our generic equation:
If \(\lambda=2 \to (1,2,3)\)
If \(\lambda=6\to(5,6,3)\)
- \(\begin{pmatrix}3+t \\ 2-6t \\ -5-2t\end{pmatrix}\)
- \(\therefore\begin{pmatrix}3+t \\ 2-6t \\ -5-2t\end{pmatrix}=\begin{pmatrix}a \\ b \\ 0\end{pmatrix}\)
- This means that \(-5-2t=0\)
- Rearrange for \(t\): \(t=-\frac{5}{2}\)
- Solve simultaneously
- \(a=3+t\)
- \(b=2-6t\)
- \(a=\frac{1}{2}\)
- \(b=17\)
- Directional part of \(d_{1}\) = \(\begin{pmatrix}6 \\ -2 \\ 4\end{pmatrix}\)
- Directional part of \(d_{2}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}\)
- \(d_{1}=2d_{2}\)
- Since they are multiples of one another, they go in the same direction.
- We choose a value of \(\lambda = 0\): \(r_{1}=\begin{pmatrix}2 \\ 5 \\ -3\end{pmatrix}\)
- We now need to find a value of \(\mu\) that gives us \(\begin{pmatrix}2 \\ 5 \\ -3\end{pmatrix}\)
- \(\begin{pmatrix}2 \\ 5 \\ -3\end{pmatrix}=\begin{pmatrix}8+3\mu \\3-\mu \\1+2\mu\end{pmatrix}\)
- Choose one to solve: \(8+3\mu=2\)
- \(\mu=-2\)
- It works for the other two equation \(\therefore\) it works.
If \(\mathbf{a}=\begin{pmatrix}a_{1} \\a_{2} \\a_{3}\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}b_{1} \\b_{2} \\b_{3} \\\end{pmatrix}\) and \(\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}\) is equation of a straight line, the Certesian form is \(\(\frac{x-a_{1}}{b_{1}}=\frac{y-a_{2}}{b_{2}}=\frac{z-a_{3}}{b_{3}}\)\)
- \(a_{1}=2\)
- \(a_{2}=-5\)
- \(a_{3}=0\)
- \(b_{1}=3\)
- \(b_{2}=1\)
- \(b_{3}=4\)
- \(\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}2 \\ -5 \\ 0\end{pmatrix}+\lambda \begin{pmatrix}3 \\ 1 \\ 4\end{pmatrix}\)
\(\(\mathbf{a}=\begin{pmatrix}4 \\0 \\-2\end{pmatrix}+\alpha \begin{pmatrix}-1 \\1 \\4\end{pmatrix}\)\)
\(\(\mathbf{b}=\begin{pmatrix}1 \\3 \\10\end{pmatrix}+\beta \begin{pmatrix}-2 \\-1 \\2\end{pmatrix}\)\)
\(\(\mathbf{c}=\begin{pmatrix}2 \\-1 \\0\end{pmatrix}+\gamma \begin{pmatrix}1 \\2 \\2 \end{pmatrix}\)\)
1. \(\begin{pmatrix}4-\alpha \\\alpha \\-2+4\alpha\end{pmatrix}=\begin{pmatrix}1-2\beta \\3-\beta \\10+2\beta\end{pmatrix}\)
\(4-\alpha=1-2\beta\) (1)
\(\alpha=3-\beta\) (2)
\(-2+4\alpha=10-2\beta\) (3)
\((1)+(2): 4=4-3\beta\)
\(\beta=0\)
\(\alpha=3\)
Therefore \(\mathbf{a}\) and \(\mathbf{b}\) intersect at point \((1, 3, 10)\)
This would now be done for pair \(\mathbf{a}\) and \(\mathbf{c}\) as well as pair \(\mathbf{b}\) and \(\mathbf{c}\)