3.11 Vector equation of a lineIB Maths: Applications and Interpretation HL: Revision notes

A line in two or three dimensions is fixed by a point on it and a direction. Its vector equation is $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b},\qquad\lambda\in\mathbb{R},$ where

• $\mathbf{r}$ is the position vector of a general point on the line,
• $\mathbf{a}$ is the position vector of a known point on the line,
• $\mathbf{b}$ is a direction vector, parallel to the line,
• $\lambda$ is a parameter: each value of $\lambda$ gives one point on the line.

Example: $\mathbf{r}=\begin{pmatrix}2 \\ -1 \\ 4\end{pmatrix}+\lambda\begin{pmatrix}3 \\ 2 \\ -1\end{pmatrix}$. At $\lambda=0$ the point is $(2,-1,4)$ and at $\lambda=2$ it is $(8,3,2)$. Negative values of $\lambda$ give points on the other side of $\mathbf{a}$.

From a point and a direction: use $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$ directly.

From two points $A$ and $B$: the direction is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ (end minus start), and either point can be the base: $\mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB}.$ Example: $A(1,2,-3)$ and $B(4,6,9)$ give $\overrightarrow{AB}=\begin{pmatrix}3 \\ 4 \\ 12\end{pmatrix}$, so $\mathbf{r}=\begin{pmatrix}1 \\ 2 \\ -3\end{pmatrix}+\lambda\begin{pmatrix}3 \\ 4 \\ 12\end{pmatrix}$.

The equation is not unique: any point on the line can be $\mathbf{a}$, and any non-zero multiple of $\mathbf{b}$ can be the direction. In 2D the same form is used with 2-component vectors, e.g. $\mathbf{r}=\begin{pmatrix}2 \\ 5\end{pmatrix}+\lambda\begin{pmatrix}9 \\ -12\end{pmatrix}$ (the direction can be simplified to $\begin{pmatrix}3 \\ -4\end{pmatrix}$).

A point lies on the line if one value of $\lambda$ gives all its coordinates.

Example: does $(7,10,21)$ lie on $\mathbf{r}=\begin{pmatrix}1 \\ 2 \\ -3\end{pmatrix}+\lambda\begin{pmatrix}3 \\ 4 \\ 12\end{pmatrix}$?

• $x$: $1+3\lambda=7$ gives $\lambda=2$.
• $y$: $2+4(2)=10$ ✓.
• $z$: $-3+12(2)=21$ ✓.

All three agree, so it is on the line. If the same $\lambda$ failed for even one coordinate, the point would not be on the line.

You can also find an unknown coordinate: $(7,10,p)$ on the line gives $p=21$. And where the line meets a coordinate plane such as $z=0$, set that component equal to 0 and solve for $\lambda$.

Splitting $\mathbf{r}=\begin{pmatrix}x_0 \\ y_0 \\ z_0\end{pmatrix}+\lambda\begin{pmatrix}l \\ m \\ n\end{pmatrix}$ into components gives the parametric equations $x=x_0+\lambda l,\qquad y=y_0+\lambda m,\qquad z=z_0+\lambda n.$ Example: $\mathbf{r}=\begin{pmatrix}2 \\ -1 \\ 4\end{pmatrix}+\lambda\begin{pmatrix}3 \\ 2 \\ -1\end{pmatrix}$ becomes $x=2+3\lambda$, $y=-1+2\lambda$, $z=4-\lambda$.

To convert back, read off the point from the constant terms and the direction from the coefficients of $\lambda$: $x=4-2\lambda$, $y=1+\lambda$, $z=3\lambda$ gives $\mathbf{r}=\begin{pmatrix}4 \\ 1 \\ 0\end{pmatrix}+\lambda\begin{pmatrix}-2 \\ 1 \\ 3\end{pmatrix}$. A component with no constant has constant 0.

A moving object often has position $\mathbf{r}=\mathbf{r}_0+t\mathbf{v}$, with time $t$ in place of $\lambda$. Then

• $\mathbf{r}_0$ is the starting position,
• $\mathbf{v}$ is the velocity, and the speed is $|\mathbf{v}|$,
• the path of the object is the line.

Example: a drone has $\mathbf{r}=\begin{pmatrix}-20 \\ 10 \\ 30\end{pmatrix}+t\begin{pmatrix}4 \\ -2 \\ -3\end{pmatrix}$, with the $z$-axis vertical. Speed $=\sqrt{16+4+9}=5.39$ m s$^{-1}$. It reaches the ground when $z=0$: $30-3t=0$, so $t=10$ s, at $(20,-10,0)$.

State answers in context, with units, and remember that $t$ is usually restricted to $t\ge0$.