IB Maths AA 3.11 Notes

Equations of a line

A line can be written in vector form as:

$\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}$

Here:

• $\mathbf{a}$ is the position vector of a fixed point on the line
• $\mathbf{b}$ is a direction vector
• $\lambda$ is a scalar parameter

As $\lambda$ changes, the point $\mathbf{r}$ moves along the line.

A line passes through the point $(1,2,3)$ and has direction vector $\begin{pmatrix}4\\-1\\2\end{pmatrix}$. Write down its vector equation.

So $\mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}4\\1\\-2\end{pmatrix}$.

We can also write an equation in parametric form. From $\mathbf{r}=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}+\lambda\begin{pmatrix}l\\m\\n\end{pmatrix}$ we get the parametric equations:

$x=x_0+\lambda l$

$y=y_0+\lambda m$

$z=z_0+\lambda n$

Find the parametric equations for the line $\mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}4\\1\\-2\end{pmatrix}$.

So, $x=1+4\lambda$, $y=2-\lambda$, $z=3+2\lambda$.

Lastly, we can write a line in Cartesian form. If the direction vector is $(l,m,n)$, then the Cartesian form is

$\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$

For the same line we used before, the Cartesian form is $\frac{x-1}{4}=\frac{y-2}{-1}=\frac{z-3}{2}$.

To write the equation of a line, you need:

• one point on the line
• a direction vector

The direction vector can come from:

• two points on the line
• a parallel line
• physical information such as velocity

Find the equation of the line through $A(2,1,-1)$ and $B(5,3,4)$.

First find the direction vector: $\overrightarrow{AB}=(5,3,4)-(2,1,-1)=\begin{pmatrix}3\\2\\5\end{pmatrix}$.

So the vector equation is $\mathbf{r}=\begin{pmatrix}2\\1\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\3\end{pmatrix}$.