3 The vector equation of a line

Consider the straight line $APB$ shown in Figure 48. This is a line in three-dimensional space.

Figure 48

Points $A$ and $B$ are fixed and known points on the line, and have position vectors $\underset{̲}{a}$ and $\underset{̲}{b}$ respectively. Point $P$ is any other arbitrary point on the line, and has position vector $\underset{̲}{r}$ . Note that because $\overrightarrow{AB}$ and $\overrightarrow{AP}$ are parallel, $\overrightarrow{AP}$ is simply a scalar multiple of $\overrightarrow{AB}$ , that is, $\overrightarrow{AP}=t\overrightarrow{AB}$ where $t$ is a number.

Task!

Referring to Figure 48, write down an expression for the vector $\overrightarrow{AB}$ in terms of $\underset{̲}{a}$ and $\underset{̲}{b}$ .

Answer

Task!

Referring to Figure 48, use the triangle law for vector addition to find an expression for $\underset{̲}{r}$ in terms of $\underset{̲}{a}$ , $\underset{̲}{b}$ and $t$ , where $\overrightarrow{AP}=t\overrightarrow{AB}$ .

Answer

The answer to the above Task, $\underset{̲}{r}=\underset{̲}{a}+t\left(\underset{̲}{b}-\underset{̲}{a}\right)$ , is the vector equation of the line through $A$ and $B$ . It is a rule which gives the position vector $\underset{̲}{r}$ of a general point on the line in terms of the given vectors $\underset{̲}{a},\underset{̲}{b}$ . By varying the value of $t$ we can move to any point on the line. For example, referring to Figure 48,

$\text{when}t=0,\text{the equation gives}\underset{̲}{r}=\underset{̲}{a},\text{ which locates point }A,$

$\text{when}t=1,\text{the equation gives}\underset{̲}{r}=\underset{̲}{b},\text{ which locates point }B.$

If $0<t<1$ the point $P$ lies on the line between $A$ and $B$ . If $t>1$ the point $P$ lies on the line beyond $B$ (to the right in the figure). If $t<0$ the point $P$ lies on the line beyond $A$ (to the left in the figure).

Task!

Write down the vector equation of the line which passes through the points with position vectors $\underset{̲}{a}=3\underset{̲}{i}+2\underset{̲}{j}$ and $\underset{̲}{b}=7\underset{̲}{i}+5\underset{̲}{j}$ . Also express the equation in column vector form.

Answer

Task!

Using column vector notation, write down the vector equation of the line which passes through the points with position vectors $\underset{̲}{a}=5\underset{̲}{i}-2\underset{̲}{j}+3\underset{̲}{k}$ and $\underset{̲}{b}=2\underset{̲}{i}+\underset{̲}{j}-4\underset{̲}{k}$ .

Answer

3.1 Cartesian form

On occasions it is useful to convert the vector form of the equation of a straight line into Cartesian form. Suppose we write

$\underset{̲}{a}=\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \end{array}\right),\underset{̲}{b}=\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill b_3\hfill \end{array}\right),\underset{̲}{r}=\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right)$

then $\underset{̲}{r}=\underset{̲}{a}+t\left(\underset{̲}{b}-\underset{̲}{a}\right)$ implies

$\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right)=\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill a_3\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill b_1-a_1\hfill \\ \hfill b_2-a_2\hfill \\ \hfill b_3-a_3\hfill \end{array}\right)=\left(\begin{array}{c}\hfill a_1+t\left(b_1-a_1\right)\hfill \\ \hfill a_2+t\left(b_2-a_2\right)\hfill \\ \hfill a_3+t\left(b_3-a_3\right)\hfill \end{array}\right)$

Equating the individual components we find

$x=a_1+t\left(b_1-a_1\right),\text{ or equivalently }t=\frac{x-a_1}{b_1-a_1}$

$y=a_2+t\left(b_2-a_2\right),\text{ or equivalently }t=\frac{y-a_2}{b_2-a_2}$

$z=a_3+t\left(b_3-a_3\right),\text{ or equivalently }t=\frac{z-a_3}{b_3-a_3}$

Each expression on the right is equal to $t$ and so we can write

$\frac{x-a_1}{b_1-a_1}=\frac{y-a_2}{b_2-a_2}=\frac{z-a_3}{b_3-a_3}$

This gives the Cartesian form of the equations of the straight line which passes through the points with coordinates $\left(a_1,a_2,a_3\right)$ and $\left(b_1,b_2,b_3\right)$ .

Example 22

1. Write down the Cartesian form of the equation of the straight line which passes through the two points $\left(9,3,-2\right)$ and $\left(4,5,-1\right)$ .
2. State the equivalent vector equation.

Solution

1. $\frac{x-9}{4-9}=\frac{y-3}{5-3}=\frac{z-\left(-2\right)}{-1-\left(-2\right)}$

   that is

   $\frac{x-9}{-5}=\frac{y-3}{2}=\frac{z+2}{1}\text{(Cartesian form)}$

2. The vector equation is $$\begin{array}{rcll}\underset{̲}{r}& =& \underset{̲}{a}+t\left(\underset{̲}{b}-\underset{̲}{a}\right)& \\ & =& \left(\begin{array}{c}\hfill 9\hfill \\ \hfill 3\hfill \\ \hfill -2\hfill \end{array}\right)+t\left(\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 5\hfill \\ \hfill -1\hfill \end{array}\right)-\left(\begin{array}{c}\hfill 9\hfill \\ \hfill 3\hfill \\ \hfill -2\hfill \end{array}\right)\right)& \\ & =& \left(\begin{array}{c}\hfill 9\hfill \\ \hfill 3\hfill \\ \hfill -2\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill -5\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right)& \end{array}$$

Exercises

1. 1. Write down the vector $\overrightarrow{AB}$ joining the points $A$ and $B$ with coordinates $\left(3,2,7\right)$ and $\left(-1,2,3\right)$ respectively.
   2. Find the equation of the straight line through $A$ and $B$ .
2. Write down the vector equation of the line passing through the points with position vectors $\underset{̲}{p}=3\underset{̲}{i}+7\underset{̲}{j}-2\underset{̲}{k}$ and $\underset{̲}{q}=-3\underset{̲}{i}+2\underset{̲}{j}+2\underset{̲}{k}$ . Find also the Cartesian equation of this line.
3. Find the vector equation of the line passing through $\left(9,1,2\right)$ and which is parallel to the vector $\left(1,1,1\right)$ .

Answer