Vector Equation of a Line with Examples

To find the vector equation of a line, we have to start by finding the position vector of a point on the line ($latex \overrightarrow{OA}$) and the direction vector ($latex \overrightarrow{AB}$) of the line.

Since point $latex A$ has coordinates $latex (1,~ 2,~ 3)$, and point $latex B$ has coordinates $latex (4, ~6, ~9)$, its position vector will be :

$$\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

$$\overrightarrow{OB} = \begin{pmatrix} 4 \\ 6 \\ 9 \end{pmatrix}$$

Now we have to find the direction vector ($latex \vec{AB}$) of the line. This can be found by taking the difference between the position vectors of points A and B:

$$\overrightarrow{AB} = \overrightarrow{OB}~ - ~\overrightarrow{OA}$$

$$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \\ 9 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$

$$\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \\ 6 \end{pmatrix}$$

Now, we can write the vector equation of the line:

$latex \vec{r} = \overrightarrow{OA} + t~\overrightarrow{AB}$

$$\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \\ 6 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} 1+3t \\ 2+4t \\ 3+6t \end{pmatrix}$$

In this case, we know the position vector of point C on the line and the direction vector ($latex \vec{d}$) of the line.

The position vector of point C with coordinates $latex (2,~ -1, ~4)$ is:

$$\overrightarrow{OC} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$$

The direction vector of the line is given by:

$$\vec{d} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$$

Now, we can write the vector equation of the line:

$latex \vec{r} = \overrightarrow{OC}+ t~\vec{d}$

$$\vec{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} 2+4t \\ -1+5t \\ 4-2t \end{pmatrix}$$

Similar to the previous example, we know the position vector of point R on the line and the direction vector ($latex \vec{e}$) of the line.

The position vector of point A with coordinates $latex (5,~ -2, ~3)$ is:

$$\overrightarrow{OA} = \begin{pmatrix} 5 \\ -2 \\ 3 \end{pmatrix}$$

In addition, we have that the direction vector of the line is:

$$\vec{e} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$$

Then, we can write the vector equation of the line using this information:

$latex \vec{r} = \overrightarrow{OA}+ t~\vec{e}$

$$\vec{r} = \begin{pmatrix} 5 \\ -2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} 5+2t \\ -2-3t \\ 3+t \end{pmatrix}$$

We need to start by finding the position vectors ($latex \overrightarrow{OP}$) and ($latex \overrightarrow{OQ}$).

Since point $latex P$ has coordinates $latex (-1,~ 3,~ 2)$, and point $latex Q$ has coordinates $latex (3, ~7, ~-1)$, we have:

$$\overrightarrow{OP} = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}$$

$$\overrightarrow{OQ} = \begin{pmatrix} 3 \\ 7 \\ -1 \end{pmatrix}$$

Now, we are going to find the direction vector ($latex \vec{PQ}$) of the line. This vector is equal to the difference between the position vectors of points P and Q:

$$\overrightarrow{PQ} = \overrightarrow{OQ} ~-~ \overrightarrow{OP}$$

$$\overrightarrow{PQ} = \begin{pmatrix} 3 \\ 7 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix}$$

$$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 4 \\ -3 \end{pmatrix}$$

Now, we can write the vector equation of the line:

$latex \vec{r} = \overrightarrow{OP} + t~\overrightarrow{PQ}$

$$\vec{r} = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} + t \begin{pmatrix} 4 \\ 4 \\ -3 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} -1+4t \\ 3+4t \\ 2-3t \end{pmatrix}$$

We start by finding the position vector of point U with coordinates $latex (-3,~ 5, ~2)$ is:

$$\overrightarrow{OU} = \begin{pmatrix} -3 \\ 5 \\ 2 \end{pmatrix}$$

In this case, the direction vector of the line is:

$$\vec{f} = \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}$$

Now, we write the vector equation of the line using this information:

$latex \vec{r} = \overrightarrow{OU}+ t~\vec{f}$

$$\vec{r} = \begin{pmatrix} -3 \\ 5 \\ 2 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} -3-t \\ 5+4t \\ 2+3t \end{pmatrix}$$

The position vectors ($latex \overrightarrow{OS}$) and ($latex \overrightarrow{OT}$) are:

$$\overrightarrow{OS} = \begin{pmatrix} 6 \\ 4 \\ -1 \end{pmatrix}$$

$$\overrightarrow{OT} = \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix}$$

Now, we are going to find the direction vector ($latex \vec{ST}$) of the line:

$$\overrightarrow{ST} = \overrightarrow{OT} ~-~ \overrightarrow{OS}$$

$$\overrightarrow{ST} = \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix} - \begin{pmatrix} 6 \\ 4 \\ -1 \end{pmatrix}$$

$$\overrightarrow{AB} = \begin{pmatrix} -6 \\ -6 \\ 4 \end{pmatrix}$$

Then, the vector equation of the line is:

$latex \vec{r} = \overrightarrow{OS} + t~\overrightarrow{ST}$

$$\vec{r} = \begin{pmatrix} 6 \\ 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} -6 \\ -6 \\ 4 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} 6+-6t \\ 4-6t \\ -1+4t \end{pmatrix}$$

The position vectors of the given points are:

$$\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$

$$\overrightarrow{OB} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$$

The direction vector ($latex \vec{AB}$) of the line is:

$$\overrightarrow{AB} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$$

$$\overrightarrow{AB} = \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix}$$

Then, the vector equation of the line is:

$latex \vec{r} = \overrightarrow{OA} + t~\overrightarrow{AB}$

$$\vec{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + t \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} 1 \\ 2-t \\ -1+3t \end{pmatrix}$$

To determine if P lies on the line, we need to find the value of $latex t$. Comparing the components of the vector equation and the position of the point, we have:

$latex 1=1$

$latex 2-t=3~~(t=-1)$

$latex -1+3t=1~~(t=\frac{2}{3})$

The last two equations are not compatible, so the point P is not on the line.

The vector equation for ($latex \overrightarrow{AB}$) was found in the previous example. Then, we have:

$$\vec{r} = \begin{pmatrix} 1 \\ 2-t \\ -1+3t \end{pmatrix}$$

The line $latex \overrightarrow{CD}$ is given by $latex \vec{r}=\overrightarrow{OC}+s~\overrightarrow{CD}$. Therefore, we have:

$$\vec{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + s \begin{pmatrix} -1 \\ 2 \\ -6 \end{pmatrix}$$

$$\vec{r} = \begin{pmatrix} -s\\ 1+2s \\ 2-6s \end{pmatrix}$$

The lines $latex \overrightarrow{AB}~$ and $latex ~\overrightarrow{CD}$ intersect if the parameters $latex s$ and $latex t$ satisfy the following:

$$ \begin{pmatrix} 1 \\ 2-t \\ -1+3t \end{pmatrix} = \begin{pmatrix} -s\\ 1+2s \\ 2-6s \end{pmatrix}$$

From the first row, we have $latex 1=-s$. Then, for the lines to intersect, we must have $latex s=-1$.

From the second row:

$latex 2 − t = 1 + 2s$

$latex 2 − t = −1$

$latex t = 3$

Substituting $latex s=-1$ and $latex t=3$ in the third row $latex −1 + 3t = 2 − 6s$, we get $latex 8 = 8$. Then, the values of $latex s$ and $latex t$ satisfy all three equations.

The point of intersection is found by substituting $latex s=-1$ or $latex t=3$ into $latex \overrightarrow{AB}~$ or $latex ~\overrightarrow{CD}$ respectively.

The point of intersection is $latex (1,~ −1, ~8)$.