\(a=\begin{bmatrix}3\\-4\end{bmatrix}, b=\begin{bmatrix}-7\\8\end{bmatrix}\)

This question aims to find the parametric equation of the line through two given vectors.

A parametric equation is an equation that incorporates a parameter that is an independent variable. In this equation, the dependent variables are the continuous functions of the parameter.Â Two or more parameters can also be used if needed.

Generally, a line can be regarded as a set of points in the space satisfying the conditions, such as the lines having a specific point that can be defined by a position vector denoted by $\vec{r}_0$. Also, let $\vec{v}$ be the vector on a line. This vector will be parallel to a vector $\vec{r}_0$ and $\vec{r}$, which is a position vector on the line.

As a result, if $\vec{r}$ corresponds to a point on a line having the coordinates which are the components of $\vec{r}$ possess the form $\vec{r}=\vec{r}_0+t\vec{v}$. In this equation, $t$ is said to be a parameter and is a scalar that can possess any value. This generates different points on that line. So this equation is said to be a vector equation of the line.

## Expert Answer

Given that:

\(a=\begin{bmatrix}3\\-4\end{bmatrix}, b=\begin{bmatrix}-7\\8\end{bmatrix}\)

Now, the parametric equation of the line through two given vectors is:

$x=a+tb$

$x=\begin{bmatrix}3\\-4\end{bmatrix}+t\begin{bmatrix}-7\\8\end{bmatrix}$

which is the required equation.

## Example 1

Find the vector equation of the line containing the vectors $\vec{r}=\langle 0,1,2\rangle$ and $\vec{v}=\langle -2,1,3\rangle$. Also, write the parametric equations of the line.

### Solution

Since, $\vec{r}=\vec{r}_0+t\vec{v}$

$\vec{r}=\langle 0,1,2\rangle+t\langle -2,1,3\rangle$

$\vec{r}=\langle 0,1,2\rangle+\langle -2t,t,3t\rangle$

$\vec{r}=\langle -2t,1+t,2+3t\rangle$

Hence, the parametric equations of the line are:

$x=-2t, \, y=1+t$ and $z=2+3t$

## Example 2

Write the vector, parametric and symmetric form of the equation of the line through the points $(-1,3,5)$ and $(0,-2,1)$.

### Solution

For the vector form, find:

$\vec{v}=\langle -1-0,3+2,5-1\rangle=\langle -1,5,4\rangle$

So the vector form is:

$\vec{r}=\langle -1,3,5\rangle+t\langle -1,5,4\rangle$

$\vec{r}=\langle -1-t,3+5t,5+4t\rangle$

Parametric equations are:

$x=-1-t$

$y=3+5t$

$z=5+4t$

The symmetric form of the equation of line is:

$\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$

Here, $x_0=-1,y_0=3,z_0=5$ and $a=-1,b=5,c=4$

So that:

$\dfrac{x-(-1)}{-1}=\dfrac{y-3}{5}=\dfrac{z-5}{4}$

$\dfrac{x+1}{-1}=\dfrac{y-3}{5}=\dfrac{z-5}{4}$