# What is the vector representation, parametric equations and rectangular equations for the line through the points P(3,2,1) and Q(-1,2,4)?

Jan 4, 2017

$\vec{r} = \left\langle3 , 2 , 1\right\rangle + \lambda \left\langle- 4 , 0 , 3\right\rangle$, or, $\vec{r} = \left(\begin{matrix}3 \\ 2 \\ 1\end{matrix}\right) + l a m \mathrm{da} \left(\begin{matrix}- 4 \\ 0 \\ 3\end{matrix}\right)$

Parametric Equation:

$\left.\begin{matrix}x = 3 - 4 l a m \mathrm{da} \\ y = 2 \\ z = 1 + 3 l a m \mathrm{da}\end{matrix}\right.$

Cartessia Equation:

 (3-x)/4 = (z-1)/3 " "; y=2

#### Explanation:

We have:

$\vec{O P} = \left\langle3 , 2 , 1\right\rangle$
$\vec{O Q} = \left\langle- 1 , 2 , 4\right\rangle$

So a line passing through $P$ and $Q$ will be in the direction;

$\vec{P Q} = \vec{O Q} - \vec{O P}$
$\text{ } = \left\langle- 1 - 3 , 2 - 2 , 4 - 1\right\rangle$
$\text{ } = \left\langle- 4 , 0 , 3\right\rangle$

The vector equation of a straight line (using $\vec{O P}$, but equally we could use $\vec{O Q}$) is given by;

$\setminus \setminus \setminus \setminus \setminus \vec{r} = \vec{a} + \lambda \vec{d}$
$\therefore \vec{r} = \left\langle3 , 2 , 1\right\rangle + \lambda \left\langle- 4 , 0 , 3\right\rangle$

Or, in column notation

$\vec{r} = \left(\begin{matrix}3 \\ 2 \\ 1\end{matrix}\right) + l a m \mathrm{da} \left(\begin{matrix}- 4 \\ 0 \\ 3\end{matrix}\right)$

For parametric equations we jus extract the $x$, $y$ and $z$ components, leaving lmda is the parameter:

$\left.\begin{matrix}x = 3 - 4 l a m \mathrm{da} \\ y = 2 + 0 l a m \mathrm{da} \\ z = 1 + 3 l a m \mathrm{da}\end{matrix}\right. \implies \left.\begin{matrix}x = 3 - 4 l a m \mathrm{da} \\ y = 2 \\ z = 1 + 4 l a m \mathrm{da}\end{matrix}\right.$

For Cartesian equations we use the parameters form and eliminate the parameter:

$\left.\begin{matrix}x = 3 - 4 l a m \mathrm{da} \\ y = 2 \\ z = 1 + 4 l a m \mathrm{da}\end{matrix}\right. \implies \left.\begin{matrix}l a m \mathrm{da} = \frac{3 - x}{4} \\ y = 2 \\ l a m \mathrm{da} = \frac{1 - z}{3}\end{matrix}\right.$

 :. (3-x)/4 = (z-1)/3 " "; y=2 #