# Given 2x -3y + z - 6 = 0, how do you get a vector equation from this scalar or a parametric equations?

Sep 1, 2016

$\pi : \vec{r} = \left(\begin{matrix}0 \\ 0 \\ 6\end{matrix}\right) + s \left(\begin{matrix}1 \\ 0 \\ - 2\end{matrix}\right) + t \left(\begin{matrix}0 \\ 1 \\ 3\end{matrix}\right)$

#### Explanation:

if

• $\vec{n}$ is the normal vector to the plane $\pi$,

• ${r}_{o}$ is a specific point lying on $\pi$,

• $\vec{r}$ is any other point on $\pi$

then $\left(\setminus \vec{r} - {\vec{r}}_{o}\right)$ is a line lying along $\pi$

....and so we can say using the scalar dot product that

$\left(\setminus \vec{r} - {\vec{r}}_{o}\right) \cdot \vec{n} = 0$, which is the vector equation of the plane $\pi$

we can re-arrange that to $\setminus \vec{r} \cdot \vec{n} - {\vec{r}}_{o} \cdot \vec{n} = 0$

and, in Cartesian with $\vec{r} = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$, we can match it up to your plane as follows using the scalar dot product

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \cdot \left(\begin{matrix}2 \\ - 3 \\ 1\end{matrix}\right) - 6 = 0$

for a parametric form, we will need a point ${\vec{p}}_{o}$ on plane $\pi$, and 2 non-parallel vectors, $\vec{u}$ and $\vec{v}$, that run along $\pi$. We can then identify any point on $\pi$ as

$\vec{r} = {\vec{p}}_{0} + s \vec{u} + t \vec{v}$ where s and t are the paremeters.

generating a ${\vec{p}}_{o}$ is simple. we just take $2 x - 3 y + z - 6 = 0$ and set $x = y = 0$ so that ${\vec{p}}_{o} = \left(\begin{matrix}0 \\ 0 \\ 6\end{matrix}\right)$

next we want $\vec{u}$ and $\vec{v}$ to be orthogonal to $\vec{n}$

Again using the scalar dot product, that means $\vec{u} \cdot \vec{n} = \vec{v} \cdot \vec{n} = 0$

So for $\vec{u}$ we have $2 {u}_{1} - 3 {u}_{2} + {u}_{3} = 0$

We can just arbitrarily choose ${u}_{1} = 1 , {u}_{2} = 0 \implies {u}_{3} = - 2$ giving $\vec{u} = \left(\begin{matrix}1 \\ 0 \\ - 2\end{matrix}\right)$

For $\vec{v}$ we have $2 {v}_{1} - 3 {v}_{2} + {v}_{3} = 0$ but we must ensure that $\vec{v}$ is not parallel to $\vec{u}$

So this time we can just arbitrarily choose ${v}_{1} = 0 , {v}_{2} = 1 \implies {v}_{3} = 3$ giving $\vec{v} = \left(\begin{matrix}0 \\ 1 \\ 3\end{matrix}\right)$

Now we can check our work!! We would expect the result $\vec{u} \times \vec{v} = \vec{n}$ as $\vec{u}$ and $\vec{v}$ lie in $\pi$ and $\vec{n}$ is it's normal vector

$\vec{u} \times \vec{v} = \det \left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ 1 & 0 & - 2 \\ 0 & 1 & 3\end{matrix}\right)$

$= 2 \hat{x} - 3 \hat{y} + 1 \hat{z}$ BINGO!!!!

one form of parameterised plane $\pi$ is therefore

$\pi : \vec{r} = \left(\begin{matrix}0 \\ 0 \\ 6\end{matrix}\right) + s \left(\begin{matrix}1 \\ 0 \\ - 2\end{matrix}\right) + t \left(\begin{matrix}0 \\ 1 \\ 3\end{matrix}\right)$