Question

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

Answer 1

Hint: We recall that the vector equation any plane is given by \[\left( \overrightarrow{r}-\overrightarrow{{{r}_{0}}} \right)\cdot \overrightarrow{n}=0\] where $\overrightarrow{r}$ is the variable vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ , $\overrightarrow{{{r}_{0}}}$ is the position vectors of any point on the plane and $\overrightarrow{n}$ is the unit normal vector to the given plane $P:2x-3y+z-6=0$. We first find the parametric equation with parameters $s,t$ and non-parallel vectors $\overrightarrow{u},\overrightarrow{v}$ as $\overrightarrow{r}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}$.\[\]

Complete step by step answer:
We know that are give in the question the equation of plane in Cartesian form which we denote as
\[P:2x-3y+z-6=0\]
Here the scalars are $2,-3,6$. We know that the vector equation any plane is given by
 \[\begin{align}
  & \left( \overrightarrow{r}-\overrightarrow{{{r}_{0}}} \right)\cdot \overrightarrow{n}=0 \\
 & \Rightarrow \overrightarrow{r}\cdot \overrightarrow{n}-\overrightarrow{{{r}_{0}}}\cdot \overrightarrow{n}=0 \\
\end{align}\]
Here $\overrightarrow{r}$ is the variable vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ , $\overrightarrow{{{r}_{0}}}$ is the position vectors of any point on the plane and $\overrightarrow{n}$ is the unit normal vector to the given plane. Since we have to use the parametric equation for $\overrightarrow{r}$to find the vector equation for $P$ we use two parameters $s,t$ and two vectors parallel with the plane $\overrightarrow{u},\overrightarrow{v}$ not parallel to each other and have;
\[\overrightarrow{r}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}\]
Here $\overrightarrow{p}$ is any point vector on the plane $P$. We can put $x=0,y=0$ and find the p of the plane put in the equation of $P$ to have$2\cdot 0-3\cdot 0+z-6=0\Rightarrow z=6$. So we can write the position vector of $\overrightarrow{p}$ with unit orthogonal vectors $\hat{i},\hat{j},\hat{k}$
\[\overrightarrow{p}=0\cdot \hat{i}+0\hat{j}+6\hat{k}=6\hat{k}\]
Since $\overrightarrow{u}$ with component say $\overrightarrow{u}={{u}_{1}}\hat{i}+{{u}_{2}}\hat{j}+{{u}_{3}}\hat{k}$ is perpendicular with normal vector $\overrightarrow{n}$ we have $2{{u}_{1}}-3{{u}_{2}}+{{u}_{3}}=0$. We can choose arbitrarily ${{u}_{1}}=1,{{u}_{2}}=0$ to have ${{u}_{ 3}}=-2$.So we have
\[\overrightarrow{u}=1\cdot \hat{i}+0\hat{j}+\left( -2 \right)\hat{k}=\hat{i}-2\hat{k}\]
Similarly e $\overrightarrow{v}$ with component say $\overrightarrow{v}={{v}_{1}}\hat{i}+{{v}_{2}}\hat{j}+{{v}_{3}}\hat{k}$ is perpendicular with normal vector $\overrightarrow{n}$ we have $2{{v}_{1}}-3{{v}_{2}}+{{v}_{3}}=0$. We can choose arbitrarily ${{v}_{1}}=0,{{v}_{2}}=1$ to have ${{v}_{ 3}}=3$.So we have
\[\overrightarrow{s}=0\cdot \hat{i}+1\cdot \hat{j}+3\hat{k}=\hat{j}+3\hat{k}\]
Let us check where $\overrightarrow{n}$ will be perpendicular to $\overrightarrow{u},\overrightarrow{v}$ . We take the cross product of $\overrightarrow{u},\overrightarrow{v}$ to have;
\[\overrightarrow{u}\times \overrightarrow{v}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   1 & 0 & -2 \\
   0 & 1 & 3 \\
\end{matrix} \right|=\hat{i}\left( 0-\left( -2 \right) \right)-\hat{j}\left( 3-0 \right)+\hat{k}\left( 1-0 \right)=2\overrightarrow{i}-3\hat{j}+\hat{k}\]
We know that the equation of a plane and the normal have the same coefficients. Here $P$ has coefficients $\left( 2,-3,4 \right)$ and its normal $\overrightarrow{n}$ will also have $\left( 2,-3,4 \right)$. So $\overrightarrow{u}\times \overrightarrow{v}$ is parallel to $\overrightarrow{n}$ and hence perpendicular to $\overrightarrow{u},\overrightarrow{v}$. So the required variable vector is
\[\begin{align}
  & \overrightarrow{r}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v} \\
 & \Rightarrow \overrightarrow{r}=6\hat{k}+s\left( \hat{i}-2\hat{k} \right)+t\left( \hat{j}+3\hat{k} \right) \\
\end{align}\]
So one possible vector equation of plane with $\overrightarrow{n}=2\hat{i}-3\hat{j}+\hat{k}$, any position vector $\overrightarrow{{{r}_{0}}}$ on $P$ and variable vector in parametric form $\overrightarrow{r}=6\hat{k}+s\left( \hat{i}-2\hat{k} \right)+t\left( \hat{j}+3\hat{k} \right)$ is
\[\overrightarrow{r}\cdot \overrightarrow{n}-\overrightarrow{{{r}_{0}}}\cdot \overrightarrow{n}=0\]

Note:
We note that the Cartesian equation of normal to the plane $ax+by+cz+d=0$ is given by $a\left( x-{{x}_{0}} \right)+b\left( y-{{y}_{0}} \right)+c\left( z-{{z}_{0}} \right)+d=0$ where $\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ is any point lying on the plane. We can alternatively write the parametric equation as $x=s,y=t,z=6-\dfrac{s}{3}+\dfrac{2t}{3}$. The perpendicular condition of two vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}=0$ and the vector perpendicular to both of them is given by$\overrightarrow{a}\times \overrightarrow{b}$.