Finding the scalar parametric equation of the line

Question

I am trying to find the scalar parametric equation of the line which passes through the point $ A =[1,2,1]$ and is perpendicular to the plane $2z-y+2x=2$. But I am not sure how to go about doing this?

Firstly I am trying to get three answers in terms of $x=... y=...$ and $ z=...$

So I assume the normal vector is: $n=<2,-1,2>$

Then this becomes $2(x-1)-(y-2)+2(z-1)=0$

But then I am not sure what the question is really asking? Like what does $x$ equal? Any help would be greatly appreciated!

Answer 1

You have the normal to the plane $(2, -1, 2)$. This is the direction vector of your line. The line passes through $(1,2,1)$ so the line equation is $$ (x,y,z)=(1,2,1)+\lambda(2, -1, 2) $$ where $\lambda$ is the parameter. Then you have 3 individual equations, one each for $x,y$ and $z$ in terms of your parameter.

For example, $x=1+2\lambda$.

Answer 2

There is no scalar equation of a line in $\mathbb{R}^3$, but there is a set of parametric equations $$\left\{ \begin{align} x&=1+2t\\ y&=2-t\\ z&=1+2t \end{align}\right.$$