Line parallel to another line on a plane

Question

I have the following plane $H= \begin{cases} x=1-3a+b\\ y=a+2b & \\ z=2-a+b\end{cases} $

and line : $l=\begin{cases} x=t+2\\ y=2t-1 & \\ z=t\end{cases} $ on H.

How can I find a line lying on the plane H which is parallel to l?

Answer 1

Let $\vec{n}(p,q,r)$ be a vector normal of the plane.

Thus, $$-3p+q-r=0$$ and $$p+2q+r=0$$ and we see that $$\vec{n}(3,2-7)$$ is valid.

Thus, $$3(x-1)+2y-7(z-2)=0$$ or $$3x+2y-7z+11=0$$ is an equation of the plane (it's unnecessary of course).

Since $(1,2,1)(3,2,-7)=0$, we see that the line is indeed parallel to the plane,

which gives the answer: $$(1,0,2)+s(1,2,1).$$

Answer 2

Hint: Pick up a point from plane like $p(1,0,2)$ and make a line with $p$ and vector $(1,2,1)$ of $\ell$.