Can we convert the equation in an equivalent one with one parameter?

Question

We are given lines of the form \begin{align*}&x=\begin{pmatrix}1 \\ 3 \\ 0 \\ 2\end{pmatrix}+a\cdot \begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix} \\ &y=\begin{pmatrix}2 \\ -1 \\ -2 \\ 1\end{pmatrix}+b\cdot \begin{pmatrix}1 \\ 0 \\ -1 \\ 1\end{pmatrix}+c\cdot \begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\end{align*} Is it possible to convert the second one in a form of the first one, i.e. with just one parameter?

Answer 1

No.

It would only be possible if the vectors following b and c were linearly dependent, i.e. multiples of each other.

Answer 2

You may write it without any parameter at all. Given $$E\colon \vec y=\vec a+b\vec u+c\vec v,$$ a vector $\vec y$ satisfies the equation iff the volume of the parallelepiped spanned by $\vec u$, $\vec v$ and $\vec y-\vec a$ equals zero, that is, the Gram-determinant of the three vectors vanishes: $$\det\begin{pmatrix} \langle \vec u,\vec u\rangle & \langle \vec u,\vec v\rangle &\langle \vec u,\vec y-\vec a\rangle\\ \langle \vec v,\vec u\rangle & \langle \vec v,\vec v\rangle & \langle \vec v,\vec y-\vec a\rangle\\ \langle \vec y-\vec a,\vec u \rangle & \langle \vec y-\vec a,\vec v\rangle &\langle \vec y-\vec a,\vec y-\vec a\rangle \end{pmatrix}=0.$$

EDIT: If you have orthonormal $\vec u$ and $\vec v$ plugging in any vector $\vec y$ in the above determinant will give you the distance from $\vec y$ to the $2$-dimensional plane $E$.