Hmm, I thought that if the space is 3 dimentional there should be 3 vectors in the base, and that's where I got e3.

I was actually trying to find the vector equation of this line, from this formula:

\(\displaystyle

\vec{r} \times \vec{a} = \vec{b}

\)

Vector \(\displaystyle \vec{a}\) must be parrallel to the line, and I thought that the substraction of position vectors of points that are in this line must be parralel to this line, and that's how I got this: \(\displaystyle \vec{a} = \vec{r2} - \vec{r1} = \vec{e2} - \vec{e1} \) .

Then, since point r1 is in the line, I plugged it into the equation, which gave me this: \(\displaystyle

\vec{r1} \times (\vec{e2} - \vec{e1}) = \vec{b}\)

\(\displaystyle \vec{b} = \vec{r1} \times (\vec{e2} - \vec{e1}) \)

\(\displaystyle \vec{b} = \vec{e1} \times (\vec{e2} - \vec{e1})

\)

And then I got vector b from det of a matrix 3x3 with such columns: (e1, 1, 0), (e2, 0, 1), (e3, 0, -1) - I'm sorry, I don't know how to make matrix in LaTex. Anyway it turned out that vector b in this equation is: \(\displaystyle \vec{b} = -\vec{e2} + \vec{e3}\) . Is that wrong?