Linear system: 3 variables, 2 equations w/o all variables

Question

The question asks to find the direction numbers for the line of intersection of the planes: $$ x + y + z = 1 , x + z =0 $$

I'm comfortable solving these sorts of linear systems when both equations include each variable. However here I'm slightly stuck.

If I parameterize $y = t$ for instance I have:

$$x = -z$$ $$-z + t + z = 1$$ $$t=1$$

From this I would guess that the directional numbers would be $(-1, 1, 1)$ (or $(1,1,-1)$ depending on if you substitute for $x$ or $z$). However the book lists that it is: $(1,0,-1)$. How did they get $0$ for $y$?

Answer 1

Since $x + z = 0$ we have that $y = 1$(in second equation). You can see that y does not depend on t.

Therefore, if $x = t$, then $z = -t$ and $y = 1 + 0*t$.

So you have $<1, 0, -1>$

Answer 2

Build the augmented matrix

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1\\ 1 & 0 & 1 & 0\end{array}\right]$$

and then use Gauss-Jordan elimination to obtain the RREF of the augmented matrix

$$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{array}\right]$$

Hence, the solution space is the line parametrized by

$$\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 1\\ 0\end{bmatrix} + t \begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}$$