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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 $y$?

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    The solution you have solves the system. So maybe there is a typo in the book. Also, you can write $+0y $ in the second equation to make it something that is more familiar. – 2017-02-14
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    You have not calculated the direction, but you have found two points, each of which lying on both planes. The "direction" is the vector that points from one of those points to the other one (or a multiple of this vector). Take $\frac{1}{2}((1,1,-1)-(-1,1,1))$ and you get the solution in the book. – 2017-02-14

2 Answers 2

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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 $<1, 0, -1>$

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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

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

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