Find the line of intersection of the planes The planes are x+2y+3z=1 and x-y+z=1. My guess would be to set them equal to each other, since they are both equal to 1, we could write that as x+2y+3z=x-y+z. This simplifies to 3y+2z=0, it doesn't seem like this would be our answer though.
Update: I now understand that the cross product of the two normal vectors gives us the direction vector for the line that we're interested in finding but I don't know how to find a point on the line.
 A: The method is explained here.
A: You have $x+2y+3z=1$ and $x-y+z=1$ iff $3y+2z=0$ and $x=1-2y-3z$ iff $y=-{2 \over 3} z$ and $x=1-2y-3z=1-{5 \over 3} z$.
So, the line is given by $\{(1-{5 \over 3} z, -{2 \over 3} z, z) \}_{z \in \mathbb{R}}$.
A: Your planes are equivalent to the linear system
$$\left[
\begin{array}{ccc}
1 & 2 & 3 \\
1 & -1 & 1 \\
\end{array}\right]
\left[
\begin{array}{c}
x \\ y \\ z \\
\end{array}\right]
=
\left[
\begin{array}{c}
1 \\ 1 \\
\end{array}\right]
$$
If the normal vectors are nonzero and are not parallel then you can always perform row operations so that $\left[\begin{array}{c}1\\0\\\end{array}\right]$ is in one column and $\left[\begin{array}{c}0 \\1\\\end{array}\right]$ is in another one.  Then you can pick an arbitrary value for the variable corresponding to the third column and determine the corresponding values for the other two.
$$\left[
\begin{array}{ccc|c}
1 & 2 & 3 & 1\\
1 & -1 & 1 & 1\\
\end{array}\right]\\
\left[
\begin{array}{ccc|c}
1 & 2 & 3 & 1\\
0 & -3 & -2 & 0\\
\end{array}\right]\\
\left[\begin{array}{ccc|c}
1 & 0 & \frac53 & 1\\
0 & -3 & -2 & 0\\
\end{array}\right]\\
\left[\begin{array}{ccc|c}
1 & 0 & \frac53 & 1\\
0 & 1 & \frac23 & 0\\
\end{array}\right]\\
$$
Picking $z=0$ gives $(1,0,0)$.  Picking $z=3$ gives $(-4,-2,3)$.
