I'm given the following problem:
Find the parametric equation of the line of intersection of the two given planes: $y-6z = 7$ and $9x - 8y = 5$
My attempted solution follows:
I take the cross product of two vectors normal to the given planes, $<0,1,-6> \times <9,-8,0>$, and obtain the vector $<-48,-54,-9>$. I then set $z=0$, and solve the given simultaneous equations to determine a point along the line of intersection, $(\frac{61}{9}, {7}, 0)$. I can then write the parametric equation of intersection as:
$x = \frac{61}{9} - 48t$, $y = 7 - 54t$, $z = -9t$, using parameter $t$
Unfortunately, however, it seems I've erred in my solution. Is this the proper technique?
The cross product will give you the direction of the line of intersection, yes!
However the line may or may not pass through $z=0$ (e.g. it could be parallel to the XY plane) so this method doesn't guarantee a solution.
The simplest way to get a solution would just be to solve the system of equations, by substitution:
$y-6z = 7$ implies $y = 6z + 7$.
Plugging this into the $9x -8y=5$, you get $9x-42z-56 = 5$ which is the line you are looking for.