Let there be 2 planes: $x-y+z=2, 2x-y-z=1$ Find the equation of the line of the intersection of the two planes, as well as that of another plane which goes through that line.
Attempt to solve: the line equation was according to what I calculated (1,0,1) +t(2,3,1), and I have no idea as for the second task..
The equation of the line that is the intersection of the two planes is the solution set of the system:
$x - y + z = 2$ and $2x - y - z = 1$. Let $z = t$ be any real number, then:
$x - y = 2 - t$,and $2x - y = 1 + t$. So $x = 1 + t - (2 - t) = 2t - 1$, and $y = x - (2 - t) = 2t - 1 - (2 - t) = 3t - 3$. So $(x,y,z) = (2t - 1, 3t - 3, t) = (-1, -3, 0) + t(2, 3, 1)$.
For finding the equation of another plane that goes through this line, there can be infinitely many of them. To find just one example of such a plane, we can choose a point that lies on it. Take this point $A = (0, 0, 1)$. Let $t = 0$ we have point $B = (-1, -3, 0)$, and let $t = 1$, we have point $C = (1, 0, 1)$. Now we have $3$ points $A, B, C$ we can write an equation of a plane that goes thru them, and this is the plane you ask for.