I've been trying to find a way to work this out for hours now with no luck.
The question is:
Find the equation of the plane(s) passing through the intersection of
the planes $x+3y+6=0$ and $3x-y-4z=0$ and whose perpendicular distance
from the origin is unity.
What I've tried so far:
I found the direction of the line passing through the intersection by cross multiplying the normal vectors of the two given planes.
The direction of the line at the intersection is: $<-12, 4, -10>$
I'm thoroughly confused right now and have no idea what to do from here on. Finding the equation of the line at the intersection is something I can do, but I have no idea how to find the equation of a plane at the intersection which is also at a distance of $1$ from the origin. Any help would be much appreciated.
Best Answer
Any plane passing through the intersection of the given planes can be written in the form $x+3y+6 + k(3x-y-4z) = 0$. The perpendicular distance of this from the origin is $$\frac{6}{\sqrt{(1+3k)^2+(3-k)^2+16k^2}} = \pm 1$$ and hence we have $36k^2 = 36$ and $k = \pm 1$. Thus the planes are $2x+y-2z+3 = 0$ and $-x+2y+2z+6=0$