How to find the equation of the plane passing through the intersection of two other planes and whose perpendicular distance from the origin is given?

Question

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.

Answer 1

When the equation of the plane is written in the form $\vec n \cdot \vec r = d$, where $\vec n$ is the unit normal vector to the plane, directed away from the origin, and $\vec r$ is a point on the plane, then $d$ is the distance from the plane to the origin.

So in this case, the equation is $\vec n \cdot \vec r = 1$. But note the requirement that $\vec n$ be a unit vector. So one equation for $\vec n$ is that $\|\vec n\|^2 = 1$.

Since the line of intersection lies within our plane, the direction vector for that line has to point along the plane, and therefore must be normal to $\vec n$. So a second equation for $\vec n$ is $\vec n \cdot \vec v = 0$, where $\vec v = (-12, 4,-10)$.

Finally, there are numerous lines with this same direction vector. But there is only one that we are interested in: the line that is the intersection of the other two planes. To get this line and not any of the others, we need a point on this line. This is fairly easy to find. Pick a value for one of the variables, then solve the system of the two given planar equations to find the other two variables. Once you have this point, call it $\vec r$, recall that since it is a point on the plane you are looking for, it satisfies the planar equation. Which gives you a third equation in $\vec n$: $$\vec n \cdot \vec r = 1$$ $$\|\vec n \|^2 = 1$$ $$\vec n \cdot \vec v = 0$$ This is three equations in the three unknowns of the components of $\vec n$. So solve for them and you are done.

Answer 2

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 $-x+2y+2z+6=0$