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.

  • $\begingroup$ The shortest distance from the plane $Ax+By+Cz+D=0$ to the origin is $|D|/\sqrt {A^2+B^2+C^2}$. $\endgroup$ – DanielWainfleet Mar 18 '17 at 2:49

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

  • $\begingroup$ (+1) Very cool approach! Definitely not something I would've tried $\endgroup$ – Dylan Jun 2 '18 at 8:19
  • $\begingroup$ Not quite true. The plane $3x-y-4z=0$ can’t be written in this form. Fortunately, it’s not a possible solution. Also, there appears to be a typo in the equation of the second plane: its constant term should also be $3$, not $6$. $\endgroup$ – amd Feb 15 at 8:55

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.


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