# Intersection of four planes

I have some difficulties with a problem, where i have been given four planes: $$x+ay+az=a$$ $$x+a^2y=a^3$$ $$x+a^2y+az=a$$ $$x+ay+2az=a,$$ and need to find the points in $$M$$, where any point belongs to $$M$$, if it contains the coordinates $$4$$ and $$-4$$, and there is an $$a$$ in which it lies in all four planes.

I know how to reduce the system of equations to reduced row echelon form, and have also found values of $$a$$, where they all intersect, but i can't seem to find a way to solve this specific problem. Any help would be appreciated.

• I am still a little bit confused. What exactly is $M$, and what do you mean by: "it contains the coordinates $4$ and $-4$,"? I assume the planes lie in $\mathbb{R}^{3}$. Commented Oct 17, 2019 at 8:55
• $M$ is a set of points, where all the points in $M$ need to have $4$ and $-4$ as two of their coordinates. There also needs to exist an $a$ in which a point in $M$ lies in all four planes. And yes, the planes lie in $\mathbb{R}^3$. Commented Oct 17, 2019 at 9:16

If I understood your question correctly, I believe you are looking for any points satisfying the following conditions:

(i) There must be at least one $$\:4\:$$ and one $$\:-4\:$$ among the coordinates; and

(ii) There must be at least one $$\:a\:$$ such that the point lies on all four planes.

We start by simplifying the conditions.

a) If $$\:a=0\:$$, then all equations become $$\:x=0\:$$, which yields the four points $$\:(0,\pm 4, \pm 4)\:$$.

b) If $$\:a \neq 0\:$$, then combining the first and last equations gives $$\:z=0\:$$, and combining the second and third equations gives $$\:a=\pm 1\:$$. If $$\:a=1\:$$, all four equations become $$\:x+y=1\:$$; if $$\:a=-1\:$$, we get the system $$\:x-y=-1\:$$ and $$\:x+y=-1\:$$. However, neither case allows for a solution containing a $$\:4\:$$ and a $$\:-4\:$$.

So under the stated conditions, the only solutions would be the four points $$\:(0,\pm 4, \pm 4)\:$$.

However, if the conditions were loosened slightly so that the first condition became

(i) There must be at least one $$\:4\:$$ or one $$\:-4\:$$ among the coordinates

then case a) would yield all the points on the four lines $$\:x=0, y=\pm 4\:$$ and $$\:x=0, z=\pm 4\:$$;

and case b) with $$\:a=1\:$$ would yield the points: $$\:(5,-4,0),(4,-3,0),(-3,4,0)\:$$ and $$\:(-4,5,0)\:$$.

Although I do not understand the question in detail, it is clear by direct computation that the system has exactly the following solutions over a field of characteristic zero. Either we have $$z=0,\; a=1,\; x+y=1,$$ or we have $$a=x=0,$$ or we have $$z=0,\; x=-1,\; y=0.$$ In particular, the last solution holds for arbitrary $$a\in K$$.