Suppose you have the coordinates of three points $A,B,C$. You want to construct a plane $\pi$ such that
$ d(\pi, A) = r_1 $
$ d(\pi, B) = r_2 $
$ d(\pi , C ) = r_3 $
where $d(\pi,v) $ is the perpendicular distance function, which returns the distance between point $v$ and the plane $\pi$.
What are the conditions on $r_1, r_2, r_3$ so that a solution exists? And what is the equation of this plane in terms of the coordinates of $A,B,C$ and the values of $r_1, r_2, r_3 $?
My attempt:
Starting with the first point $A$, a plane that is $r_1$ away from $A$, must be of the form
$ n \cdot ( r - (A + r_1 n ) ) = 0 \tag{1} $
where $n$ is a unit vector. The envelope of all such planes is a sphere centered at $A$ and with radius $r_1$.
Applying the second condition, we find that
$ r_2 = | n \cdot ( B - A) - r_1 | \tag{2}$
And applying the third condition, we get
$ r_3 = | n \cdot (C - A) - r_1 | \tag{3}$
First, let's assume that,
$ | n \cdot(B - A) - r_1 | = n \cdot(B - A) - r_1 \tag{4} $
This implies that
$ n \cdot (B - A) \ge r_1 \tag{5} $
And from $(2)$ and $(4)$,
$ n \cdot (B - A) = r_1 + r_2 \tag{6} $
Since $ n \cdot (B - A) $ is bounded from above by $ \| B - A \| $ and from below by $- \| B - A \| $
Then under the assumption of $(4)$, we must have
$ - \| B - A \| \le r_1 + r_2 \le \| B - A \| \tag{7} $
The left inequality is automatically satisfied because $r_1 + r_2$ is a positive number.
Next, if we assume that
$ | n \cdot(B - A) - r_1 | = r_1 - n \cdot(B - A) \tag{8} $
This can only happen if
$ n \cdot (B - A) \lt r_1 \tag{9} $
Now from $(8)$ and $(2)$,
$ n \cdot (B - A) = r_1 - r_2 \tag{10} $
Hence, solutions will exist in this case if and only if
$ - \| B - A \| \le r_1 - r_2 \le \| B - A \| \tag{11}$
Repeating the same steps for point $C$ gives two more conditions:
$ r_1 + r_3 \le \| C - A \| \tag{12} $
$ - \| C - A \| \le r_1 - r_3 \le \| C - A \| \tag{13}$
So we have four cases for the signs of $ ( n \cdot (B - A) - r_1 )$ and $( n \cdot (C - A) - r_1) $. Taking this cases one by one, and applying the above inequalities, and assuming they are satisfied, then we can have a maximum of $4$ systems of equations, each having two possible solutions. Therefore a maximum of $8$ distinct planes will solve the problem.
I would greatly appreciate it if someone could go over the above analysis, and verify these findings.
Thank you all.