Few days ago I solved a task: find all planes which equidistant to $A(3,5,-1)$, $B(7,5,3)$, $C(9,-1,5)$ and $D(5,3,-3)$?
I understand that it's easy to solve, but the main problem open all modules after using formula of distance between point and plane.
My teacher said me it's could be solved easier. Actually I don't know how. Any ideas?
Suppose our plane is: $a x + b y + c z + d = 0$
Distance of a point from a plane is $d = \frac {|ax_0 + by_0 + c z_0 + d|}{\sqrt{a^2+b^2 + c^2}}$
If all the points are equidistant:
$|a x_n + b y_n + c z_n + d|$ is equal for all $4$ points.
$A = \begin{bmatrix} 3&5&-1\\7&5&3\\9&-1&5\\5&3&-3\end{bmatrix}$
To find each plane
$B_1 = \begin{bmatrix} 7&5&3\\9&-1&5\\5&3&-3\end{bmatrix}$
$B_1$ is $A$ with the first row removed.
$B\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$
$\begin{bmatrix}a\\b\\c\end{bmatrix} = B_1^{-1}\begin{bmatrix}1\\1\\1\end{bmatrix}$
to find d:
$A\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1\\1\\1\\2d-1\end{bmatrix}$
Now do the same, each time removing a different row.
$(a,b,c,d)\\ (1, 0, -1, 6)\\ (1,1,0,10)\\ (1,2,-1,14)\\ (5,1,-2,28)$
There are the 4 found with the technique above.
$(A_2 - A_1) \times (A_4-A_3), (A_3 - A_1) \times (A_4-A_2), (A_4 - A_1) \times (A_3-A_2)$
and the remaining planes:
$(1,-1,-1,2)\\ (2,1,-1,14)\\ (2,-1,-1,-16)$