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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?

openspace
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1 Answers1

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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:

$(1,-1,-1,2)\\ (2,1,-1,14)\\ (2,-1,-1,-16)$

Doug M
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  • Yes , my solution is the same. But , it's not easy to consider all modules. – openspace Dec 22 '16 at 22:50
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    Since any 3 points form a plane, there is a parallel plane between each collection of 3 points, that is equidistant to the 4th point. And since there are 4 ways to collect 3 points there are 4 planes.... – Doug M Dec 22 '16 at 23:11
  • I think your idea isn't fully true. Consider plane z = 0, points (1,0,2) (-1,0,2) (1,0,-2) and (-1,0,-2) plane (z = 0) equidistant for all points , but it's also intersect any plane with any 3 points. So there must be more solutions. And I guess there will be 7 solutions. – openspace Dec 23 '16 at 00:37
  • @openspace In "computational geometry" (this is the name of the big area in which such questions are treated), one generally avoids degenerate cases (such as 4 coplanar points as in the example you just gave). – Jean Marie Dec 23 '16 at 01:04
  • @JeanMarie I mean the next: consider plane $x-y-z-2 = 0$ and this plane is equidistant from this 4 points, but this solution doesn't hold this plane. – openspace Dec 23 '16 at 01:12
  • The 4 points form a tetrahedron, and I identified the planed paralllel to each face, by you are correct, between each pair of opposite edges, there fits a plane... so you found one $x-y-z-2 = 0$ – Doug M Dec 23 '16 at 01:37
  • @DougM so I guess tere must be also all planes between opposite edges , there 3 opposite pairs of edges, so there 7 solutions ? – openspace Dec 23 '16 at 01:41
  • @DougM how can I get the last three planes? My first idea consider tetrahedron and find middle points of some edges and find plane ,which consist of this points. – openspace Dec 23 '16 at 02:04
  • I put an update. Find the vector between one pair of point. Find the vector between the other pair, find the cross product. That gives you the normal to the plane. Then solve for $d.$ – Doug M Dec 23 '16 at 02:25