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I want to find the numbers $a$, $b$, $c$, $d$ of the plane $ax + by +cz +d = 0$ which make the distance from the point $M(1,2,3)$ to the plane equal to 2. I tried 

SetSystemOptions[
  "ReduceOptions" -> "ExhaustiveSearchMaxPoints" -> {10000, 100000}];

Clear[n, M];
n = {a, b, c};
M = {1, 2, 3};
TraditionalForm[Simplify[a  x +  b y +  c z + d == 0]] /. 
 Solve[{Abs[n . M + d]/Norm[n] == 2, 
   1 <= a <= 10, -5 <= b <= 10, -5 <= c <= 10, a > b > c}, {a, b, c, 
   d }, Integers]

The set of equations in Mathematica's output contains duplicates, for example: $8 x+6 y-40=0$ and $4 x+3 y-20=0$. How can I eliminate the duplicates?

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3 Answers 3

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One possible way is to filter your list afterwards and delete duplicates of answers. Two answers are equal when their coefficients {a,b,c,d} are linear dependent. So when you have two vectors which are linear dependent, their normalized versions v1 and v2 are either equal or they are ones negative: v1==-v2.

You can use DeleteDuplicates with the above rule:

SetSystemOptions[
    "ReduceOptions" -> "ExhaustiveSearchMaxPoints" -> {10000, 100000}];

Clear[n, M];
n = {a, b, c};
M = {1, 2, 3};
sol = Solve[{Abs[n . M + d]/Norm[n] == 2, 
      1 <= a <= 10, -5 <= b <= 10, -5 <= c <= 10, a > b > c}, {a, b, c,d }, Integers]

TraditionalForm[Simplify[a x + b y + c z + d == 0]] /. 
 DeleteDuplicates[sol, 
  With[{v1 = Normalize[{a, b, c, d} /. #1], 
     v2 = Normalize[{a, b, c, d} /. #2]},
    v1 == v2 || v1 == -v2] &]

This gives only 32 of your 38 original solutions and the two equations you mentioned are reduced to the second one.

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I would recommend Reduce and EuclideanDistance function. 

r = Reduce[ a x + b y + c z + d == 0 && EuclideanDistance[{1, 2, 3}, {x, y, z}] == 2,
            {a, b, c, d}, Integers]

 (a | b | c | d) ∈ Integers && (
    (z == 1 && y == 2 && x == 1 && d == -a - 2 b - c) || 
    (z == 3 && ((y == 0 && x == 1 && d == -a - 3 c) ||
                (y == 2 && ((x == -1 && d == a - 2 b - 3 c) || 
                            (x == 3 && d == -3 a - 2 b - 3 c))) ||
                (y == 4 && x == 1 && d == -a - 4 b - 3 c))) ||
    (z == 5 && y == 2 && x == 1 && d == -a - 2 b - 5 c))

Then e.g. having set an appropriate SetSystemOptions

Normal @ Solve[1 <= a <= 5 && -5 <= b <= 5 && -5 <= c <= 5 && a > b > c && r,
               {a, b, c, d}, Integers]

then migt be a need for e.g. Factor and DeleteDuplicates.

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  • $\begingroup$ please write for me DeleteDuplicates in your answer. I can't write. Thank you. $\endgroup$
    – minthao_2011
    Jan 28, 2013 at 16:33
  • $\begingroup$ @minthao_2011 My answer was only step toward another possible way to proceed. I'll try to improve it a bit later, perhaps tomorow. Accepting another answer can discourage others because they might think you've got quite a good answer and it is less rewarding for anyone who could provide a better solution. $\endgroup$
    – Artes
    Jan 28, 2013 at 18:36
  • $\begingroup$ @minthao_2011 My approach is If[ Factor[#] === #, #, Last@Factor[#[[1]]] == 0] & /@ YourList // DeleteDuplicates // TraditionalForm. It works well. Nontheless I don't understand why you are looking only for integer solutions. Real solutions with help of Reduce look nice e.g. Reduce[{Abs[n.M + d]/Norm[n] == 2, 1 <= a <= 10, -5 <= b <= 10, -5 <= c <= 10, a > b > c}, {a, b, c, d}, Reals]. My solution describes all integer points lying on all planes (of integer coefficients) intersecting the sphere, so it was a bit different than your original search. But it is not clear what you wanted. $\endgroup$
    – Artes
    Feb 4, 2013 at 17:22
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You only need add GCD[a,b,c,d]==1 in your code. 

SetSystemOptions[
  "ReduceOptions" -> "ExhaustiveSearchMaxPoints" -> {10000, 100000}];
Clear[n, M];
n = {a, b, c};
M = {1, 2, 3};
TraditionalForm[Simplify[a x + b y + c z + d == 0]] /. 
 Solve[{Abs[n.M + d]/Norm[n] == 2, 
   1 <= a <= 10, -5 <= b <= 10, -5 <= c <= 10, a > b > c, 
   GCD[a, b, c, d] == 1}, {a, b, c, d}, Integers]
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