I'm working with a family of planes, given by $2a_1 x_1 +2 a_2 x_2 - x_3 +a_1^2+a_2^2=0$. I would like to use mathematica to plot these planes for many values of a and b, say a range of

$-10\leq a_1\leq10$ and $-10 \leq a_2 \leq 10$. However, when I do plot it using ContourPlot3D, the answer if very distorted, as many of the planes extend and obstruct vision. Is there a way to view this without all of the "junk" to go along with it? I was thinking the intersection of all such planes might give something meaningful, maybe something resembling a n-gon of some sort, or a paraboloid.

To give context, the family of planes should describe an envelope of a function.

EDIT: What I tried doing was the following:

Show[Table[ ContourPlot3D[ 2 a x + 2 b y - z + a^2 + b^2, {x, -5, 10}, {y, -5, 10}, {z, -10, 10}], {a, -3, 3}, {b, -3, 3}]]

Notice that this gives a really gross plot.. You can't really see the function that the envelope is tangent to.

To give an example, Show@Table[{ContourPlot[ x + a^2 y - 2 a == 0, {x, 0, 5}, {y, 0, 5}]}, {a, 0, 3, .1}]

Is a graph of the family of lines of the form $x_1+a^2 x_2 - 2a = 0$ describing the envelope of some function.

  • $\begingroup$ Is $a=a_1$ and $b=a_2$? $\endgroup$
    – mickep
    Oct 27, 2014 at 7:07
  • $\begingroup$ woops, yes, will fix. $\endgroup$
    – DaveNine
    Oct 27, 2014 at 7:08
  • $\begingroup$ Welcome! How about posting the (yet unsatisfactory) code so we don't have to imagine and make up everything from scratch? Your description of the desired result is very vague at the moment, which might discourage answers. $\endgroup$
    – Yves Klett
    Oct 27, 2014 at 7:10
  • $\begingroup$ Done, hope that it helps. $\endgroup$
    – DaveNine
    Oct 27, 2014 at 7:14
  • $\begingroup$ Another idea that I had was to just plot x-y, y-z, and x-z planes, but it just didn't look as nice. We have to be careful using Table, as the amount of memory Mathematica uses increases very much so with the number of planes being graphed. $\endgroup$
    – DaveNine
    Oct 27, 2014 at 7:15

1 Answer 1



 Plot3D[2 a1 x + 2 a2 y + a1^2 + a2^2, {x, -5, 5}, {y, -10, 10}], 
 {a1, -10, 10}, {a2, -10, 10}]

enter image description here

Then you canuse the Table

 Plot3D[2 a1 x + 2 a2 y + a1^2 + a2^2, {x, -5, 5}, {y, -10, 10}], 
 {a1, 0, 10, 5}, {a2, 0, 10, 5}]]

enter image description here

  • $\begingroup$ This actually does it, thanks. I don't know why I didn't think to move the third variable over and plot it more easily. If we do Show[Table[ Plot3D[2 a1 x + 2 a2 y + a1^2 + a2^2, {x, -5, 10}, {y, -5, 10}, PlotRange -> {-10, 10}], {a1, -10, 10, 1}, {a2, -10, 10, 1}]] we can see what the enveloped function is..it looks like a paraboloid. $\endgroup$
    – DaveNine
    Oct 27, 2014 at 7:26
  • $\begingroup$ @DaveNine, If you fix the value of variable $x,y$, change the value of $a_1 \quad and \quad a_2$ ,you can achieve the equation :$$z=a_1^2+a_2^2+2x a_1+2y a_2$$ obviously,which is a paraboloid equation in relation to $a_1 \quad and \quad a_2$ $\endgroup$
    – xyz
    Oct 27, 2014 at 7:39

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