2
$\begingroup$

I want to find a plane that passes through points {1,0,0} and {0,1,0} and is tangent to surface $z(x,y)=x^{2}+y^{2}$.

Solve[{a, b, c}.{1, 0, 0} == d && a*0 + b*1 + c*0 == d && 
  a*x0 + b*y0 + c*z0 == d && z0 == x0^2 + y0^2 && 
  VectorAngle[{a, b, c}, {-2 x0, -2 y0, 1}] == 
   0,(*MatrixRank[{2x0,2y0,1},{a,b,c}]\[Equal]1*){a, b, c, d, x0, y0, 
  z0}]

But I can't get the answer I want with the above code(the answer is $z=0$ and $2x+2y-z=2$). What should I do?

$\endgroup$

5 Answers 5

4
$\begingroup$
Clear["`*"];
f = x^2 + y^2 - z;
Solve[{Grad[f, {x, y, z}].({x, y, z} - {1, 0, 0}) == 0 , 
  Grad[f, {x, y, z}].({x, y, z} - {0, 1, 0}) == 0, f == 0}, {x, y, z}]
Grad[f, {x, y, z}].({X, Y, Z} - {x, y, z}) == 0 /. % // Simplify
$\endgroup$
5
$\begingroup$

This is pretty easy to do with the extant region functionality:

eq = {u, v, u^2 + v^2};
plane = InfinitePlane[eq, Transpose[D[eq, {{u, v}}]]];

sols = plane /. Solve[RegionMember[plane, #] & /@ {{1, 0, 0}, {0, 1, 0}}, {u, v}]
   {InfinitePlane[{0, 0, 0}, {{1, 0, 0}, {0, 1, 0}}], 
    InfinitePlane[{1, 1, 2}, {{1, 0, 2}, {0, 1, 2}}]}

If you want to see the plane equations themsleves, you need an extra step:

Simplify[RegionMember[#, {x, y, z}], {x, y, z} ∈ Reals] & /@ sols
   {z == 0, 2 + z == 2 (x + y)}
$\endgroup$
3
$\begingroup$

I expect two solutions!

First we consider the plane p0,p1,p2

p1 = {1, 0, 0};
p2 = {0, 1, 0};
p0 = {x0, y0, x0^2 + y0^2};

with normal

en= Cross[p0 - p1, p0 - p2];

The normal is forced to be pependicular to the tangentplane at p0

p0/. Solve[{n.D[p0, x0] == 0, n.D[p0, y0] == 0}, {x0, y0}, Reals]
(*{{0, 0, 0}, {1, 1, 2}}*)  

GraphicsRow[{Show[{Plot3D[x^2 + y^2, {x, -3, 3}, {y, -3, 3},Mesh -> None,BoxRatios -> {1, 1, 1}]
, Graphics3D[{Point[{p0, p1, p2}], InfiniteLine[{p1, p2}], InfinitePlane[{p1, p2, p0}]} /. sol[[1]]]}],
Show[{Plot3D[x^2 + y^2, {x, -3, 3}, {y, -3, 3}, Mesh -> None,BoxRatios -> {1, 1, 1}]
, Graphics3D[{Point[{p0, p1, p2}], InfiniteLine[{p1, p2}], InfinitePlane[{p1, p2, p0}]} /. sol[[2]]]}]}]

enter image description here

$\endgroup$
3
$\begingroup$

Try this:

Gxyz = z - x^2 - y^2;
p = {x, y, z};
p1 = {1, 0, 0};
p2 = {0, 1, 0};
n = Grad[Gxyz, p]
equ1 = n.(p - p1) == 0
equ2 = n.(p - p2) == 0
equ3 = Gxyz == 0
sol = Solve[{equ1, equ2, equ3}, p]
n0 = n /. sol

gr1 = ContourPlot3D[Gxyz == 0, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}];
gr2 = ContourPlot3D[n0[[1]].(p - p1) == 0, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}];
gr3 = ContourPlot3D[n0[[2]].(p - p1) == 0, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}];
Show[gr1, gr2, gr3]
$\endgroup$
2
$\begingroup$
Solve[{Resolve[
   ForAll[{x, y}, x^2 + y^2 + a*x + b*y + c >= 0] && 
    Exists[{x, y}, x^2 + y^2 + a*x + b*y + c == 0], 
   Reals], {a, b, -1}.{1, 0, 0} + c == 0, {a, b, -1}.{0, 1, 0} + c == 
   0}, {a, b, c}]
a*x + b*y - z + c == 0 /. %
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.