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I know this question may seem duplicate, but having searched all the possibilities, I could not find exact answer to this problem. Having a surface, for example $x^2+y^2-z^2=-1$, and a plane, such as $x+2y+3z=0$, what are the major techniques to evaluate the distance between surface and plane? Any complete explanation will be strongly appreciated.

Note: In fact, I want to evaluate shortest distance by using Lagrange Multipliers. So if there is not any concrete method, any help towards "shortest distance" by using Lagrange method would also be kindly appreciated.

BlizzardWalker
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3 Answers3

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I will say standard way to work is : express points in the manifolds as variables, and express the distance between two points as the function of variables. In your example, point in the manifold $p_m$ can be expressed as : $p_m(x_m,y_m)=(x_m,y_m,\pm\sqrt{x_m^2+y_m^2+1})$ and similarily point in the plane $p_b=p_b(x_b,y_b)=(x_b,y_b,\frac{x_b+2y_b}{3})$. Therefore the distance of two point is the function (I assume your distance is given as Euclidean distance) : $$d(x_m,y_m,x_b,y_b)=\sqrt{(x_m-x_b)^2+(y_m-y_b)^2+(\pm\sqrt{x_m^2+y_m^2+1}-\frac{x_b+2y_b}{3})^2}.$$ Now this is the optimization problem, so one can use adepquate methods depending on the problem.

Jingeon An-Lacroix
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  • Is it still possible to find distance without finding the third variable ? In particularly, I am curious if we used all three variables as they were, not substituting $z$ via $x$ and $y$. – BlizzardWalker Apr 28 '20 at 16:51
  • @DarkKnight I understand what you want, but I don't think there is 'standar' way of doing that. Maybe for some special manifolds that you already know it's precise properties (e.g. sphere), but otherwise since manifold must be parametrizable, so method stated above is always possible. – Jingeon An-Lacroix Apr 28 '20 at 17:00
  • One generally works with the square of the distance to simplify the calculations. – amd Apr 28 '20 at 17:28
  • @amd Yeah, of course but that’s included in the solving optimization problem. – Jingeon An-Lacroix Apr 28 '20 at 17:39
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If $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are points in space, the distance between them is minimized when the function $$ f((x_1,y_1,z_1),(x_2,y_2,z_2)) = (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1- z_2)^2 $$ is minimized. But if $(x_1,y_1,z_1)$ is on the surface, and $(x_2,y_2,z_2)$ is on the plane, you have two constraints given by \begin{align*} g_1(x_1,y_1,z_1) &= x_1^2 + y_1^2 -z_1^2 \\ g_2(x_2,y_2,z_2) &= x_2 + 2y_2 + 3z_2 \end{align*} The Lagrange Multiplier equations give six equations in the eight variables \begin{align*} 2(x_1 - x_2) &= \lambda_1 2x_1 \\ 2(y_1 - y_2) &= \lambda_1 2y_1 \\ 2(z_1 - z_2) &= -\lambda_1 2z_1 \\ -2(x_1 - x_2) &= \lambda_2 \\ -2(y_1 - y_2) &= 2\lambda_2 \\ -2(z_1 - z_2) &= 3\lambda_2 \end{align*} Then there are the two constraint equations $g_1(x_1,y_1,z_1) = -1$ and $g_2(x_2,y_2,z_2) = 0$.

Matthew Leingang
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Only minimum distance between the plane and the surface will be unique and it will be required

The direction ratios of the normal to the surface $F(x,y,z)=0$ is given as $$\vec n =\left(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right)=(2x,2y,-2z)$$ $n\\vec n$ needs to be parallel to the vector normal to the plane $\vec n'=(1,2,3)$, so we have $$\vec n =k \vec n' \implies x=k/2, y=k, z=-3k/2$$ Next $x,y,z)$ lie onthe surface $$x^2+y^2-z^2=-1 \implies k^2/4+k^2-9k^2/4=-1 \implies $k\=\pm 1 $$ So the required minimum distance is the perpendicular distance of the plane from the point $(1/2,1,-3/2)$ which is $$d=\frac{|1/2+2-9/2|}{\sqrt{1+4+9}}=\sqrt{\frac{2}{7}}$$

Z Ahmed
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