distance between intersection of two planes and origin

Question

I have a following problem. I want to find the smallest possible distance between the line intersection of two planes given by:

\begin{equation} x + 2y−2z = 3 \text{ and } 2x + y + 2z = 6 \end{equation}

I already found line of intersection which is given by: \begin{equation} \begin{aligned} x &= 3 + 2t \\ y &= − 2t \\ z &= − t \end{aligned} \end{equation}

But I could not figure out how to use this to find the minimum point so I thought of using Lagrangian method which would look like following:

\begin{equation} dist = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^ 2} \end{equation}

subject to constraints given by the equations of the two planes but the problems seems too unwieldy and difficult. if anyone can help I would really appreciate it.

Answer 1

You can make the equations more tractable by using the square of the distance as the objective function instead. However, there’s also a fairly straightforward solution method based on geometric considerations.

The minimal distance to the line is measured along a direction perpendicular to the line, so the nearest point on the line to the origin is the intersection of the line with its perpendicular plane through the origin. You have $(2,-2,-1)^T$ as a direction vector for the line, so an equation of this plane is $2x-2y-z=0$. You will end up with a simple linear equation in $t$ to solve.

Answer 2

Let us see what happen with the Lagrange Multipliers formulation

$$ \Pi_1\to (p-p_1)\cdot \vec n_1 = 0\\ \Pi_2\to (p-p_2)\cdot \vec n_2 = 0 $$

with $p = (x,y,z)$ are the two restriction planes so

$$ L(p,\lambda_1,\lambda_2) = \frac 12p\cdot p+\lambda_1(p-p_1)\cdot \vec n_1+\lambda_2(p-p_2)\cdot \vec n_2 $$

and the stationary conditions are

$$ L_p = p +\lambda_1\vec n_1+\lambda_2\vec n_2 = 0\\ L_{\lambda_1} = (p-p_1)\cdot \vec n_1 = 0\\ L_{\lambda_2} = (p-p_2)\cdot \vec n_2 = 0\\ $$

then substituting we have two equations and two unknowns $\lambda_1,\lambda_2$

$$ \lambda_1\vec n_1\cdot\vec n_1+\lambda_2\vec n_2\cdot\vec n_1 = p_1\cdot \vec n_1\\ \lambda_1\vec n_1\cdot\vec n_2+\lambda_2\vec n_2\cdot\vec n_2 = p_2\cdot\vec n_2 $$

finally we have after the solutions $\bar\lambda_1,\bar\lambda_2$

$$ p^* = \bar\lambda_1\vec n_1+\bar\lambda_2\vec n_2 $$

so $d = \lVert p^*\rVert$

NOTE

Here the minimization of $\lVert p\rVert$ is equivalent to the minimization of $\frac 12\lVert p\rVert^2 = \frac 12p\cdot p$ and also

$$ \vec n_1 = (1,2,-2)\\ p_1 = (3,0,0)\\ \vec n_2 = (2,1,2)\\ p_2 = (0,6,0) $$