using Lagrange multipliers to determine shortest distance between a point and straight line

Question

Find the shortest distance between $(0,2,0)$ and the straight line
$x=y=z$ .

Let the required point on the straight line be $(x,y,z)$ . Now we set up the Lagrangian as $$L=x^2+(y-2)^2+z^2+lambda(x-y)+mu(y-z)+nu(z-x)$$ where $lambda,mu,nu$ are Lagrange multipliers. The extremizing equations are $$frac{partial L}{partial x}=frac{partial L}{partial y}=frac{partial L}{partial z}=0$$ which gives $$2x+lambda-nu=2(y-2)+mu-lambda=2z-mu+nu=0$$ Solving for $x,y,z$ and substituting in $x=y=z$ we get $$begin{pmatrix}2&-1&-1 \ 1&-2&1 \ 1&1&-2end{pmatrix}begin{pmatrix}lambda\ mu \ nuend{pmatrix}=begin{pmatrix}-4\ -4 \ 0end{pmatrix}$$ But I keep getting the coefficient matrix as singular matrix, so it doesn't give us the unique value of Lagrange multipliers which help to minimize the distance required.
What is possibly going wrong there? Any help is appreciated.

Christian Blatter · Accepted Answer

The line is a $1$-dimensional manifold in ${mathbb R}^3$. In order to define it you need $3-1=2$ constraints; say $x-y=0$ and $x-z=0$. Try the problem again with just these equations, and $2$ Lagrange multipliers.
As noted in a comment there are simpler ways to solve this problem, e.g., by parametrizing the line as $tmapsto(t,t,t)$.

Raffaele · Answer

The distance of the point $(0,2,0)$ from the given line $(t,t,t)$ is given by minimizing the function $f(t)=sqrt{t^2+(t-2)^2+t^2}; ; f(t) = sqrt{3 t^2-4 t+4}$
$$f'(t)=frac{6 t-4}{2 sqrt{3 t^2-4 t+4}}$$
$$f'(t)=0to t=frac{2}{3}$$
And we have $$fleft(frac{2}{3}right)=2 sqrt{frac{2}{3}}$$

using Lagrange multipliers to determine shortest distance between a point and straight line

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