I need to find the shortest distance, in D-dimensional Euclidean space ($\mathbb{R}^D$) from a point $\textbf{x}_0$ to a hyperplane $H: \textbf{w}^T \textbf{x} + b = 0$, using the method of Lagrange multipliers. The answer should be an expression in terms of $\textbf{w}, b$ and $\textbf{x}_0$.

Note: I am aware that a few similar questions exist, such this one. I am creating a new question because I need to know how the derivation steps work in order to get a solution in a specific form. I know how to solve this problem in three dimensions, but not with linear algebra. Any help would be appreciated.


Consider the Lagrange function $$L(\mathbf x,\lambda)=\|\mathbf x-\mathbf x_0\|^2+2\lambda(\mathbf w^T\mathbf x+b)$$ The Lagrange multiplier is multiplied by $\,2\,$ to simplify the computations (this is legal).

Since $$L(\mathbf x,\lambda)=\|\mathbf x\|^2-2(\mathbf x_0-\lambda \mathbf w)^T\mathbf x+2\lambda b+\|\mathbf x_0\|^2$$ one has $$\frac {\partial L}{\partial \mathbf x}=2\mathbf x-2(\mathbf x_0-\lambda \mathbf w)$$ using formulas (69) and (131) in the pdf quoted by @user25004.

Solving $\,\dfrac {\partial L}{\partial \mathbf x}=0\,$, one obtains $$\mathbf x=\mathbf x_0-\lambda \mathbf w$$ which, substituted in the equation of the hyperplane, gives $$\lambda=\frac {\mathbf w^T\mathbf x_0+b}{\|\mathbf w\|^2}$$ so the shortest distance is $$\|\mathbf x_0-\lambda \mathbf w-\mathbf x_0\|=|\lambda|\|\mathbf w\|=\frac {|\mathbf w^T\mathbf x_0+b|}{\|\mathbf w\|}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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