Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Use Lagrange multipliers to determine the shortest distance from a point $\,x \in R^n\,$ to a plane $\{y\mid b^Ty = c\}.$

I don't even know where to start!

share|cite|improve this question

To use Lagrange multipliers, you must write this problem as a constrained minimisation problem. Let $\Vert \cdot \Vert$ be the norm associated to the canonical dot product in $\mathbb{R}^{n}$ and $\mathcal{P} = \lbrace y \in \mathbb{R}^{n}, \, {}^t b y = c \rbrace$. If I'm not mistaken, your problem is :

$$ \min \limits_{y \in \mathcal{P}} \Vert x - y \Vert^{2} $$

which can also be written :

$$ \min \limits_{y \in \mathbb{R}^{n}} \Vert x - y \Vert^{2} \\ \text{s.t.} \, {}^t b y = c $$

Then, you can introduce Lagrange multipliers...

share|cite|improve this answer

Assume that you know something about optimization, your described problem then can be rewritten as $${\rm{minimize}} \rho(x,y)=\|x_{n\times 1}-y_{n\times 1}\|\\ {\rm{such\,that}}\,\, b^Ty=c$$ where $\|.\|$ is your preferred distance measure, e.g. euclidean distance.

To solve this optimization problem, then you need the Lagrange Multiplier to encode this constraint $b^Ty=c$ into the original minimization objective function $\rho(x,y)=\|x_{n\times 1}-y_{n\times 1}\|$ and to form a new one as follows $$J(y)= \|x_{n\times 1}-y_{n\times 1}\|+\lambda(b^Ty-c)$$ where $\lambda$ is the Lagrange multiplier.

Suppose $y^*$ is the solution that minimizes $J(y)$ above, i.e. $$y^*=\arg\min_{y\in R^n}J(y)$$ then $\rho(x,y^*)$ tells you the shortest distance from a known point $x$ to the plane.

Note: $y^*$ is dependent on the selected distance measure $\|.\|$. In other words, if you consider a different distance measure, then the resulting $y^*$ is also different. You may see this page for more information about different distance measures (norm in mathematics).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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