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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!

amWhy
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Natalie
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2 Answers2

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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...

pitchounet
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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).

pitfall
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