HCR - Optimization Problem

We can also see this as an optimization problem:

$$\begin{array}{l}\min ||x_p-x_h||\\ x_p\in \Sigma \end{array}$$

Then:

$$\begin{array}{l}||x_p-x_h||=\sqrt{(x_p-x_h{)}^2}\\ \\ \begin{array}{lc}\Rightarrow \min ||x_p-x_h||& =\min (x_p-x_h{)}^2\\ & |\\ & =\min (x_p-x_h{)}^2\\ & |\\ & =\alpha \cdot \min (x_p-x_h{)}^2,\alpha >0\end{array}\end{array}$$

→ We can reformulate the problem as searching for:

$$\min \frac{1}{2}(x_p-x_h{)}^2$$

Being vectors $x_p$ and $x_h$ we can write:

$$\min \frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T$$

And also we rewrite the constraint: $x_p\in \Sigma$, ($ax+by+cz+d=0$) as:

$$[a,b,c]\cdot x_p+d=0$$

Then the problem becomes:

$$\begin{array}{l}\min \frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T\\ [a,b,c]\cdot x_p+d=0\end{array}$$

Lagrangian Approach:

Using the HCR - Lagrangian Theorem we can rewrite the problem as:

$$Q(x_p,\lambda )=\frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T+{\lambda }^T\cdot \left([a,b,c]\cdot x_p+d\right)$$

and search for its minimum, so:

$$(x_p,\lambda ):\{\begin{array}{l}\frac{\partial Q(x_p,\lambda )}{\partial x_p}=0\\ \frac{\partial Q(x_p,\lambda )}{\partial \lambda }=0\end{array}$$

Then we have:

$$(x_p,\lambda ):\{\begin{array}{l}(x_p-x_h)+{\lambda }^T\cdot [a,b,c]=0\\ [a,b,c]\cdot x_p+d=0\end{array}$$

So, with a bit of substitutions we obtain:

$$x_p=-\frac{[a,b,c]}{||[a,b,c]|{|}^2}\cdot d$$

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