A Generalization of the Distance Formula

Geometry Level pending

Everybody knows that the distance of a point (3D vector) $q$ from the plane $n \cdot ( r - r_0 ) = 0$ is given by

$d = \dfrac{ | n \cdot (q - r_0) | }{ \sqrt{ n \cdot n } }$

In this problem, I generalize the formula to $\mathbb{R}^n$ , and to multiple hyperplanes.

So consider the point $q = (3, 3, 4, 4, 5) \in \mathbb{R}^5$ , we want to find the minimum distance between $q$ and a point $x =(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5$ that lies on the intersection of the following hyperplanes

$x_1 + 2 x_2 + 3 x_3 + 4 x_4 + 5 x_5 = 10$

$3 x_1 + 2 x_2 - 4 x_3 + 5 x_4 + 6 x_5 = 20$

Report your answer rounded to 3 decimal places.

Details and Assumptions:

The distance between $x$ and $q$ is defined as

$d = \sqrt{ (x - q) \cdot (x - q ) } = \sqrt{ (x - q)^T (x - q) }$

The answer is 7.254.

1 solution

Hosam Hajjir
Apr 24, 2017

The problem can be stated as follows:

$\text{Minimize } (x- q)^T (x - q) , \text{subject to } A^T x = b$

Direct application of the Lagrangian multipliers method yields the desired result.

Define $g(x) = (x - q)^T (x - q) + m^T(A^T x - b)$

where $m$ is the Lagrangian multipliers vector. Then, the gradient with respect to $x$ is

$\nabla_x {g} = 2 (x - q) + A m = 0 \cdots \cdots (1)$

and the gradient with respect to $m$ is

$\nabla_m {g} = A^T x - b = 0 \cdots \cdots (2)$

From (1), we obtain,

$x = q - (1/2) A m \cdots \cdots (3)$

Substituting this into (2),

$A^T (q - (1/2) A m) - b = 0$

$-(1/2) A^T A m = b - A^T q$

Therefore,

$m = -2 (A^T A)^{-1} (b - A^T q)$

Substituting this into (3),

$x - q = -(1/2) A m = A (A^T A)^{-1} (b - A^T q)$

Finally,

$\begin{aligned} f_{\text{min}} &= \sqrt{(x - p)^T (x - p)} = \sqrt{ (b - A^T q)^T (A^T A)^{-T} A^T A (A^T A)^{-1} (b - A^T q)} = \\ &= \sqrt{ (b - A^T q )^T (A^T A)^{-1} (b - A^T q)} \\ &= \sqrt{ (A^T q - b )^T (A^T A)^{-1} (A^T q - b)} \\ \end{aligned}$

And this is the generalized formula. What remains is to substitute the given values to obtain $f_{\text{min}}=7.254$

A Generalization of the Distance Formula

The answer is 7.254.

1 solution

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