# Shortest Distance from Origin - nD

This note is part of the set Shortest Distance from Origin

As work through the problems in this set, you might find a pattern in the formulae of distance of point from line, distance of point from plane and so on. In this note, we will prove that pattern.

Points in one dimensional space, lines in two dimensional space, and planes in there dimensional space are all examples of hyperplanes. A hyperplane in $\mathbb{R}^{n}$ is an $n -1$ dimensional figure, and divides $\mathbb{R}^{n}$ in two halves. It is the set of all points that satisfy

$a_{0} + a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} + \cdots + a_{n} x_{n} = 0$

for some real numbers $a_{i}$. For any given point $(b_{1}, b_{2}, b_{3},\ldots, b_{n})$, prove that the shortest distance of a hyperplane in $\mathbb{R}^{n}$ from that point is given by

$d = \frac{|a_{0} + a_{1} b_{1} + a_{2} b_{2} + a_{3}b_{3} + \cdots + a_{n}b_{n}|}{\sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + \cdots +a_{n}^{2} }}$ Note by Pranshu Gaba
5 years, 1 month ago

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## Comments

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Let $\vec{N} = (a_1,a_2,a_3,a_4,\cdots, a_n)$ be a vector in $\mathbb{R}^n$.

Let $\vec{P} = (x_1,x_2,\cdots, x_n)$ be a point in $\mathbb{R}^n$ such that $\vec{P}\cdot\vec{N}$ is always constant. Let this constant be $k (k \in R)$.

Hence, the locus of point $P$ is given by:

$\displaystyle \Rightarrow \vec{P} \cdot \vec{N} = k$

$\displaystyle \Rightarrow \displaystyle \sum_{i=1}^{n}a_i x_i = k$

The above equation represents a hyperplane in $\mathbb{R}^n$ and $\vec{N}$ is the vector normal to the hyperplane.

Comparing it with the equation of given hyperplane yields $k= - a_0$.

Now consider $\vec{B} = (b_1,b_2,\cdots, b_n)$.

The distance of point $B$ from the hyperplane will be measured along the unit vector normal to the hyperplane.

To measure this distance, we find the extent of the join of $B$ and a point $R$ on the hyperplane, along the unit vector normal to the hyperplane. Hence,

$Distance = \left| \vec{BR} \cdot \hat{N} \right|$

$Distance = \left| \frac{ (\vec{R} - \vec{B}) \cdot \vec{N}} {\left| \vec{N} \right| } \right|$

$Distance = \frac{\left| \vec{R} \cdot \vec{N} - \vec{B} \cdot \vec{N} \right|} {\left| \vec{N}\right|}$

$Distance = \frac{\left| k - \displaystyle \sum_{i=1}^n a_i b_i \right|}{\sqrt\sum_{j=1}^n a_j^2}}$

$Distance = \frac{\left| a_1b_1 + a_2b_2 + a_3b_3 + \cdots + a_nb_n - k\right|}{\sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}}$

That is for a general hyperplane. For the given hyperplane, $k=-a_0$.

$\displaystyle \Rightarrow Distance = \frac{\left|a_0 + a_1b_1 + a_2b_2 + \cdots + a_nb_n \right|}{\sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}}$

- 5 years ago

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That's correct. Well done Harsh!

- 5 years ago

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Thank you!

- 5 years ago

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