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 Rn \mathbb{R}^{n} is an n−1 n -1 dimensional figure, and divides Rn \mathbb{R}^{n} in two halves. It is the set of all points that satisfy

a0+a1x1+a2x2+a3x3+⋯+anxn=0a_{0} + a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} + \cdots + a_{n} x_{n} = 0

for some real numbers aia_{i}. For any given point (b1,b2,b3,…,bn) (b_{1}, b_{2}, b_{3},\ldots, b_{n}) , prove that the shortest distance of a hyperplane in Rn\mathbb{R}^{n} from that point is given by

d=∣a0+a1b1+a2b2+a3b3+⋯+anbn∣a12+a22+a32+⋯+an2 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} }}

#Geometry

Note by Pranshu Gaba
5 years, 3 months ago

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Let N⃗=(a1,a2,a3,a4,⋯ ,an)\vec{N} = (a_1,a_2,a_3,a_4,\cdots, a_n) be a vector in Rn\mathbb{R}^n.

Let P⃗=(x1,x2,⋯ ,xn)\vec{P} = (x_1,x_2,\cdots, x_n) be a point in Rn\mathbb{R}^n such that P⃗⋅N⃗\vec{P}\cdot\vec{N} is always constant. Let this constant be k(k∈R)k (k \in R) .

Hence, the locus of point PP is given by:

⇒P⃗⋅N⃗=k\displaystyle \Rightarrow \vec{P} \cdot \vec{N} = k

⇒∑i=1naixi=k\displaystyle \Rightarrow \displaystyle \sum_{i=1}^{n}a_i x_i = k

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

Comparing it with the equation of given hyperplane yields k=−a0k= - a_0.

Now consider B⃗=(b1,b2,⋯ ,bn)\vec{B} = (b_1,b_2,\cdots, b_n) .

The distance of point BB 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 BB and a point RR on the hyperplane, along the unit vector normal to the hyperplane. Hence,

Distance=∣BR⃗⋅N^∣Distance = \left| \vec{BR} \cdot \hat{N} \right|

Distance=∣(R⃗−B⃗)⋅N⃗∣N⃗∣∣Distance = \left| \frac{ (\vec{R} - \vec{B}) \cdot \vec{N}} {\left| \vec{N} \right| } \right|

Distance=∣R⃗⋅N⃗−B⃗⋅N⃗∣∣N⃗∣Distance = \frac{\left| \vec{R} \cdot \vec{N} - \vec{B} \cdot \vec{N} \right|} {\left| \vec{N}\right|}

Distance=∣k−∑i=1naibi∣∑j=1naj2Distance = \frac{\left| k - \displaystyle \sum_{i=1}^n a_i b_i \right|}{\sqrt{\displaystyle \sum_{j=1}^n a_j^2}}

Distance=∣a1b1+a2b2+a3b3+⋯+anbn−k∣a12+a22+⋯+an2Distance = \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=−a0k=-a_0.

⇒Distance=∣a0+a1b1+a2b2+⋯+anbn∣a12+a22+⋯+an2\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}}

Harsh Khatri - 5 years, 3 months ago

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That's correct. Well done Harsh!

Pranshu Gaba - 5 years, 3 months ago

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Thank you!

Harsh Khatri - 5 years, 3 months ago
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