Q. 17 - Calculus

Question

How does the Pythagorean theorem generalize to higher dimensions? In particular, how would you compute the distance between two points in four-dimensional space? Five-dimensional space? n-dimensional space? 

Step-by-Step Solution

Answer

 Pythagoras' Theorem states that the square of the hypotenuse (c) of a right-angled triangle equals the sum of the squares of the other two sides (a and b) which can be applied to any number of dimensions. 

Step 1. Generalization of the Pythagorean theorem

According to Pythagoras' Theorem, the square of the hypotenuse (c) of a right-angled triangle equals the sum of the squares of the other two sides (a and b).

                                          that is    \(c^2=a^2+b^2\)

but, this can be applied to any number of dimensions.

for the nth dimension:

we can write:               \(c^2={a_1}^2+{a_2}^2+.......+{a_n}^2\)

Therefore, we can generalize Pythagoras' Theorem by including any number of dimensions after 2D, 3D, and beyond.

Step 2. computation in four-dimensional space

let the two points be A \((a_1,b_1,c_1,d_1)\) and B \((a_2,b_2,c_2,d_2)\) 

In four-dimensional space:

The distance between points A and B is given by:

\(D = \sqrt{({a_2-a_1})^2+({b_2-b_1})^2+({c_2-c_1})^2+({d_2-d_1})^2}\)

Step 3. computation in five-dimensional space

let the two points be A \((a_1,b_1,c_1,d_1,e_1)\)  and B \((a_2,b_2,c_2,d_2,e_2)\) 

In five-dimensional space:

The distance between points A and B is given by:

\(D = \sqrt{({a_2-a_1})^2+({b_2-b_1})^2+({c_2-c_1})^2+({d_2-d_1})^2+({e_2-e_1})^2}\)

Step 4. computation in n-dimensional space

let the two points be A \((a_1,a_2,a_3,......a_n)\)  and B \((b_1,b_2,b_3,......b_n)\) 

The distance between points A and B is given by:

\(D = \sqrt{({b_1-a_1})^2+({b_2-a_2})^2+.........({b_n-a_n})^2}\)