Question

Show that for any two vectors x and y which are orthogonal to each other
(that is (x, y) = 0) we have the following relation:

(1/4)||x+y||^2 =(1/4)||x-y||^2

Answer 1

http://openstudy.com/users/estudier#/users/estudier/updates/4e47b3750b8b8d00ebdb8779

Answer 2

In these types of questions you need to take one side (either LHS or RHS) and then try to make them look the same.

So: if x = [x1,x2] y= [y1,y2]

Then

LHS = (1/4)(abs(x+y))^2
= sqrt((x1+y1)^2 + (x2 + y2)^2)/4
= sqrt(x1^2 + y1^2 + 2*x1*y1 + x2^2 + y2^2 + 2*x2*y2)/4
= sqrt(x1^2 + y1^2 + 0 + x2^2 + y2^2 + 0)/4 (as (x,y)=0)

RHS = (1/4)(abs(x-y))^2
= sqrt((x1-y1)^2 + (x2-y2)^2)/4
= sqrt(x1^2 + y1^2 - 2*x1y1 + x2^2 + y2^2 - 2*x2y2)/4
= sqrt(x1^2 + y1^2 - 0 + x2^2 + y2^2 - 0)/4 (as (x,y) = 0)
= LHS

Therefore the equation holds