Using the dot product in a proof

kevinlightman · #1 $Old$ October 23rd, 2009, 05:56 PM

So I have to prove that for any 2 vectors x & y:

|x - y| >= ||x| - |y||

and

|x - y|^2 + |x + y|^2 = 2(|x|^2 + |y|^2)

I am familiar with an algebraic proof for the first part however they need to be proved specifically using the dot product rule. Also in geometry what implications might the second equation have with regards to parallelograms?

CaptainBlack · #2 $Old$ October 24th, 2009, 08:49 AM

Quote:

Originally Posted by kevinlightman $View Post$

So I have to prove that for any 2 vectors x & y:

|x - y| >= ||x| - |y||

and

|x - y|^2 + |x + y|^2 = 2(|x|^2 + |y|^2)

I am familiar with an algebraic proof for the first part however they need to be proved specifically using the dot product rule. Also in geometry what implications might the second equation have with regards to parallelograms?

$|x-y|^2=(x-y).(x-y)=|x|^2-x.y-y.x+|y|^2\ \ \ \ \ ...(1)$

ans:

$||x|-|y||=|x|^2-2|x||y|+|y|^2\ \ \ \ \ ...(2)$

but $x.y\le|x||y|$ hence $(1)\ \ge\ (2)$

CB