Prove (u+v)dot(u-v)=0 iff |u|=|v|

[f25274]

Homework Statement

Show that (u+v)dot(u-v)=0 iff |u| = |v|

Homework Equations

if u= x₁, y₁
and if v= x₂, y₂
then u dot v= x₁x₂ + y₁y₂

The Attempt at a Solution

((x1+x2),(y1+y2)) dot ((x1-x2),(y1-y2))=
(x1^2-x2^2)+(y1^2-y2^2)=
if |u|=|v| then sqr(x1^2+y1^2)=sqr(x2^2+y2^2)
x1^2+y1^2=x2^2+y2^2
Now back to problem
x1^2+y1^2-x2^2-y2^2=0
let x1^2+y1^2=a
since x1^2+y1^2=x2^2+y2^2, x2^2+y2^2=a
a-a=0

i've shown the if part but how do I show the iff part?

 

[Mark44]

Homework Statement

Show that (u+v)dot(u-v)=0 iff |u| = |v|

Homework Equations

if u= x₁, y₁
and if v= x₂, y₂
then u dot v= x₁x₂ + y₁y₂

The Attempt at a Solution

((x1+x2),(y1+y2)) dot ((x1-x2),(y1-y2))=
(x1^2-x2^2)+(y1^2-y2^2)=
if |u|=|v| then sqr(x1^2+y1^2)=sqr(x2^2+y2^2)
x1^2+y1^2=x2^2+y2^2
Now back to problem
x1^2+y1^2-x2^2-y2^2=0
let x1^2+y1^2=a
since x1^2+y1^2=x2^2+y2^2, x2^2+y2^2=a
a-a=0

i've shown the if part but how do I show the iff part?

Assume that (u + v)$\cdot$(u - v) = 0.

Use the fact that (a + b)$\cdot$(c + d) = a $\cdot$ c + a $\cdot$ d + b $\cdot$ c + b $\cdot$ d.

 

[f25274]

Assume that (u + v)$\cdot$(u - v) = 0.

Use the fact that (a + b)$\cdot$(c + d) = a $\cdot$ c + a $\cdot$ d + b $\cdot$ c + b $\cdot$ d.

ohhhhhhhh
u dot u= u^2 right?

(u+v) dot (u-v) = u² + u dot -v + u dot v -v²
-x1x2-y1y2+x1x2+y1y2=0
(u+v) dot (u-v) = u^2-v^2
okay now what do I do...

 

[Mark44]

ohhhhhhhh
u dot u= u^2 right?

No. u $\cdot$ u = |u|²

(u+v) dot (u-v) = u² + u dot -v + u dot v -v²
-x1x2-y1y2+x1x2+y1y2=0
(u+v) dot (u-v) = u^2-v^2
okay now what do I do...

 

[f25274]

oh :I
ok I got it now thanks!