Question 1203665
Show that 



{{{4u * v = abs(abs(u + v))^2 − abs(abs(u − v))^2}}}


Hint. {{{abs(abs(u))^2=u*u}}}, so


{{{(u+v)^2-(u-v)^2}}}

={{{(u+v)(u+v)-(u-v)(u-v)}}}

={{{u^2 + 2uv + v^2-(u^2 - 2uv + v^2)}}}

={{{u^2 + 2uv + v^2-u^2 +2uv - v^2}}}

={{{cross(u^2) + 2uv + cross(v^2)-cross(u^2) +2uv - cross(v^2)}}}

= {{{2uv  +2uv }}}

= {{{4uv }}}




Show that u and v is orthogonal if and only if 
{{{abs(abs(u + v)) = abs(abs(u-v))}}}


If {{{u}}} and {{{v}}} are orthogonal vectors, then:


{{{abs(abs(u+v))^2=abs(abs(u))^2+abs(abs(v))^2}}}

{{{abs(abs(u-v))^2=abs(abs(u))^2+abs(abs(-v))^2=abs(abs(u))^2+abs(abs(v))^2}}}


now {{{abs(abs(u+v))^2=abs(abs(u-v))^2}}}, but the norm is ever positive therefore: 


{{{abs(abs(u+v))=abs(abs(u-v))}}}

.
now, if {{{abs(abs(u+v))=abs(abs(u-v))}}} we have:

{{{abs(abs(u+v))^2=abs(abs(u))^2+2u*v+abs(abs(v))^2}}}

{{{abs(abs(u-v))^2=abs(abs(u))^2-2u*v+abs(abs(v))^2}}}


By the equality

{{{abs(abs(u))^2+2u*v+abs(abs(v))^2=abs(abs(u))^2-2u*v+abs(abs(v))^2 if and only if

{{{2u*v=-2u*v}}} <=> {{{4u*v=0}}} <=>{{{u*v=0}}}

this means {{{u}}} and {{{v}}} are {{{orthogonal}}}