V is an Euclidean vectorspace $u,v \in V$ $||u||=3 ,||u+v||=4, ||u-v||=6 $ defines explicitly $||v||$
My idea is the following:
i)$||u+v||=(u+v)^2=u^2+2uv+v^2 =4$
ii)$||u-v||=(u-v)^2=u^2-2uv+v^2 =6$
If I add i) and ii) I get that $2u^2+2v^2=10 \implies u^2+v^2=5$ and from the description of the exercise we know that $||u||=3 \implies u^2=9$ which means that $v^2=-4$ which seems to be a contradiction to me. Am I right? Is my calculation correct? Thank you in advance.
The identity you need is $$ \|u+v\|^2 +\|u-v\|^2 = 2\|u\|^2 + 2 \|v\|^2. $$ This is almost what you have shown. You should just put the $\|.\|$ and the scalar products to be perfectly correct. $$ \|u+v\|^2 = \|u\|^2 + 2\langle u,v\rangle +\|v\|^2. $$