Quote:

Originally Posted by

**Showcase_22** I did mistype the next question just by mentioning that a z is in the question. The rest of it is right though. I did it like this:

$\displaystyle ||u||^2+2||u||||3v||+||3v||^2+||3u||^2-2||3u||||v||+||v||^2$

=1+6+9+9-6+1=20

As Moo wrote it, it is not always true that $\displaystyle \|u+v\|^2=\|u\|^2+2\|u\|\|v\|+\|v\|^2$. However, it is true that $\displaystyle \|u+v\|^2=\|u\|^2+2u\cdot v+\|v\|^2$ because of the definition of the norm ($\displaystyle \|u\|=\sqrt{u\cdot u}$ where $\displaystyle \cdot$ is the dot product) :

$\displaystyle \begin{aligned}\|u+v\|^2&=(u+v)\cdot(u+v)\\

&=u\cdot u+u\cdot v+v\cdot u+v\cdot v\\

&=\|u\|^2+2u\cdot v+\|v\|^2\end{aligned}$.

Using this identity we get

$\displaystyle

\|u+3v\|^2=\underbrace{\|u\|^2+\|3v\|^2}_{10}+6u\c dot v

$ and $\displaystyle \|3u-v\|^2=\underbrace{\|3u\|^2+\|v\|^2}_{10}-6u\cdot v$

thus $\displaystyle \|u+3v\|^2+\|3u-v\|^2=20$.