Trying to prove:

Show that there are no vectors u and v such that $\Vert u\Vert =1$, $\Vert v\Vert =2$, and $u\cdot v=3$

Don't know where to go from here:

$\Vert u\Vert =2\u27f9\Vert u{\Vert}^{2}=4\u27f9u\cdot u=1$

$\Vert v\Vert =1\u27f9\Vert v{\Vert}^{2}=1\u27f9v\cdot v=4$

Not sure if this is the right direction to take, but we have:

$\begin{array}{r}u\cdot v=v\cdot v-u\cdot u=3\\ =(v+u)\cdot (v-u)\end{array}$

Show that there are no vectors u and v such that $\Vert u\Vert =1$, $\Vert v\Vert =2$, and $u\cdot v=3$

Don't know where to go from here:

$\Vert u\Vert =2\u27f9\Vert u{\Vert}^{2}=4\u27f9u\cdot u=1$

$\Vert v\Vert =1\u27f9\Vert v{\Vert}^{2}=1\u27f9v\cdot v=4$

Not sure if this is the right direction to take, but we have:

$\begin{array}{r}u\cdot v=v\cdot v-u\cdot u=3\\ =(v+u)\cdot (v-u)\end{array}$