Prove vectors with specific length and dot product cannot exist

Question

Trying to prove:

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

Don't know where to go from here:

$$\|u\| = 2 \Longrightarrow \|u\|^2 = 4 \Longrightarrow u\cdot u = 1 $$

$$\|v\| = 1 \Longrightarrow \|v\|^2 = 1 \Longrightarrow v\cdot v = 4 $$

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

\begin{align*} u\cdot v = v\cdot v - u\cdot u = 3 \\ = (v + u)\cdot (v - u) \end{align*}

Answer 1

$ u \cdot v = \|u\| \|v\| \cos \theta $

So

$ 3 = (1)(2) \cos \theta $

which implies

$\cos \theta = \dfrac{3}{2} $

which is not possible.

Answer 2

Another way to realize the answer: \begin{align*} \|u + v\|^{2} & = \langle u + v, u + v\rangle\\\\ & = \langle u,u\rangle + 2\langle u,v\rangle + \langle v,v\rangle\\\\ & = \|u\|^{2} + 2\langle u,v\rangle + \|v\|^{2}\\\\ & = 1^{2} + 2\times 3 + 2^{2}\\\\ & = 1 + 6 + 4\\\\ & > (1 + 2)^{2}\\\\ & = (\|u\| + \|v\|)^{2} \end{align*}

which violates the triangle inequality.