Show that there are no vectors "u" and "v" in R^3 such that ||u||=1, ||v||=2, and =3.

Question

Show that there are no vectors "u" and "v" in R^3 such that ||u||=1, ||v||=2, and <v,u>=3.

loser66 · Answer

Suppose yes, then
$cos (u, v) =\dfrac{<u, v>}{||u||||v||}= \dfrac{3}{2}$ 
does it make sense? You give out the conclusion.

chrisplusian · Answer

I honestly don't follow that way.

loser66 · Answer

why not? $-1\leq cos (\theta) \leq 1$ 
That is the restriction of  a cos function, and the result is out of the range of cos. Hence the "supposing" does not hold .

loser66 · Answer

I use contradiction method and it is the easiest way to use all of the given information. But if you don't get it. I am sorry.

chrisplusian · Answer

Thank you for the attempt, and I don't doubt what you have done. I just think there has to be a different way to show it. Can you think of a different way to do it?

loser66 · Answer

@oldrin.bataku

anonymous · Answer

$$\langle v,u\rangle=\|u\|\|v\|\cos\theta=2\cos\theta$$however $\cos\theta$ has a maximum value of $1$ so $2\cos\theta\le 2$, yet $3\not\le 2$ so there can be no such vectors

anonymous · Answer

geometrically, we have that $\langle v,u\rangle\le\|v\|\|w\|$, and $3\not\le2$

chrisplusian · Answer

Is there a way to expand the dot product to show this?