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

1. Loser66

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

2. anonymous

I honestly don't follow that way.

3. Loser66

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 .

4. Loser66

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.

5. anonymous

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?

6. Loser66

@oldrin.bataku

7. anonymous

$$\langle v,u\rangle=\|u\|\|v\|\cos\theta=2\cos\theta$$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

8. anonymous

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

9. anonymous

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