Linear Algebra: $\vec{u} \cdot \vec{w}< 0$ Say the angle between $\vec{u}$ and $\vec{v}$ is obtuse and the angle between $\vec{v}$ and $\vec{w}$ is acute. Then $\vec{u} \cdot \vec{w}< 0$
And the answer is:
False. For example,


*

*$\vec{u}=[1, 0],$

*$\vec{v}=[−1, 2],$

*$\vec{w}= [1, 1].$


But that answer doesn't really teach me much.. I obviously understand that the relationship does not hold when those values are plugged into the inequality. But how did this answer come about?
 A: I did it with my fingers. Index finger is $\vec{u}$, middle finger is $\vec{v}$. Make the angle obtuse. Now $\vec{w}$ has to come away from $\vec{v}$ by some acute angle. So it can't get close enough to $\vec{u}$ to have a positive dot product with it.
Positive dot product means pointing basically with each other. Negative dot product means pointing basically against each other.
A: You could start from $U=(1,0)$. Then, since you want $V$ to form an obtuse angle, you set $V=(\cos t,\sin t)$ with $t$ starting from $\pi/2$. Also you want the angle between $V$ and $W$ to be acute so you set $W=(\cos(t+s),\sin(t+s))$ with $s\in[-\pi/2,\pi/2]$. 
Now search the maximum of $f(t,s)=V(t)\cdot W(t+s)$
This is not exactly what "I" prefer. I rather try first some example to understand what's going on, then an example like the given one easily shows up.
A: First, the idea of this exercise is probably to make you remember the identity
$$ u \cdot v = \|u\|\|v\| \cos \alpha$$
where $\alpha$ is the angle between $u$ and $v$. If this is $<0$ it must be that $\cos \alpha < 0$. Which can only happen if $\alpha \in (\pi / 2, 3\pi /2)$.
Obtuse angle means between $\pi / 2$ and $\pi$ and acute means between $0$ and $\pi /2$.
The final step to reaching the solution is to put it all together: if $u,v$ are obtuse and $v,w$ is acute then is it possible to have an angle between $u,w$ that is in $(\pi / 2, 3\pi /2)$?
