Can three vectors have dot product less than $0$? 
Can three vectors in the $xy$ plane have $uv<0$ and $vw<0$ and $uw<0$?

If we take $u=(1,0)$ and $v=(-1,2)$ and $w=(-1,-2)$
$$uv=1\times(-1)=-1$$
$$uw=1\times(-1)+0\times(-2)=-1$$
$$vw=-1\times(-1)+2\times(-2)=-3$$
is there anyway to show that without examples, just working with $u=(u_1,u_2)$, $v=(v_1,v_2)$ and $w=(w_1,w_2)$, I mean analytically?
 A: Sure. Given two vectors $u,v \in \mathbb{R}^2$, we have
$$ u \cdot v = ||u||||v|| \cos (\theta)$$
where $\theta$ is the angle between them. If we take $u,v,w$ to be three unit vectors, each at angle $120$ degrees / $\frac{2\pi}{3}$ radians from each other, then the dot product will always be $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$.  
A: The dot product between two vectors is negative when the angle $\theta$ between them is greater than a right angle, since 
$$
u \cdot v = |u||v|\cos(\theta).
$$
So when three vectors point more or less to the vertices of an equilateral triangle all three dot products will be negative.
A: Let $v_1 = (x_1,y_1) $ , $v_2 = (x_2,y_2)$ and $v_3 =(x_3,y_3)$.
Then $v_1 \odot v_2 = (x_1,y_1)(x_2,y_2) = x_1x_2 + y_1y_2$
$v_1 \odot v_3 = (x_1,y_1)(x_3,y_3) = x_1x_3 + y_1y_3$
$v_2 \odot v_3 = (x_2,y_2)(x_3,y_3) = x_2x_3 + y_2y_3 $
So you want :
$x_1x_2 + y_1y_2 < 0$
$x_1x_3 + y_1y_3 < 0$
$x_2x_3 + y_2y_3 < 0$
Which is a $3 \times 3$ system of inequalities that has solutions in for $x_1,x_2,x_3,y_1,y_2,y_3 \in \mathbb R$.
