When the dot product $(-a+b)\cdot c$ turns to be negative? Let $a, b, c\in(\mathbb{R}^3)^*$ be 3 distinct vectors. Which condition I have to impose on $a, b, c$ so that
$$(-a+b)\cdot c <0$$
holds true?
I know that dot product is distributive over vector addition, so
$$(-a+b)\cdot c <0\iff -a\cdot c + b\cdot c<0.$$
Also, I know that the dot product between 2 vectors is negative if, denoted by $\theta$ the angle between them, it is $\frac{\pi}{2}<\theta < \frac{3\pi}{2}$.
Nonetheless, I can not find a condition involving all the 3 vectors which guarantees the desired inequality.
Could someone please help me with that?
Thank you in advance.
 A: This is a condition. You can see the geometric interpretation if you rewrite your expression first as $$b\cdot c<a\cdot c$$
and then divide both sides by the magnitude of $c$:
$$b\cdot\hat c<a\cdot\hat c$$
Here $c=|c|\hat c$. You project the vectors $a$ and $b$ on the $\hat c$ axis. The expression just means that the projection of $b$ is at a lower value than the projection of $a$. Or equivalent, $b$ is in the half space given by the plane perpendicular to $c$, going through $a$, closer to $-\infty$ along $\hat c$. So any point $B$ that is to the lower-left of the $AE$ line obeys that condition.
EDIT:
In the picture below, let $O$ be the origin. Then $a=OA$, $b=OB$, $c=OC$. Then $a\cdot\hat c=OE$, and $b\cdot\hat c=OF$. The inequation says that $E$ is more to the right than $F$, if we define $C$ to be to the right of $O$

A: Your condition is $\vec{b} \cdot \vec{c} < \vec{a} \cdot \vec{c}$.
Strictly speaking, that is the answer, but in terms of magnitudes and angles (if that is what you are asking), this means:
$$
\begin{align}
bc\cos(\theta_{bc}) &< ac\cos(\theta_{ac}) \\
b\cos(\theta_{bc}) &< a\cos(\theta_{ac})
\end{align}
$$
Other than that, I don't see further restriction.
