I have three 2D unit vectors in the origin, $\textbf{a}$, $\textbf{b}$ and $\textbf{c}$.
How can we determine if the equations
$$\alpha\textbf{a}+\beta\textbf{b}+\gamma\textbf{c}=0$$
have positive solutions for $\alpha$, $\beta$ and $\gamma$?
More intuitively: how to determine if there exist three vectors $-$ given by the three directions $\textbf{a}$, $\textbf{b}$ and $\textbf{c}$ $-$ with a zero resultant?
Best Answer
GReyes is correct.
Because we have three vectors in a two-dimensional space, they cannot be linearly independent. That means there are coefficients $A, B, C$ such that $$A\,\mathbf a + B\,\mathbf b + C\,\mathbf c = 0$$ If no two of $\mathbf a, \mathbf b ,\mathbf c$ are parallel, all three coefficients must be non-zero. We can always mulitply the entire equation by $-1$ if needed, so we can assume that either two or all three coefficients are positive. If one of the coefficients (W.L.O.G. $B$) is negative, bring that term to the other side and divide to get an expression $$\mathbf b = A'\mathbf a + C'\mathbf c$$ where the two coefficients $A' = \frac A{-B}, C' = \frac C{-B}$ are positive.
Suppose a vector $\mathbf v = \alpha\,\mathbf a + \beta\,\mathbf b + \gamma\,\mathbf c$ with positive $\alpha, \beta, \gamma$. But we can substitute for $\mathbf b$ to get $v =(\alpha + \beta A')\mathbf a + (\gamma + \beta C')\mathbf c$, which still has positive coefficients. Thus any vector you can get by including $\mathbf b$, you also get without $\mathbf b$. When you can write one of the vectors as a positive sum of the other two, that vector becomes redundant.
On the other hand, when all three of $A, B, C$ are positive, then every vector in the plane can be written as a positive combination of them. Once again we can write $$\mathbf b = A'\mathbf a + C'\mathbf c$$ except in this case both $A'$ and $C'$ are negative. Since the plane is two-dimensional, any vector $v = \alpha \mathbf a + \gamma \mathbf c$ for some $\alpha, \gamma$. Either coefficient might be negative. Choose $\beta$ to be greater than $\dfrac{\alpha}{A'}, \dfrac{\gamma}{C'},$ and $0$. So $\alpha - \beta A' > 0, \gamma - \beta C' > 0$. Then $$\mathbf v = (\alpha - \beta A')\mathbf a + \beta\mathbf b + (\gamma - \beta C') \mathbf c$$ is an expression $v$ with all three coefficients positive. This includes the case when $v = 0$.
$A, B, C$ will be positive exactly when $0$ is inside the triangle formed by $\mathbf a, \mathbf b, \mathbf c$.