Check if a point lies beneath vectors convex to it?

Question

I am trying to check if a point lies beneath vectors convex to it by trying to deduce which of the four cases below fits any given scenario:

I am hoping to isolate B, in that there's some general way to deduce whether a given scenario is case B or not (without the need to compute angles between vectors) i.e. since only $\vec{a}$ and $\vec{b}$ in case B are convex to the point $p$

Answer 1

Just find the coordinates of $p - h$ with respect to the frame whose axes are vectors $\vec{a}, \vec{b}$ and whose origin is $h$

The associated linear system can be written as

$ [ \vec{a} , \vec{b} ] [x_1, y_1]^T = p - h $

Then, it follows that

$ [x_1, y_1]^T = [ \vec{a}, \vec{b} ]^{-1} (p - h) $

Since you're interested in Case $B$, then you want $x_1 \gt 0 , y_1 \gt 0$.

For example, if point $h = [2, 3]^T , \vec{a} = [1, -3]^T , \vec{b} = [2, 6]^T $, and $p$ is given as $ [ 5, 7 ]^T$, then

$ \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \begin{bmatrix} 1 && 2 \\ -3 && 6 \end{bmatrix}^{-1} \begin{bmatrix} 3 \\ 4 \end{bmatrix} $

And this solves to

$ \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \dfrac{1}{12} \begin{bmatrix} 6 && -2 \\ 3 && 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} =\dfrac{1}{12} \begin{bmatrix} 10 \\ 13 \end{bmatrix} $

Hence, $x_1 \gt 0 , y_1 \gt 0 $, so this is Case $B$.