Where is the positive half-space of this determinant?

Question

With reference to the figure below, the determinant $[a,b,c,d]$ should be positive according to a right-hand rule for planes, but when I calculate the determinant for various points, I get the wrong sign. The way I've interpreted this right-hand rule, is to point my right-hand fingers from a to b, then bend the fingers towards c, and my thumb points to the positive half-space. What is wrong with this?

Given four three-dimensional points $a=(a_x,a_y,a_z),b=(b_x,b_y,b_z),c=(c_x,c_y,c_z),d=(d_x,d_y,d_z)$, the determinant is defined as:

\begin{equation}\begin{bmatrix}a,b,c,d\end{bmatrix} := \begin{vmatrix}a_x&a_y&a_z&1\\b_x&b_y&b_z&1\\c_x&c_y&c_z&1\\d_x&d_y&d_z&1\end{vmatrix} \end{equation}

\begin{equation} = \begin{vmatrix}a_x-d_x&a_y-d_y&a_z-d_z\\b_x-d_x&b_y-d_y&b_z-d_z\\c_x-d_x&c_y-d_y&c_z-d_z\end{vmatrix} \end{equation}

Answer 1

The vectors in the drawing are $b-a,$ $c-a$ and $d-a.$ The corresponds to the subtraction of the first row from the other ones in the $4\times 4$ matrix: $$ \begin{vmatrix} a_x & a_y & a_z & 1 \\ b_x-a_x & b_y-a_y & b_z-a_z & 0 \\ c_x-a_x & c_y-a_y & c_z-a_z & 0 \\ d_x-a_x & d_y-a_y & d_z-a_z & 0 \\ \end{vmatrix} $$ The $1$ in the upper right corner is on a "minus position" of the plus/minus-checkerboard we use for determinant calculation. Therefore this is $$ \begin{vmatrix} a_x & a_y & a_z & 1 \\ b_x-a_x & b_y-a_y & b_z-a_z & 0 \\ c_x-a_x & c_y-a_y & c_z-a_z & 0 \\ d_x-a_x & d_y-a_y & d_z-a_z & 0 \\ \end{vmatrix} = -\begin{vmatrix} b_x-a_x & b_y-a_y & b_z-a_z \\ c_x-a_x & c_y-a_y & c_z-a_z \\ d_x-a_x & d_y-a_y & d_z-a_z \end{vmatrix} $$

Answer 2

When considering that the determinant should equal $\vec{ad}\cdot (\vec{ab} \times \vec{ac})$, it becomes clear which half space is the positive one, using the standard vector cross-product right-hand rule.

Solving both the determinant and the triple product show that the right-hand rule or the determinant was incorrectly defined and yields the wrong sign.