You can do this in $\mathbb{R}^2$ using the determinant. If your ordered pair of vectors is $V = \langle a,b \rangle$ and $W = \langle c,d \rangle$ then you have a $+$ angle if $ad-bc>0$ and a $-$ angle if $ad-bc<0$.
However, there is no consistent way to do this in $\mathbb{R}^n$ if $n \ge 3$. The reason for this is that what one is really measuring is distance on the $n-1$-dimensional sphere $S^{n-1}$. Distance, in general, is an unsigned quantity: it is always positive or zero; negative distance does not make sense.
The difference between $n \ge 3$ and $n=2$ is that $S^1$ is one-dimensional and it has a natural sense of "counter-clockwise orientation".
You can do this in $\mathbb{R}^2$ using the determinant. If your ordered pair of vectors is $V = \langle a,b \rangle$ and $W = \langle c,d \rangle$ then you have a $+$ angle if $ad-bc>0$ and a $-$ angle if $ad-bc<0$.
However, there is no consistent way to do this in $\mathbb{R}^n$ if $n \ge 3$. The reason for this is that what one is really measuring is distance on the $n-1$-dimensional sphere $S^{n-1}$. Distance, in general, is an unsigned quantity: it is always positive or zero; negative distance does not make sense.
The difference between $n \ge 3$ and $n=2$ is that $S^1$ is one-dimensional and it has a natural sense of "counter-clockwise orientation".