# dot product negative angle

I have two two-dimensional unit vectors a and b. I'm trying to get their angle related to their order. arc cosine of the dot product returns the absolute value of the angle. if b is before a I want to get negative angle.
e.g. if a is on 1 o'clock and b is on 2 o'clock I want to get answer of 30 degrees but if a is on 2 o'clock and b is on 1 o'clock I want to get answer of -30 degrees. answers should be in radians, I used degrees for simpler explanation.
How can I do this? a method for finding who is before who will be good enough. this is for an application, so an answer of "just look at them" won't work, but a mathematical relation will be good (e.g. some equation that is negative only if b is before a).

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You might be interested in Determinants (en.wikipedia.org/wiki/Determinant): the determinant of two 2d vectors is the product of the norms and the sine between the two vectors; the sine can give you the sign. – nik Jun 13 '11 at 12:23
If this is for a computer program, then your programming language might have a function called atan2(x,y). Check for its documentation to make sure, but IIRC that returns the angle (with sign!) between the vectors $xi+yj$ and the positive $x$-axis. If $a=(x_1,y_1)$ and $b=(x_2,y_2)$ then you get your result by carefully comparing the outputs of atan2($x_1,y_1$) and atan2($x_2,y_2$). Emphasis on CAREFUL. The angles wrap around the circle, so you need to make up your mind, on how you treat the case, when one angle is -175 degrees and the other 175. Add multiples of $2\pi$ when appropriate! – Jyrki Lahtonen Jun 13 '11 at 12:26

Treat them as 3D vectors with $z=0$ and find the cross product.

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 Isn't there an easier way? – Dani Jun 13 '11 at 12:22 @Dani, it boils down to two multiplications and a comparison. You're not going to get much easier than that. – Peter Taylor Jun 13 '11 at 12:25 yes, I've been able it to that. didn't know cross product is such an easy operation. – Dani Jun 13 '11 at 12:28