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597

answers:

3

How would you find the signed angle theta from vector a to b?

And yes, I know that theta = arccos((a.b)/(|a||b|)).

However, this does not contain a sign (i.e. it doesn't distinguish between a clockwise or counterclockwise rotation).

I need something that can tell me the minimum angle to rotate from a to b. A positive sign indicates a rotation from +x-axis towards +y-axis. Conversely, a negative sign indicates a rotation from +x-axis towards -y-axis.

assert angle((1,0),(0,1)) == pi/2.
assert angle((0,1),(1,0)) == -pi/2.
A: 

Perhaps atan2(a.y,a.x) - atan2(b.y,b.x) is what you are looking for?

Moron
This was close, but the points need to be reversed.
Chris S
+6  A: 

If you have an atan2() function in your math library of choice:

signed_angle = atan2(b.y,b.x) - atan2(a.y,a.x)
Sparr
Perfect, thank you.
Chris S
What about a = (-1,1) and b = (-1,-1), where the answer should be pi/2? You should check if the absolute value is bigger than pi, and then add or subtract 2*pi if it is.
Derek Ledbetter
@DerekGood catch. I actually discovered this myself while implementing the solution.
Chris S
+2  A: 

What you want to use is often called the “perp dot product”, that is, find the vector perpendicular to one of the vectors, and then find the dot product with the other vector.

if(a.x*b.y - a.y*b.x < 0)
    angle = -angle;

You can also do this:

angle = atan2( a.x*b.y - a.y*b.x, a.x*b.x + a.y*b.y );
Derek Ledbetter