How can I find the angle between two vectors, including directional information?

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Sebastian Echeverri 2015-2-24

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回答： shantanu kumar 2022-12-19

Hello, I am a graduate student, and I am working on a script that tracks the position of animals during a courtship. I have position data in the form of XY coordinates from two points on each animal's body taken from top down filming. I use these two points to create a vector that defines the animal's orientation. My script needs to calculate the angle between these two vectors, but also include directional information - IE, go from -180 through 0 to 180 degrees, depending on where the vectors are placed (see image).

This is the code that I currently have. It gives me the desired angle (I believe), but is NOT directional. 60 degrees to either side spits out as 60 degrees no matter which it is.

angle_maleToFemale_radians = acos(dot(maleFemaleVector,femaleVector)/(norm(maleFemaleVector)*norm(femaleVector))); angle_maleToFemale_degrees(index) = radtodeg(angle_maleToFemale_radians); angle_maleToFemale_degrees(index) = 180 - angle_maleToFemale_degrees(index);

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Roger Stafford 2015-2-24

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编辑：Roger Stafford 2015-2-24

在 MATLAB Online 中打开

If v1 = [x1,y1] and v2 = [x2,y2] are the components of two vectors, then

a = atan2d(x1*y2-y1*x2,x1*x2+y1*y2);

gives the angle in degrees between the vectors as measured in a counterclockwise direction from v1 to v2. If that angle would exceed 180 degrees, then the angle is measured in the clockwise direction but given a negative value. In other words, the output of 'atan2d' always ranges from -180 to +180 degrees.

One further observation: Besides the greater range of 'atan2d' as compared with 'acosd', the former does not suffer the inaccuracies that occur with 'acosd' for angles near zero and 180 degrees.

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Roger Stafford 2015-2-26

在 MATLAB Online 中打开

There are two ways in which the formulas you have used are not in accordance with accepted practices in vector analysis. First, if you consider that a vector points from point A to point B, then the vector is expressed as PB-PA. You have the opposite. As it turns out, that alone would not affect the answer, however. Secondly, the angle between vectors is considered as measured from the first vector counterclockwise to the second vector with the convention that if the angle goes beyond 180, then 360 is to be subtracted from it, making it negative. To be compatible with the diagram you showed, that would make the vector from P1 to P3 the first vector and that from P2 to P1 the second vector. Unfortunately the formula I gave you is for the opposite order. That will affect your results.

Here is code that corresponds to the diagram you have shown. I use va to denote the first vector and vb the second one.

   va = p3-p1; % <-- Consider this the first vector
   vb = p1-p2; % <-- and this the second vector
   a = atan2d(va(1)*vb(2)-va(2)*vb(1),va(1)*vb(1)+va(2)*vb(2));

Note that the expression "va(1)*vb(2)-va(2)*vb(1)" is the z-component of the cross product of vectors va (first) by vb (second) where the z-axis is pointing upward from your diagram. This corresponds to writing the expression

atan2d( cross(va,vb) , dot(va,vb) )

where again we understand that this refers to the z-component of the cross product and that the two vectors lie in the x-y plane. This should help you generalize this method for any 2D diagram you might create.

(End of lecture in vector analysis)