How can I determine the angle between two vectors in MATLAB?

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MathWorks Support Team
评论： Bruno Luong 2024-10-1
How can I determine the angle between two vectors in MATLAB?
I have two vectors. Is there a MATLAB function that can determine the angle between them?

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MathWorks Support Team
编辑：MathWorks Support Team 2020-5-27
There is no in-built MATLAB function to find the angle between two vectors. As a workaround, you can try the following:
CosTheta = max(min(dot(u,v)/(norm(u)*norm(v)),1),-1);
ThetaInDegrees = real(acosd(CosTheta));
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Akihumi
Akihumi 2020-5-27
Hi, did you miss out a bracket for the min? I got an error and only resolve it with the following code instead.
CosTheta = max(min(dot(u,v)/(norm(u)*norm(v)),1),-1);
ThetaInDegrees = real(acosd(CosTheta));
Bruno Luong
Bruno Luong 2024-9-28
编辑：Bruno Luong 2024-9-28
This is actually incorrect for complex vectors
CosTheta = max(min(dot(u,v)/(norm(u)*norm(v)),1),-1);
ThetaInDegrees = real(acosd(CosTheta));
The correct code is
CosTheta = max(min(real(dot(u,v))/(norm(u)*norm(v)),1),-1);
ThetaInDegrees = acos(CosTheta)

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更多回答（2 个）

James Tursa
James Tursa 2015-7-9
编辑：James Tursa 2019-1-5
This topic has been discussed many times on the Newsgroup forum ... if I looked hard enough I'm sure I could find several Roger Stafford posts from many years ago on this. E.g., here is one of them:
https://www.mathworks.com/matlabcentral/answers/161515-how-do-i-find-the-angle-between-a-vector-and-a-line
The basic acos formula is known to be inaccurate for small angles. A more robust method is to use both the sin and cos of the angle via the cross and dot functions. E.g.,
atan2(norm(cross(u,v)),dot(u,v));
An extreme case to clearly show the difference:
>> a = 1e-10 % start with a very small angle
a =
1e-10
>> u = 4*[1 0 0] % arbitrary non-unit vector in X direction
u =
4 0 0
>> v = 5*[cos(a) sin(a) 0] % vector different from u by small angle
v =
5 5e-10 0
>> acos(dot(u,v)/(norm(u)*norm(v))) % acos formulation does not recover the small angle
ans =
0
>> atan2(norm(cross(u,v)),dot(u,v)) % atan2 formulation does recover the small angle
ans =
1e-10
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Bruno Luong
Bruno Luong 2022-12-3
@Felix Fischer If you want to find angles of multiple vector pairs put in matrix, use vecnorm rather than norm.

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Bruno Luong
Bruno Luong 2024-9-28
编辑：Bruno Luong 2024-9-28
There is a good formula from Kahan, chap 12 of this Mindless paper, for given x and y two vectors of length(m) - in R^m, the angle theta between x and y can be computed as
nx = norm(x);
ny = norm(y);
xx = x*ny;
yy = y*nx;
a = xx-yy;
b = xx+yy;
theta = 2*atan(sqrt(sum(a.^2)/sum(b.^2)))
The advantage of this method is good stability and in case of
nx = norm(x) = ny = norm(y) (= 1, not required)
the code can be reduced to
a = x-y;
b = x+y;
theta = 2*atan(sqrt(sum(a.^2)/sum(b.^2)))
or more compactly
theta = 2*atan(sqrt(sum((x-y).^2)/sum((x+y).^2)))
% or
theta = 2*atan(norm(x-y)/norm(x+y))
The number of arithmetic operations is less than the atan2 formula in James Tursa's answer (only applied in R^3) which is numericall more stable than TMW answer using only dot product.
Note that this implementation does not have issue when b is all-0 vector. But in case both x and y are 0s - so as a and b, Kahan method returns NaN rather than 0 as with atan2. IMO NaN is mathemetically more coherent result.
Beside this degenerated case the result is in interval [0,pi].
NOTE: For complex vectors replace any statements of the form sum(u.^2) by sum(u.*conj(u)); with u being a, b, or x-y, x+y.
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Bruno Luong
Bruno Luong 2024-10-1
Ran in:
Same comparison and observe the robustness
a = 1e-10 % start with a very small angle
a = 1.0000e-10
u = 4*[1 0 0] % arbitrary non-unit vector in X direction
u = 1×3
4 0 0
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<mw-icon class=""></mw-icon>
v = 5*[cos(a) sin(a) 0] % vector different from u by small angle
v = 1×3
5.0000 0.0000 0
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acos(dot(u,v)/(norm(u)*norm(v))) % acos formulation does not recover the small angle
ans = 0
atan2(norm(cross(u,v)),dot(u,v)) % atan2 formulation does recover the small angle
ans = 1.0000e-10
nu = norm(u);
nv = norm(v);
xx = u*nv;
yy = v*nu;
a = xx-yy;
b = xx+yy;
theta = 2*atan(sqrt(sum(a.^2)/sum(b.^2)))
theta = 1.0000e-10

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