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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?

MathWorks Support Team
on 27 May 2020 at 4:00

Edited: MathWorks Support Team
on 27 May 2020 at 13:42

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));

Pierre-Pascal
on 11 Jan 2016

So why doesn't matlab give us a function for that instead of having us look endlessly on forums?

William Chamberlain
on 27 Jul 2016

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James Tursa
on 9 Jul 2015

Edited: James Tursa
on 5 Jan 2019

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:

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

Johannes Kalliauer
on 14 Jan 2020

Thanks, sometimes even imaginäry Angles occur if using acos-function.

James Tursa
on 3 Feb 2020

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Gabor Bekes
on 15 Sep 2016

Edited: Gabor Bekes
on 15 Sep 2016

This does the same thing, also capable of determining the angle of higher (than one) dimensional subspaces.

subspace(vector1,vector2)

Aras
on 3 May 2018

This method needs to be used carefully because it provides an angle between 0 and π/2 radians, instead of between 0 and π.

E.g., the angle between vectors [1, 0] and [-1, 0] are given as 0, while the result is expected to be π, considering their opposite directions.

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Daniel Vasilaky
on 9 Jul 2015

Edited: Walter Roberson
on 15 Sep 2015

acosd(CosTheta)

will give you the same answer.

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Boris Povazay
on 17 Jun 2018

Just a note on how to vectorize the whole thing: (semicolons purposely omitted to see the intermediate results)

u = [1 2 0];

v = [1 0 0];

C=cross(u,v)

NC=norm(C)

D=dot(u,v)

ThetaInDegrees = atan2d(NC,D)

Rep=5

uf = repmat(u,5,1)

vf = repmat(v,5,1)

vC=cross(uf,vf,2) %vectorized

vNC=vecnorm(vC,2,2) % since only z-rotation is allowed anyway, this is equivalent to: vNC=vC(:,3)

vD=dot(uf,vf,2)

vThetaInDegrees = mean(atan2d(vNC,vD))

or in short (the hard to read variant)

VThetaInDegrees =atan2d( vecnorm(cross(Vu,Vv,2),2,2) , dot(Vu,Vv,2) )

Boris Povazay
on 17 Jun 2018

One more thing to mention: this calculation takes the norm and therefore is not the solution to the question: angle between vectors! - It is rather the angle between unoriented vectors. The solution to the question rather should result in [-180:+180] to distinguish the orientation of the angle. - It is mentioned here.

-> So how can this be rewritten without loosing the orientation due to the norm?

Jan
on 17 Jun 2018

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theodore panagos
on 29 Oct 2018

Coordinates of two vectors xb,yb and xa,ya .

angle(vector.b,vector.a)=pi/2*((1+sgn(xa))*(1-sgn(ya^2))-(1+sgn(xb))*(1-sgn(yb^2)))

+pi/4*((2+sgn(xa))*sgn(ya)-(2+sgn(xb))*sgn(yb))

+sgn(xa*ya)*atan((abs(xa)-abs(ya))/(abs(xa)+abs(ya)))

-sgn(xb*yb)*atan((abs(xb)-abs(yb))/(abs(xb)+abs(yb)))

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Adana Mason
on 28 Nov 2019

(a) Find parametric equations for the line of intersection of the planes and (b) find the angle

between the planes. 3𝑥 − 2𝑦 + 𝑧 = 1, 2𝑥 + 𝑦 − 3𝑧 = 3.

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