Calculate angles correctly between two vectors using the dot product.

Hi all,

As shown below in the figure, I have two vectors, and , and I want to calculate the angle with the norm, and, between each of these vectors with respect to vector L using the dot product.

Here is a simple code to calculate these angles:

close all;
clear all;

%define the reference vector
L = [-0.5 -0.5];

%v1
v1 = [-0.6788 0.3214];
%theta1 with the x axis
theta1 = acos((v1(1)*L(1)+v1(2)*L(2))/(sqrt(v1(1)^2+v1(2)^2)*sqrt(L(1)^2+L(2)^2)));
%theta1 with the norm
theta1norm = pi/2 - theta1;

%v2
v2 = [0.3214 -0.6788];
%theta2 with the x axis
theta2 = acos((v2(1)*L(1)+v2(2)*L(2))/(sqrt(v2(1)^2+v2(2)^2)*sqrt(L(1)^2+L(2)^2)));
%theta1 with the norm
theta2norm = pi/2 - theta2;

My question here is that the product produces both angles to be equal, i.e., . However, and as shwon in the figure, it is clear that , i.e., and . How can I caulcate these angles correctly using the dot product?

Thanks.

[Dyuman Joshi]

"However, and as shwon in the figure, it is clear that Theta2n = Theta1N + pi/2"

How exactly is that clear or shown?

"How can I caulcate these angles correctly using the dot product?"

Utilize these functions - dot, norm

Answer 1

[Hassaan]

% Close all figures and clear variables
close all;
clear all;

% Define the reference vector
L = [-0.5 -0.5];

% Define the first vector v1 and calculate the angle with L
v1 = [-0.6788 0.3214];
% Dot product of v1 and L
dot_v1_L = dot(v1, L);
% Norms of v1 and L
norm_v1 = norm(v1);
norm_L = norm(L);
% Angle theta1 with the x axis
theta1 = acos(dot_v1_L / (norm_v1 * norm_L));
% Angle theta1 with the norm
theta1norm = pi/2 - theta1;

% Define the second vector v2 and calculate the angle with L
v2 = [0.3214 -0.6788];
% Dot product of v2 and L
dot_v2_L = dot(v2, L);
% Norms of v2
norm_v2 = norm(v2);
% Angle theta2 with the x axis
theta2 = acos(dot_v2_L / (norm_v2 * norm_L));
% Angle theta2 with the norm
theta2norm = pi/2 - theta2;

% Convert angles to degrees
theta1_degree = radtodeg(theta1);
theta2_degree = radtodeg(theta2);
theta1norm_degree = radtodeg(theta1norm);
theta2norm_degree = radtodeg(theta2norm);

% Display the results
disp(['Theta1: ', num2str(theta1_degree), ' degrees']);

Theta1: 70.3367 degrees

disp(['Theta2: ', num2str(theta2_degree), ' degrees']);

Theta2: 70.3367 degrees

disp(['Theta1 from the norm: ', num2str(theta1norm_degree), ' degrees']);

Theta1 from the norm: 19.6633 degrees

disp(['Theta2 from the norm: ', num2str(theta2norm_degree), ' degrees']);

Theta2 from the norm: 19.6633 degrees

The functions dot and norm are used to calculate the dot product and the magnitude of the vectors, respectively, and acos computes the arccosine of the given value to obtain the angle in radians. The function radtodeg converts the angle from radians to degrees.

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Answer 2

[James Tursa]

Also see this related link for a robust method using atan2 that can recover small angles:

https://www.mathworks.com/matlabcentral/answers/101590-how-can-i-determine-the-angle-between-two-vectors-in-matlab?s_tid=srchtitle

Answer 3

[Torsten]

I think an intuitive way is to compute the angles between the positive x-axis and the respective vector counterclockwise first (the result will be between 0 and 360) and then make the necessary subtractions.

theta1 = cart2pol(L(1)/norm(L),L(2)/norm(L))*180/pi
if theta1 <=0
  theta1 = 360 + theta1;
end
theta2 = cart2pol(v1(1)/norm(v1),v1(2)/norm(v1))*180/pi
if theta2 <=0
  theta2 = 360 + theta2;
end
theta3 = cart2pol(v2(1)/norm(v2),v2(2)/norm(v2))*180/pi
if theta3 <=0
  theta3 = 360 + theta3;
end

Answer 4

[Sinazo]

To evaluate the scalar product and angle between two vectors using MATLAB. Let vector A = 3ûₓ + 4ûᵧ and vector B = 12ûₓ + 5ûᵧ. Use these vectors to perform the scalar product and angle calculations.

[Torsten]

A = [3;4];
B = [12;5];
sp = A.'*B;
nA = norm(A);
nB = norm(B);
ang = acos(sp/(nA*nB))*180/pi

ang = 30.5102

A1 = [1;0];
sp1 = A1.'*B;
ang_lower = acos(sp1/(1*nB))*180/pi

ang_lower = 22.6199

B2 = [0;1];
sp2 = A.'*B2;
ang_upper = acos(sp2/(nA*1))*180/pi

ang_upper = 36.8699

ang_lower+ang+ang_upper

ans = 90.0000