How to calculate angle between two vectors?

The attached plot is as you can see an array of points in 3D. The lines link points from the XY plane with points along the Z-axis. They are suppose to be vectors, however when I plot with quiver3 I get a vector facing normal to the XY surface, instead of what I would like it pointing towards the points toward the Z-axis, if that makes sense.

Below you can see a part of the code:

total_p=200;

for i = 1 : nout
    for j = 1 : pout
       
        if i == 1
           ko = nr1;
           r = r1;
           
       elseif i == 2
           ko = nr2;
           r = r2;
           
       elseif i == 3
           ko = nr3;
           r = r3;
           
       elseif i == 4
           ko = nr4;
           r = r4;
        end
      
       teta = 2*pi()/ko;
       
       for k = 1 : ko 
        
           F{j,i}(k,1) = r*cos(k*teta);
           F{j,i}(k,2) = r*sin(k*teta);
           F{j,i}(k,3) = 0 - (j-1)*(s+a)*1e3; 
           for l = 1 : total_p
              
             rout{j,i}(l,1)=sqrt(((F{j,i}(k,1)-calc_area(l,1))^2)+((F{j,i}(k,2)-calc_area(l,2))^2)+((F{j,i}(k,3)-calc_area(l,3))^2));
             %teta2{j,i}(l,1)= rad2deg(atan(rout{j,i}(l,1)/0.1));
                
           end
           figure(1)
           scatter3(F{j,i}(k,1),F{j,i}(k,2),F{j,i}(k,3),'k.')
           hold on
           daspect([1,1,1])
       end
       clear r ko teta1
    end
end

In the end I wish to calculate the magnitude of vectors between points of F and and calc_area.

where calc_area has points x,y=0 and z=5 to 100; I attempted to calculate the magnitude and saved it in rout cell. Then I have teta2, where I am calculating the angle, however it yields around 90 degrees. For the red line this should be very small... and for the blue line around 30 degrees from visual inspection. What am I doing wrong here?

Thank you for any help.

[William Rose]

@Ors, quiver3() automatically scales the lengths of the vectors it draws it an attempt to prevent vectors from overlapping. Turn off the automatic scalling by adding option 'off':

quiver3(X,Y,Z,U,V,W,'off')

where "The vectors X, Y, and Z represent the location of the base of each arrow, and U, V, and W represent the directional components of each arrow. By default, the quiver3 function shortens the arrows so they do not overlap. Call axis equal to use equal data unit lengths along each axis. This makes the arrows point in the correct direction.". quiver3 help

The angle you appear to seek is θ, the complement of the angle between the vector r and the z-axis. Compute this by calculating the dot product of the vector r with .

Recall that , where ϕ is the angle between r and .

Then (since ).

Finally, .

Answer 1

[William Rose]

@Ors, My comment above was intended to be an answer. Please accept this answer if it is satisfactory. Here is some code.

F=[-20,200,0;100,-100,0;-200,20,0]; %points in X-Y plane
calc_area=[0,0,5;0,0,50;0,0,100];   %points on z-axis
r=calc_area-F;                      %vectors
plot3(calc_area(:,1),calc_area(:,2),calc_area(:,3),'r.') %plot red points on z-axis
hold on; grid on; axis equal;       %plot details
xlabel('X'),ylabel('Y');zlabel('Z') %axis labels
plot3(F(:,1),F(:,2),F(:,3),'k.');   %plot black points in the X-Y plane
xlim([-250,250]); ylim([-250,250]); zlim([0 150])  %set axis limits
X=F(:,1); Y=F(:,2); Z=F(:,3);       %starting points for arrows 
U=r(:,1); V=r(:,2); W=r(:,3);       %arrow lengths in each dimension
quiver3(X,Y,Z,U,V,W,'off')          %draw arrows on the plot

When you run the code, you will be able to rotate the 3D plot, by clicking onthe 3-D rotation icon above the plot, then clicking and dragging within the plot area.

Compute the angles of the vectors above the X-Y plane:

for i=1:3
    phiDeg(i)=acosd(dot(r(i,:),[0,0,1])/norm(r(i,:)));
end
thetaDeg=90-phiDeg

thetaDeg = 1×3

1.4250 19.4712 26.4512

The code for the angles implements the equations I gave in my comment above. I used acosd() to get the angle phi, in degrees.

Good luck.

[Ors]

Thank you so much William! :D

Although both, your comment and answer are more than satisfactory I am curious

why my method did not work. This is not urgent though, however I would really appreciate it.

[William Rose]

@Ors,

In your code fragment, I do not see where you try to calculate the angles between the vectors from F to calc_area and the normal to the X-Y plane. Therefore I cannot comment on why your method did not work.

In my script, the angles between the vectors from F to calc_area is called "theta()". Your "teta" is not analagous to my "theta". Your teta is an angle in the x-y plane, which is used as a multiplier to compute the x-y coordinates of the points F, as seen belwow in this fragment from the code you posted:

teta = 2*pi()/ko;
for k = 1 : ko 
    F{j,i}(k,1) = r*cos(k*teta);
    F{j,i}(k,2) = r*sin(k*teta);
    F{j,i}(k,3) = 0 - (j-1)*(s+a)*1e3; 
    %...
end

The code above creates a set of ko points F{}() which are spaced at angle in the X-Y plane.

[Ors]

yes, sorry... its commented out and is teta2

should have mentioned that :)

[William Rose]

@Ors,

YOu have

teta2{j,i}(l,1)= rad2deg(atan(rout{j,i}(l,1)/0.1));

It apears that rout is the length of a 3D vector from a point F to a point calc_area.

You could simplify the equation to

teta2=rad2deg(atan(num/den));

This will work IF num=length of the projection of the vector onto the x-y plane, and if den=length of the projection of the vector onto the z axis. Neither of those conditions appear to be true. That is why this method does not give the correct answer.

[Ors]

I see, thank you William :D