# How can I get then angle between a line connecting two points in the plot and a vector (with only direction specified)?

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Sooraj Narayan on 10 Jul 2019
Commented: Bjorn Gustavsson on 10 Jul 2019
As you can see above, I have x and y co-ordinates for n number of Wind Turbines in a wind farm. Suppose there is a vector (wind speed direction) with direction in degrees, how do i find the angle between the line connecting all the other points to any point i,j in the x-y plane and the direction vector (whose onlyu data known is the direction?
xy=[0 0
0 800
0 1600
0 2400
500 0
500 800
500 1600
500 2400
1000 0
1000 800
1000 1600
1000 2400
1500 0
1500 800
1500 1600
1500 2400
2000 0
2000 800
2000 1600
2000 2400];
The above matrix shows the x-y coordiantes of 20 points in the x-y plane.
I need the angle between the line connecting point 1 to all other 19 points and a vector with a direction (say 45 degrees).

Ruben Costa on 10 Jul 2019
Hi, my aproach is here!
xy=[0 0
0 800
0 1600
0 2400
500 0
500 800
500 1600
500 2400
1000 0
1000 800
1000 1600
1000 2400
1500 0
1500 800
1500 1600
1500 2400
2000 0
2000 800
2000 1600
2000 2400]; %This is your matrix with the points
angle=[]; %this is where the angles will be stored
for i=2:size(xy,1)
nline=i-1;
angle(nline) = (atan((xy(i,2)-xy(1,2))/(xy(i,1)-xy(1,1))) - atan((2-0)/(2-0))) * 180/pi;
end
- (xy(i,2)-xy(1,2))/(xy(i,1)-xy(1,1)) ---> this is the value of the tangent of the line created with the point (0,0) and other point of the matrix xy
- (2-0)/(2-0)) ---> this is the value of the tangent of the line created with the point (0,0) and point (2,2) because i knew that this line would have 45º
Then I calculated the arctan( the angles) of both and subtracted to know the angle between them. Then I only made the conversion from radians to degrees!
Hope it helps!

### More Answers (1)

Bjorn Gustavsson on 10 Jul 2019
Best way to go about this is to convert the "wind"-direction to a wind-vector, then use:
w_angle = atan2(norm(cross(wv,b)), dot(wv,b));
Where b is the vector between your points.
HTH