Angle between 2 lines

How can i calculate the angle between such two lines.is the dotvector method useful in such case?

line([50 50],[1 10000],'Color','G');
line([newBoxPolygon(1, 1) newBoxPolygon(4, 1)],[newBoxPolygon(1, 2) newBoxPolygon(4, 2)],'Color','G');

i tried this method

u=[newBoxPolygon(1, 1) newBoxPolygon(4, 1),newBoxPolygon(1, 2) newBoxPolygon(4, 2)]
v=[[50 50],[1 10000]]
CosTheta = dot(u,v)/(norm(u)*norm(v));
ThetaInDegrees = acos(CosTheta)*180/pi

but i'm getting an error can you please help

[Jos (10584)]

What is the exact error you're getting?

[Elie]

There is no logical or compiler error , the error is in the angel it self, because the angle between the 2 lines is less than that splitted by the function above

Answer 1

[Jos (10584)]

You want to enter the direction vectors U and V into the formula

dU = [diffUx diffUy]
dV = [diffVx diffVy]
cosTheta = dot(dU,dV) ./ (norm(dU) * norm(dV))

[Elie]

what do you mean by diffUx and diffUy .... ?

[Jos (10584)]

I should have made this more clear.

Your original U is a matrix specifying two points (X1,Y1) and (X2,Y2). These points specify a line for which you need to know the direction vector (dU). dU can be calculated by taking the difference between the X-coordinates of the two points (diffUx) and the difference between the Y-coordinates (diffUy). So,

dU = [(X2-X1) (Y2-Y1)]

[Elie]

Hey i tried your advice but im getting the following error "A and B must be same size" This is my code . what is the error

%Line1
x1=(newBoxPolygon(1, 1));y1=(newBoxPolygon(1, 2));
x2=[newBoxPolygon(1, 1) newBoxPolygon(4, 1)];y2=[newBoxPolygon(1, 2) newBoxPolygon(4, 2)];

%The theoritcal line
x1_1=1; x1_2=1000;
y1_1=900; y1_2=900;
dU = [(x2-x1) (y2-y1)];
dV = [(x1_2-x1_1) (y1_2-y1_1)];
cosTheta = dot(dU,dV) ./ (norm(dU) * norm(dV));
disp(cosTheta);

[Jos (10584)]

You are setting up the vectors the wrong way: the direction vectors dU and dV should be 2 element arrays. dU is, but dV is not.

If you remove the semi-colons you see what I mean.