Angle between 2 lines

Question

Elie 2014-2-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/116754-angle-between-2-lines

评论： Jos (10584) 2014-2-24

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

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Jos (10584) 2014-2-19

What is the exact error you're getting?

Elie 2014-2-19

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

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Answer 1

Jos (10584) 2014-2-19

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/116754-angle-between-2-lines#answer_124895

编辑：Jos (10584) 2014-2-19

在 MATLAB Online 中打开

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))

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Elie 2014-2-21

在 MATLAB Online 中打开

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);