MATLAB: Finding varible in equation from vectors.

linearloopMATLABvectorization

I have

x = [1 2 3]
y = [1 2 1]
m = [1 -1 0]

For the equation y = mx +c (linear expression) I want to find c (y intercept) for each. i.e here

           1=1*1 + c (equation 1)
           2=-1*2 + c (equation 2)
           1=0*3 + c (equation 3)

How would I write this as a vectorized code where the values for c are stored in a matrix. Note I want to find c for each equation not for all. I can't use built in functions. I think I have to use ==? It must hold true if there was more equations or there values were changed.

Thanks,

Best Answer

Giuseppe, use

 c = y - m.*x

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MATLAB: Angle Between two vectors.

Given two vectors A and B, the dot product of the two vectors (A dot B) gives the product ABcos(ang), so to get just the angle, you want to take the dot product of two unit vectors;

Assume A = [ax, ay, az], B = [bx, by, bz]

magA = sqrt(ax^2+ay^2+az^2);  % = sqrt(A(1)^2 + A(2)^2 + A(3)^2)
magB = sqrt(bx^2+by^2+bz^2);  % = sqrt(B(1)^2 + B(2)^2 + B(3)^2)
UA = [ax/magA, ay/magA, az/magA];  % A unit vector,  = [A(1)/magA, A(2)/magA, A(3)/magA]
UB = [bx/magB, by/magB, bz/magB];  % B unit vector,  = [B(1)/magB, B(2)/magB, B(3)/magB]
cosAng = dot(UA,UB);  % = UA(1)*UB(1) + UA(2)*UB(2) + UA(3)*UB(3) 
ang = acos(cosang);  % This is the angle between vector A and Vector B, in radians

As a function;

function ang = Vangle(A,B)
    magA = sqrt(A(1)^2 + A(2)^2 + A(3)^2);
    magB = sqrt(B(1)^2 + B(2)^2 + B(3)^2);
    UA = [A(1)/magA, A(2)/magA, A(3)/magA]
    UB = [B(1)/magB, B(2)/magB, B(3)/magB]
    cosAng =  UA(1)*UB(1) + UA(2)*UB(2) + UA(3)*UB(3); 
    ang = acos(cosang); 
    return 
end

Best Answer

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