`m=2T=[1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];V1=0.956V2=2.4A=[ 1 0 0]; B=[1 1 0]; D=B-A; d1=m*A; for X=1:1:m+1 c(x,:)=d1+D*(x-1) Vndq=T*c(x,:)'; Vnq=Vndq(1,1) Vnd=Vndq(2,1) d(x)=abs(V2-Vnq)+abs(V1-Vnd) end`

in the code given minimum distance is

` d =[1.3560 0.6489 0.8582]`

and

` C = 2 0 0 2 1 0 2 2 0`

c row matrix 2 1 0 of the minimum distance 0.6489 should be obtained

## Best Answer