Re: Big Loss of Precision

[batoota123]

The cross product of two vectors is orthogonal to both.  The dot product 
> of two orthogonal vectors is zero. 
> 
> This is not an octave problem, just an ill-formed situation. 
> 
> octave:1405> printf(" %.20f\n",V); 
>    3.99120302628332668249 
>    -0.91842329903575536942 
>    1.29246846883179156151 
> 
> octave:1406> printf(" %.20f\n",R); 
>    -1518.35794621729837672319 
>    -4017.17627231122196462820 
>    4816.75939627815478161210 
> 
> octave:1407> dot(cross(R,V),R) 
> ans =    1.4901e-08 
> octave:1408> dot(cross(R,V),V) 
> ans =    3.6380e-12 
> 
> cross has 6 multiplications and 3 add/subtract 
> dot has 3 multiplications and 3 add/subtract 
> 
> Octave eps is 2.2204e-16 
> 
> machine eps is 1.11022e-16 (quadmath 1.925930e-34). 
> 
> This case has a very serious loss of precision! 
> 
> Tom Dean
... [show rest of quote]

A similar problem with similar magnitudes.  Very different outcome. 
A = [2 4 8]; norm(A) 
B = A.^4; norm(B) 
C=cross(A,B) 
dot(C,A) 
dot(C,B) 


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