# Unexpected output from Cross[ ]?

Posted 3 months ago
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 I admit that Cross[{0,1,0,0},{0,0,1,0},{0,0,0,1}]= - {1,0,0,0} (1) applied to four 4-dimensional basis vectors e1={0,1,0,0}, e2={0,0,1,0}, e3={0,0,0,1}], and e4={1,0,0,0} can successfully lead us to a calculation of the volume V V = Cross[v1,v2,v3].v4 (*nner product of Cross[v1,v2,v3] with v4*) of a parallelohedron made up of four 4-dimensional vectors vi = {xi,yi,zi,wi} = e1 xi + e2 yi +e3 zi +e4 wi (i=1,2,3,4) represented on the four 4-dimensional orthonormal unit basis vectors e1={0,1,0,0}, e2={0,0,1,0}, e3={0,0,0,1}, and e4={1,0,0,0} On the other hand, however, the most natural extension of a well known cross product Cross[{1,0,0},{0,1,0}]={0,0,1} would be Cross[{0,1,0,0},{0,0,1,0},{0,0,0,1}]= {1,0,0,0}, not -{1,0,0,0}. (2) I will be happy if someone kindly explains me why Cross of 3 unit vector be given by (2). Attachments:
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Posted 3 months ago
 I don't know, but maybe because with this definition the four vectors are positively oriented: In[8]:= v2 = {0, 1, 0, 0}; v3 = {0, 0, 1, 0}; v4 = {0, 0, 0, 1}; cross = Cross[v2, v3, v4]; Det[{v2, v3, v4, cross}] Out[12]= 1 
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Posted 3 months ago
 Thank you for your suggestive note regarding my query on Cross[e2, e3, e4]= -e1. (In this note, previous symbol "v" for unit vectors are changed to "e".)While looking at your note, I came across a geometrical interpretation of why and how Cross[e2, e3, e4] be given by -e1, It goes as follows. (The following example concerns a 3-dimensional case.) Supppose one is interested in the evaluation of the volume V of a parallelopiped defined by 3 unit vectors e1={1,0,0}, e2={0,1,0}, and e3={0,0,1}.  V = Det[(e1 e2 e3 )]= \[LeftBracketingBar] { {{ {{ {{ {1}, {0} }}, {0} }, { {{ {0}, {1} }}, {0} }} }, { {{ {0}, {0} }}, {1} }} } \[RightBracketingBar]=1 Then he likes to change the order of counting these vectors, starting with Overscript[e1, ^] (= e3), Overscripte2, ^, and then Overscript[e3, ^] (=Cross[Overscript[e1, ^],Overscript[e2, ^]] , so that the volume will be Overscript[V, ^] = Det[(Overscript[e1, ^] Overscript[e2, ^] Overscript[e3, ^] )] However, since the parallelopipet itself has not changed, except for the order of counting the 3 ridges, its volume V remains unchanged, i.e. Overscript[V, ^]=V. Further, under the right-handed screw rule, Overscript[e3, ^](=Cross[Overscript[e1, ^],Overscript[e2, ^]]= ordinary cross product Overscript[e1, ^]*Overscript[e2, ^]) =-e1, so that we should define Cross[Overscript[e1, ^],Overscript[e2, ^]] as follows: Cross[Overscript[e1, ^],Overscript[e2, ^]] = -e1. This is in accord with the following manipulation of V: V = Det[(e1 e2 e3 )]= \[LeftBracketingBar]1 0 0 0 1 0 0 0 1 \[RightBracketingBar]=\[LeftBracketingBar]0 0 -1 0 1 0 1 0 0 \[RightBracketingBar]= Det[(Overscript[e1, ^] Overscript[e2, ^] Overscript[e3, ^] )] It seems to me that the same logic hold in the 4-dimensional case, as verified in your note.Thanks again.
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