Is you question to show that or are you telling us an interesting fact about cross products?
In case you’re not familiar with it, the cross product is very handy in 3d geometry:
aXb is a vector perpendicular to a and b in direction given by right hand rule (rotate (screw) a into b thru the acute angle θ between them), and magnitude absinθ.
aXb = “det” |i, j, k ; a1, a2, a3 ; b1, b2, b3| , = |row; row; row|, (not obvious, proof in textbooks)
Ex (0,0,1) X (a1,a2,a3) = |i,j,k ; 0,0,1 ; a1,a2,a3 | = -a2i + a1j = (-a2,a1,0)
It follows that a.bXc = det |a;b;c|
i,j,k is orthonormal basis (also called e1,e2,e3). "det" not really determinant- write out as determinant and evaluate as determinant.
It would be visually clearer if I could print determinants, sorry.
ILikeSerena:
In either case it’s the determinant, written with any convention you like.
aXb = i(a2b3-a1b2) –j(a1b3-b1a3) +k(a1b2-a1b2).
It’s just short hand for (a1i + a2j + a3k) X (b1i + b2j + b3k) and noting iXi =0, iXj = k, etc. Sketch a little right hand triad of i,j,k to instantly get all the possibilities.
Geometrically a.bXc is volume of the triad a,b,c, and is usually evaluated as det |a; b; c|
This isn’t a question, just a point of information, since I get the impression the cross product is not generally handy.
For a complete lucid explanation of cross product google “cross product” wiki.