c*******d
发帖数: 353 | 1 given . as dot product, and x as cross product, we know that
(axb).(cxd) = (a.c)(b.d) - (a.d)(b.c) (1)
also known as identity of lagrange.
How to use the above formula to prove (axb)xc = (a.c)b - (b.c)a (2)?
I know how to prove (2) with kronecker's identities. But what's the
connection between (1) and (2)?
I don't get this from p.14 in 'Differential Geometry' by Kreyszig.
Also, what's the connection between d2x/dt2 and d2x/ds2? Problem 12.2
contradicts with (12.7')
Thanks, | R*********r
发帖数: 1855 | 2 对任意矢量d， ((axb)xc ).d==(axb).(cxd)=((a.c)b - (b.c)a) .d
因此(axb)xc=(a.c)b - (b.c)a | c*******d
发帖数: 353 | |
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