So what is the geometric interpretation of Lagrange's Identity and Jacobi Identity? Thanks.
motoole2
Great question!! I don't want to give away the answer that easily in this case though.
;-) Anyone care to take a stab at answering this?
billiam
As for the Lagrange Identity:
$u \times (v \times w)$ is a vector which is normal to both $u$ and $v \times w$. Since $v$ and $w$ are both normal to $v \times w$, $u \times (v \times w)$ is a vector which is normal to $u$ and lies in the same plane as $v$ and $w$ i.e. is in the subspace spanned by the two vectors $v$ and $w$.
From a linear algebra standpoint, this means the triple product can be represented as a linear combination of vectors $v$ and $w$. This sort of validates the RHS as a not so insane simplification: $\lambda_1v - \lambda_2w$ where $\lambda_1, \lambda_2$ are scalar values.
I think there's a little more to the geometric interpretation that I'm not understanding though :-\
So what is the geometric interpretation of Lagrange's Identity and Jacobi Identity? Thanks.
Great question!! I don't want to give away the answer that easily in this case though. ;-) Anyone care to take a stab at answering this?
As for the Lagrange Identity:
$u \times (v \times w)$ is a vector which is normal to both $u$ and $v \times w$. Since $v$ and $w$ are both normal to $v \times w$, $u \times (v \times w)$ is a vector which is normal to $u$ and lies in the same plane as $v$ and $w$ i.e. is in the subspace spanned by the two vectors $v$ and $w$.
From a linear algebra standpoint, this means the triple product can be represented as a linear combination of vectors $v$ and $w$. This sort of validates the RHS as a not so insane simplification: $\lambda_1v - \lambda_2w$ where $\lambda_1, \lambda_2$ are scalar values.
I think there's a little more to the geometric interpretation that I'm not understanding though :-\