I find it hard to understand Jacobi identity geometrically. I was trying to relate the geometrical meaning of cross product to this triangle but no luck so far. Any hints?
keenan
This document gives an explanation, but I'd love to see an even better one! Someone in lecture yesterday suggested considering the three edges of a tetrahedron, for instance (though at present I'm not clear on how that would work).
bpopeck
I will take a shot at a geometric interpretation of Lagrange's identity. In short, I think the identity tells us that the shadow of $u$ in the $vw$-plane rotated by 90 degrees is orthogonal to $u$.
The cross product of two vectors gives a vector orthogonal to each of them, so we know $u \times (v \times w)$ is orthogonal to $v \times w$ so it must lie in the $vw$-plane. We also know $u \times (v \times w)$ is orthogonal to $u$.
$u$ projected into the $vw$-plane is $(u\cdot v)v + (u\cdot w)w$. To rotate the projection by 90 degrees, we can negate one of the coordinates and swap them. Negate the $v$-coordinate (choose $v$ in order to preserve the orientation of the cross product, I think) and swap the coordinates to get $(u\cdot w)v - (u\cdot v)w$. This is Lagrange's identity.
I find it hard to understand Jacobi identity geometrically. I was trying to relate the geometrical meaning of cross product to this triangle but no luck so far. Any hints?
This document gives an explanation, but I'd love to see an even better one! Someone in lecture yesterday suggested considering the three edges of a tetrahedron, for instance (though at present I'm not clear on how that would work).
I will take a shot at a geometric interpretation of Lagrange's identity. In short, I think the identity tells us that the shadow of $u$ in the $vw$-plane rotated by 90 degrees is orthogonal to $u$.
The cross product of two vectors gives a vector orthogonal to each of them, so we know $u \times (v \times w)$ is orthogonal to $v \times w$ so it must lie in the $vw$-plane. We also know $u \times (v \times w)$ is orthogonal to $u$.
$u$ projected into the $vw$-plane is $(u\cdot v)v + (u\cdot w)w$. To rotate the projection by 90 degrees, we can negate one of the coordinates and swap them. Negate the $v$-coordinate (choose $v$ in order to preserve the orientation of the cross product, I think) and swap the coordinates to get $(u\cdot w)v - (u\cdot v)w$. This is Lagrange's identity.
Here is a rough diagram.