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I understand that when we apply the Lagrange's identity onto the expression we can get 0. However, in terms of geometry, the expression add 3 vectors starting from each vertex with the same direction as the altitude going across that vertex. Why the sum of these 3 vectors is the 0 vector, as we don't know the angle between them and I'm not sure the relation between their length....


@yingxiul Here's one explanation though I don't find it particularly clear! Would love to see someone write up a (much) simpler version of this explanation, with nice pictures. ;-)


With regard to the Jacobi identity: In this particular case, let the area of the triangle be A, then the magnitude of (v x w) is 2A and the direction is out of the screen. Thus u x (v x w) has magnitude 2A|u| parallel to the altitude of u. Thus u x (v x w) is u rotated -90 degrees in the plane of the triangle, multiplied by 2A. The same can be said for v x (w x u) and w x (u x v) by cyclic symmetry. Since u + v + w add up to the 0 vector, then applying a -90 degree rotation to all and then multiplying by magnitude 2A will still have a sum of the 0 vector.

Alternatively, without using the altitudes, (v x w) has magnitude 2A coming out of the screen, and by symmetry (w x u) and (u x v), i.e. v x w = w x u = u x v =: n. Thus u x (v x w) + v x (w x u) + w x (u x v) = (u + v + w) x n = 0 x n = 0.

I wonder though, how one can think of this geometrically when u, v, w aren't the sides of a triangle. If I remember correctly, the Jacobi identity applies for arbitrary vectors u, v, w, not just those constrained to u + v + w = 0.


Excellent. Yes, you're right that this argument applies only in the special case of $u + v + w = 0$, even though the Jacobi identity applies for any triple of vectors. So, still some mystery there! :-)