Question

Show that \(\ u \cdot v \times w = w \cdot u \times v \)

Answer 1

They're all vectors and the cross product is implied to be first, otherwise if you did the dot product first you'd get a scalar and you can cross a scalar with a vector! =)

Answer 2

tried the easy but long method to do it?

u = u1 i+ u2 j
v = v1 i + v2 j
w = w1 i + w2 j
?

Answer 3

Well I found a really simple way to prove it with tensor calculus and I thought I'd share it to get people interested in the subject since without it, it's completely terrible haha.

Answer 4

please do :)

Answer 5

in tensor notation the dot and cross products are \[\Large a \cdot b= a_ib^i\]\[\Large a \times b = \epsilon^{ijk}a_j b_k\] That epsilon thing with 3 indices is the Levi Civita tensor and it's skew symmetric with respect to all of its indices, so for example if I swap the last two indicies it changes the sign on it like this \[\Large \epsilon^{123}=-\epsilon^{132}\] So now let's just plug stuff in \[\Large u \cdot v \times w = u_i \epsilon^{ijk}v_jw_k\] all those terms are associative and we can move the indices around since they're skew symmetric. If we do it twice then two negative signs cancel out and these are all dummy indices, so we can just trade out k's with i's to make it look like how we originally wrote it... \[\Large w_k \epsilon^{kji}u_jv_i = w \cdot u \times v\] That's it, we did it. It's all pretty commonly found stuff in tensor calculus, so it's not even a very surprising identity!

Answer 6

\(\Large \epsilon^{123}=-\epsilon^{132}\)
this property really made this a one step solution :)

Answer 7

Yeah, it's a really useful symbol. In this form it's basically a 3D cube matrix with 27 entries without getting into the metric tensor if we assume we're looking at it in Cartesian coordinates, all the entries of it are either 0, 1, or -1. Most of the entries are really 0 though, which is actually kind of nice, and fun to show: There are only 3! entries that aren't 0 so you don't have to think of all 3^3 entries.\[\Large \epsilon^{112}=-\epsilon^{112}=0\] Any ones with double entries of the same number aregoing to be zero, since you can just permute those two symbols and get the same but negative back. The real use of this symbol is in expressing determinants, it's very powerful for showing some of the more difficult linear algebra identities in a very simple way.

Answer 8

Another simple and semi-painless proof: \[\Large a \times (b \times c) = \epsilon^{ijk}a_j \epsilon_{k lm}b^lc^m= \delta^{ijk}_{klm}a_jb^lc^m= \\ \Large \delta^{ij}_{lm}a_jb^lc^m = (\delta^i_l \delta^j_m - \delta^i_m \delta^j_l)a_jb^lc^m = \\ \Large a_jc^jb^i - a_jb^jc^i = (a \cdot c) b-(a \cdot b) c\]