Let #bbu#, #bbv#, #bb(w)# be vectors. How can we prove #bbu xx (bbv xx bb(w)) = (bbu * bb(w)) bbv - (bbu * bbv) bb(w)#?

1 Answer
George C.
Mar 7, 2018

See explanation...

Explanation:

There may be a better way, but here's a direct method:

#< v_1, v_2, v_3> xx < w_1, w_2, w_3>#

#=abs((hat(i), hat(j), hat(k)), (v_1, v_2, v_3), (w_1, w_2, w_3))#

#=< v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1>#

#< u_1, u_2, u_3> xx < v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1>#

#=abs((hat(i), hat(j), hat(k)), (u_1, u_2, u_3), (v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1))#

#=< u_2(v_1w_2-v_2w_1)-u_3(v_3w_1-v_1w_3), u_3(v_2w_3-v_3w_2)-u_1(v_1w_2-v_2w_1), u_1(v_3w_1-v_1w_3)-u_2(v_2w_3-v_3w_2)>#

#=< u_2v_1w_2-u_2v_2w_1-u_3v_3w_1+u_3v_1w_3, u_3v_2w_3-u_3v_3w_2-u_1v_1w_2+u_1v_2w_1, u_1v_3w_1-u_1v_1w_3-u_2v_2w_3+u_2v_3w_2>#

#(< u_1, u_2, u_3> * < w_1, w_2, w_3>)< v_1, v_2, v_3>#

#=(u_1w_1+u_2w_2+u_3w_3)< v_1, v_2, v_3>#

#=< u_1v_1w_1+u_2v_1w_2+u_3v_1w_3, u_1v_2w_1+u_2v_2w_2+u_3v_2w_3, u_1v_3w_1+u_2v_3w_2+u_3v_3w_3>#

#(< u_1, u_2, u_3 > * < v_1, v_2, v_3>)< w_1, w_2, w_3>#

#=(u_1v_1+u_2v_2+u_3v_3)< w_1, w_2, w_3>#

#=< u_1v_1w_1+u_2v_2w_1+u_3v_3w_1, u_1v_1w_2+u_2v_2w_2+u_3v_3w_2, u_1v_1w_3+u_2v_2w_3+u_3v_3w_3>#

So:

#(bb(u) * bb(w))bb(v) - (bb(u) * bb(v))bb(w)#

#=< color(red)(cancel(color(black)(u_1v_1w_1)))+u_2v_1w_2+u_3v_1w_3, u_1v_2w_1+color(green)(cancel(color(black)(u_2v_2w_2)))+u_3v_2w_3, u_1v_3w_1+u_2v_3w_2+color(blue)(cancel(color(black)(u_3v_3w_3)))> - < color(red)(cancel(color(black)(u_1v_1w_1)))+u_2v_2w_1+u_3v_3w_1, u_1v_1w_2+color(green)(cancel(color(black)(u_2v_2w_2)))+u_3v_3w_2, u_1v_1w_3+u_2v_2w_3+color(blue)(cancel(color(black)(u_3v_3w_3)))>#

#=< u_2v_1w_2-u_2v_2w_1-u_3v_3w_1+u_3v_1w_3, u_3v_2w_3-u_3v_3w_2-u_1v_1w_2+u_1v_2w_1, u_1v_3w_1-u_1v_1w_3-u_2v_2w_3+u_2v_3w_2>#

#=bb(u) xx (bb(v) xx bb(w))#

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