Verify using an example that a.(b.c)=(a.b).c is not true. Explain your reasoning both numerically and by using the definition of the dot product.
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For three vectors:
- a = (ax)i + (ay)j + (az)k
- b = (bx)i + (by)j + (bz)k
- c = (cx)i + (cy)j + (cz)k,
a.b = (ax)(bx) + (ay)(by) + (az)(bz) and b.c = (bx)(cx) + (by)(cy) + (bz)(cz)
a.(b.c)= a. [(bx)(cx) + (by)(cy) + (bz)(cz)]
=> [(ax)i + (ay)j + (az)k]*[(bx)(cx) + (by)(cy) + (bz)(cz)]
=> (ax)[(bx)(cx) + (by)(cy) + (bz)(cz)]i + (ay)[(bx)(cx) + (by)(cy) + (bz)(cz)]j + (az)[(bx)(cx) + (by)(cy) + (bz)(cz)]k
=> [(ax)(bx)(cx) + (ax)(by)(cy) + (ax)(bz)(cz)]i + [(ay)(bx)(cx) + (ay)(by)(cy) + (ay)(bz)(cz)]j + [(az)(bx)(cx) + (az)(by)(cy) + (az)(bz)(cz)]k ...(1)
(a.b).c =
[(ax)(bx) + (ay)(by) + (az)(bz)].c
=>[(ax)(bx) + (ay)(by) + (az)(bz)][(cx)i + (cy)j + (cz)k]
=> [(ax)(bx) + (ay)(by) + (az)(bz)](cx)i + [(ax)(bx) + (ay)(by) + (az)(bz)](cy)j + [(ax)(bx) + (ay)(by) + (az)(bz)](cz)k
=> [(ax)(bx)(cx) + (ay)(by)(cx) + (az)(bz)(cx)]i + [(ax)(bx)(cy) + (ay)(by)(cy) + (az)(bz)(cy)]j + [(ax)(bx)(cz) + (ay)(by)(cz) + (az)(bz)(cz)]k ...(2)
As can be seen (1) is not equal to (2)
This is easier to see as a numerical example:
a = i + 2j + 3k, b = 4i + 5j + 6k and c = 7i + 8j + 9k,
a.(b.c)= [1*4*7 + 1*5*8 + 1*6*9]i + [2*4*7 + 2*5*8 + 3*6*9]j + [3*4*7 + 3*5*8 + 3*6*9]k
=> 122i + 244j+ 366k
(a.b).c = [1*4*7 + 2*5*7 + 3*6*7]i + [1*4*8 + 2*5*8 + 3*6*8]j + [1*4*9 + 2*5*9 + 3*6*9]k
=> 224i+ 256j + 288k
As can be seen 122i + 244j+ 366k is not the same as 224i+ 256j + 288k.
Therefore it is proved that a.(b.c)=(a.b).c is not true
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