# 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.

### 1 Answer | Add Yours

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**