Proof that $\left(\vec a \times \vec b \right) \times \vec a = 0$ using index notation.

Question

To try to prove $\left(\vec a \times \vec b \right) \times \vec a = 0\,\,\,\, $ I used the following identity:

$$\left(\vec u \times \vec v \right) \times \vec w = (\vec w \cdot \vec u)\vec v - (\vec w \cdot \vec v) \vec u$$

but then i got $ (\vec a \cdot \vec a)\vec b - (\vec a \cdot \vec b)\vec a\,$ and using $\vec a = (1 , 3 , 5)$ and $\,\vec b = (2,0,4)$ gives me: $(\vec u \cdot \vec u)\vec v - (\vec u \cdot \vec v)\vec u\ = (48,-66,30)$ and not $(0,0,0)$, so is it false that $\left(\vec a \times \vec b \right) \times \vec a = 0$ ? or is the identity that i used wrong ?

Answer 1

Yes, it is false that $(\vec a\times\vec b)\times\vec a=0$. For a simpler example, take $\vec a=(1,0,0)$ and $\vec b=(0,1,0)$. Then $\vec a\times\vec b=(0,0,1)$ and $(0,0,1)\times\vec a=\vec b\neq0$.