I am trying to prove BAC-CAB rule. However, I am not sure about how can I guarantee that $\gamma=1$ is a solution for each vectors? If my proof is not correct, how should I proceed?
I had the followings:
Let $\vec{A}, \vec{B}, \vec{C}$ be three vectors. Prove that $\vec{A}\times(\vec{B}\times \vec{C})=(\vec{A}\cdot \vec{C})\vec{B}-(\vec{A}\cdot \vec{B})\vec{C}$
Proof: By the definition of vector product (cross product), we have $\vec{B} \times \vec{C} \perp \vec{B}$ and $\vec{B} \times \vec{C} \perp \vec{C}$ and $\vec{A} \times(\vec{B} \times \vec{C}) \perp \vec{A}$ and $\vec{A} \times(\vec{B} \times \vec{C}) \perp{\vec{B} \times \vec{C}}$. Thus, $\vec{A} \times(\vec{B} \times \vec{C})$ must lie in the same plane with $\vec{B}$ and $\vec{C}$ and so it is a linear combination of $\vec{B}$ and $\vec{C}$. Assume $\vec{A} \times(\vec{B} \times \vec{C})=\alpha \vec{B}+\beta \vec{C}$ for some scalars $\alpha, \beta$.
Since $\vec{A} \times(\vec{B} \times \vec{C}) \perp \vec{A}$, we have $\vec{A} \cdot(\alpha \vec{B}+\beta \vec{C})=0$. This yields $\alpha \vec{A} \cdot \vec{B}+\beta \vec{A} \cdot \vec{C}=0$ and so $-\frac{\beta}{\vec{A} \cdot \vec{B}}=\frac{\alpha}{\vec{A} \cdot \vec{C}}=: \gamma$. Substituting $\alpha$ and $\beta$, we have $$ \vec{A} \times(\vec{B} \times \vec{C})=(\gamma \vec{A} \cdot \vec{C}) \vec{B}+(-\gamma \vec{A} \cdot \vec{B}) \vec{C} \ldots(*) $$
Since $(*)$ is true for any $\vec{A}, \vec{B}, \vec{C}$, it is also true for $\hat{\imath}, \hat{\jmath}, \hat{k}$. We get $\hat{\imath} \times(\hat{\imath} \times \hat{\jmath})=\gamma\left[\left(\hat{\imath} \cdot \hat{\jmath}\right) \hat{\imath}- (\hat{\imath} \cdot \hat{\imath}) \hat{\jmath}\right]$ and so $-\hat{\jmath}=-\gamma \hat{\jmath}$ namely $\gamma=1$ Consequently, $ \vec{A} \times(\vec{B} \times \vec{C})=(\vec{A} \cdot \vec{C}) \vec{B}-(\vec{A} \cdot \vec{B}) \vec{C}$.
Thanks in advance for any help.