As in the title, I was studying the proof (from Vector Analysis – Louis Brand) of this identity, however I do not completely understand all of the steps.
$$a \times (b \times c) = b(a \cdot c) – c(a \cdot b)$$
I also visited an excellent related post – do the BAC-CAB identity for vector triple product have some interpretation?, but I am still stuck. I reproduce the proof in the book here. I don't really understand, how did the author deduce $\alpha=-\lambda(\textbf{v}\cdot\textbf{w}),\beta=\lambda(\textbf{u}\cdot\textbf{w})$ and the basis steps. Any inputs, suggestions or tips to understand the proof would be incredibly helpful!
Proof.
The vector $(\textbf{u} \times \textbf{v}) \times \textbf{w}$ is perpendicular to both $\textbf{u}\times\textbf{v}$ and therefore coplanar with $\textbf{u}$ and $\textbf{v}$.
$$(\textbf{u}\times\textbf{v})\times\textbf{w}=\alpha\textbf{u}+\beta\textbf{v}$$
But, since $(\textbf{u}\times\textbf{v})\times\textbf{w}$ is also perpendicular to $\textbf{w}$,
$$(\alpha\textbf{u}+\beta\textbf{v})\cdot\textbf{w}=0$$
All numbers $\alpha,\beta$ that satisfy this equation must be of the form $\alpha=\lambda(\textbf{v}\cdot\textbf{w}),\beta=\lambda(\textbf{u}\cdot\textbf{w})$, where $\lambda$ is arbitrary.
Thus, we have
$$\textbf{u}\times(\textbf{v}\times\textbf{w})=\lambda\{(\textbf{u}\cdot\textbf{w})v-(\textbf{v}\cdot\textbf{w})\textbf{u}\}$$
In order to determine $\lambda$, we use a special basis in which $\hat{i}$ is collinear with $\textbf{u}$, $\hat{j}$ is co-planar with $\textbf{u, v}$; then
$$\textbf{u}=u_{1}i,\textbf{v}=v_{1}i+v_{2}j,\textbf{w}=w_{1}i+w_{2}j+w_{3}k$$
On substituting these values, we obtain, after a simple calculation, $\lambda=1$.
We therefore have important expansion formulas,
$$(\textbf{u}\times\textbf{v})\times\textbf{w}=(\textbf{u}\cdot\textbf{w})\textbf{v}-(\textbf{v}\cdot\textbf{w})\textbf{u}$$
$$\textbf{w}\times(\textbf{u}\times\textbf{v})=(\textbf{w}\cdot\textbf{v})\textbf{u}-(\textbf{w}\cdot\textbf{u})\textbf{v}$$
Best Answer
If we let $\alpha=\lambda(v\cdot w)$, then $$(\alpha u+\beta v)\cdot w=\lambda(v\cdot w)(u\cdot w)+\beta(v\cdot w)=0\\\implies \beta=-\lambda(u\cdot w)$$ as required. $$(u\times v)\times w=\lambda((v\cdot w) u -(u\cdot w) v)\tag 1$$ (by the way, I have accidentally ended up with a different $\lambda$ to the one given in your derivation, since in that, they seem to have suddenly decided to find an expression for $u\times(v\times w)$ instead of $(u\times v)\times w)$ as we have been working with the whole time. I think my one fits in better.)
For the basis steps, we choose a basis of our choice, $$u=u_{1}i,v=v_{1}i+v_{2}j,w=w_{1}i+w_{2}j+w_{3}k$$ Compute the LHS of $(1)$, $$(u_1v_1(i\times i)+u_1v_2(i\times j))\times(w_1i +w_2j+w_3k)=u_1v_2k\times(w_1i+w_2j)\\=u_1v_2w_1j-u_1v_2w_2i$$
Then compute the RHS of $(1)$, $$(v_1w_1+v_2w_2)u_1i-u_1w_1(v_1i+v_2j)=(v_2w_2u_1)i-(u_1w_1v_2)j$$
Thus $\lambda=-1$, and hence $$(u\times v)\times w=(u\cdot w) v-(v\cdot w) u \tag 2$$ Equivalently $$w \times (u \times v) = u(v \cdot w) - v(u \cdot w)\tag3$$and hence $$a \times (b \times c) = b(a \cdot c) - c(a \cdot b)\tag4$$