# How does the vector triple product BAC-CAB Identity come about?

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,

$$(\textbf{u}\times\textbf{v})\times\textbf{w}=(\textbf{u}\cdot\textbf{w})\textbf{v}-(\textbf{v}\cdot\textbf{w})\textbf{u}$$

$$\textbf{w}\times(\textbf{u}\times\textbf{v})=(\textbf{w}\cdot\textbf{v})\textbf{u}-(\textbf{w}\cdot\textbf{u})\textbf{v}$$

• how do you know that lambda is a constant? Jun 13 at 9:45

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 $$a \times (b \times c) = b(a \cdot c) - c(a \cdot b)\tag4$$