3
$\begingroup$

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,

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

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

$\endgroup$
1
  • $\begingroup$ how do you know that lambda is a constant? $\endgroup$
    – peter
    Jun 13 at 9:45
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.