I have to prove the following:
$$[3(\vec{p}\cdot\hat{r})\hat{r}-\vec{p}]\times[3(\vec{m}\cdot\hat{r})\hat{r})-\vec{m}]=-2\vec{p}\times\vec{m}+3\hat{r}[\hat{r}\cdot(\vec{p}\times\vec{m})]$$
I am given the hint to use the BAC-CAB identity on the following triple product:
$$\hat{r}\times(\vec{p}\times\vec{m})=\vec{p}(\hat{r}\cdot\vec{m})-\vec{m}(\hat{r}\cdot\vec{p})$$
And then using that to evaluate the following in two ways, one directly and the other again using BAC-CAB:
$$\hat{r}\times[\hat{r}\times(\vec{p}\times\vec{m})]=\hat{r}\times\vec{p}(\hat{r}\cdot\vec{m})-\hat{r}\times\vec{m}(\hat{r}\cdot\vec{p})$$
$$=\hat{r}[\hat{r}\cdot(\vec{p}\times\vec{m})]-(\vec{p}\times\vec{m})$$
From here I decided to convert to levi civita and set the= two lines equal to each other:
$$\epsilon_{ijk}r_jp_kr_lm_m\delta_{lm}-\epsilon_{ijk}r_jm_kr_lp_m\delta_{lm}=r_mr_l\epsilon_{ijk}p_jm_k\delta_{li}-\epsilon_{ijk}p_jm_k$$
But from here I have no idea where to go. I can see pieces of what I want but I have no idea how to rearrange this at all to get the desired equality. Any help would be much appreciated.
Best Answer
The hint is useful. The problem's left-hand side expands to$$-3(r\cdot p)r\times m+3(r\cdot m)r\times p+p\times m,$$so you just need to multiply your BAC-CAB inference$$(r\cdot m)r\times p-(r\cdot p)r\times m=r[r\cdot p\times m]-p\times m$$by $3$, then rearrange, no Levi-Civita required.