# Difficult vector identity using Levi Civita

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.

• The two sides of the original identity are multi-linear. Each of the three vectors is a linear combination of three basis vectors. There are only $3^3=27$ equations to check. Commented Feb 6, 2021 at 2:15
• @Somos I fail to see what that has to do with proving the equality using levi civita notation. Of course we could brute force the identity, but that's not really the point is it. Commented Feb 6, 2021 at 2:33
• Could you check the initial formula? A typo can be there. For example, does it mean $(\vec{m}\cdot\hat{r})=(\vec{m},\vec{r})$ - a scalar product of two vectors? Commented Feb 6, 2021 at 7:04
• Okay, in three dimensions, there are only $3!=6$ non zero levi civita symbols. Check all 6 equations. Commented Feb 6, 2021 at 12:22
• @Somos Like I said earlier, we could brute force the identity to prove it, but that is not the point of the problem. Commented Feb 6, 2021 at 12:26

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.