Proving $\vec{x} \cdot (\vec{x} \times \vec{y}) = 0$

Question

R.T.P $\vec{x} \cdot (\vec{x} \times \vec{y}) = 0$ using index notation

what i had done was the following:

$\vec{x} \cdot (\vec{x} \times \vec{y}) = x_{i}(\vec{x} \times \vec{y})_{i}$

$= x_{i}(\epsilon_{ijk} x_{j} y_{k})$

$= \epsilon_{ijk} x_{i} x_{j} y_{k}$

where $\epsilon_{ijk} = \begin{cases} 1 : (i,j,k) \in \lbrace (1,2,3);(2,3,1);(3,1,2) \rbrace \\ -1: (i,j,k) \in \lbrace (1,3,2);(2,1,3);(3,2,1) \rbrace \\ 0: i=j;i=k;j=k \end{cases}$

i am unable to go further from here. is there an identity i could possibly use to finish the proof.

Answer 1

$\epsilon_{ijk}$ is anti-symmetric, i.e. if you swap two indices, its value changes sign. In particular, $$\epsilon_{ijk} = -\epsilon_{jik}$$ Note that $$\epsilon_{ijk}x_i x_j y_k = -\epsilon_{jik}x_jx_iy_k = -\epsilon_{jik}x_ix_jy_k $$ In the expression you've obtained, it wouldn't matter if you swapped $i$ and $j$. Think about what the summation (this is Einstein summation notation, you must be familiar since you're using it) turns out to be - I'm sure you'll convince yourself. Hence, $$\epsilon_{ijk}x_i x_j y_k = \epsilon_{jik}x_j x_i y_k = \epsilon_{jik}x_i x_j y_k$$ Combining this result with the immediately previous one, we get $$\epsilon_{ijk}x_i x_j y_k = - \epsilon_{ijk}x_i x_j y_k \implies \epsilon_{ijk}x_i x_j y_k = 0$$ and that is what we wanted to show!

Answer 2

If you exchange the indices $i$ and $j$ and then use that $\epsilon_{ijk} = -\epsilon_{jik}$, you get that $$\epsilon_{ijk} x_{i} x_{j} y_{k} = \epsilon_{jik}x_jx_iy_k = -\epsilon_{ijk} x_{i} x_{j} y_{k}$$ and therefore $2\epsilon_{ijk} x_{i} x_{j} y_{k}=0$