Indicial Notation 2: Proving $\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} = 0$

[Dustinsfl]

Trying to show that $\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} = 0$.
\begin{alignat}{3}
\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} & = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot(a_1\hat{\mathbf{e}}_i + a_2\hat{\mathbf{e}}_j+a_3\hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_1\hat{\mathbf{e}}_i + \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_2\hat{\mathbf{e}}_j + \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_3\hat{\mathbf{e}}_k\\
& = & \varepsilon_{ijk}a_jb_ka_1(\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_i) + \varepsilon_{ijk}a_jb_ka_2(\hat{\mathbf{e}}_i\cdot \hat{\mathbf{e}}_j) + \varepsilon_{ijk}a_jb_ka_3(\hat{\mathbf{e}}_i\cdot \hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_ka_1
\end{alignat}
Why is this last term 0?

 

[topsquark]

Trying to show that $\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} = 0$.
\begin{alignat}{3}
\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} & = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot(a_1\hat{\mathbf{e}}_i + a_2\hat{\mathbf{e}}_j+a_3\hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_i\hat{\mathbf{e}}_i

Again, a small error.

\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = \epsilon_{ijk}a_jb_k \hat{\mathbf{e}}_i \cdot a_i \hat{\mathbf{e}}_i

\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = \epsilon _{ijk}a_i a_j b_k

Now, a_i a_j is symmetric in i and j, but we are taking an antisymmetric product (from the epsilon) and summing it over a symmetric expression. This will always be zero. So

\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = 0

-Dan

 

[Dustinsfl]

Now, a_i a_j is symmetric in i and j, but we are taking an antisymmetric product (from the epsilon) and summing it over a symmetric expression. This will always be zero.

I see.