How do I evaluate the following vector product

Question

I need to evaluate the following $$((\vec{a}\times\vec{b})\times\vec{a})_i((\vec{a}\times\vec{b})\times\vec{a})_j.$$ I am assuming Levi civita notation would be useful, but couldn't utilise it. Does any one know how to evaluate this?

Answer 1

The key here is the identity $\epsilon_{ijk} \epsilon_{jpq} = \delta_{kp}\delta_{iq} - \delta_{kq}\delta_{ip}$. Then by the definition of the cross product,

$$ ((a\times b)\times a)_i = \epsilon_{ijk} (a \times b)_j a_k = \epsilon_{ijk}\epsilon_{jpq} a_p b_q a_k $$ Let's now use the identity I mentioned at the start to get

$$((a\times b)\times a)_i = (\delta_{kp}\delta_{iq} - \delta_{kq}\delta_{ip})a_p b_q a_k = a_k a_k b_i - a_k b_k a_i = |a|^2 b_i - (a\cdot b)a_i.$$

Answer 2

Use $$ (\vec{a} \times \vec{b}) \times \vec{a} = \vec{b} (\vec{a} \cdot \vec{a} ) - \vec{a} ( \vec{b} \cdot \vec{a} ) $$

Or

$$ [(\vec{a} \times \vec{b}) \times \vec{a}]_i = \vec{b}_i (\vec{a} \cdot \vec{a} ) - \vec{a}_i ( \vec{b} \cdot \vec{a} ) \\ [(\vec{a} \times \vec{b}) \times \vec{a}]_j = \vec{b}_j (\vec{a} \cdot \vec{a} ) - \vec{a}_j ( \vec{b} \cdot \vec{a} )$$

With their product

$$ = (\vec{b}_i (\vec{a} \cdot \vec{a} ) - \vec{a}_i ( \vec{b} \cdot \vec{a} ))(\vec{b}_j (\vec{a} \cdot \vec{a} ) - \vec{a}_j ( \vec{b} \cdot \vec{a} ))\\ = \vec{b}_i (\vec{a} \cdot \vec{a} )(\vec{b}_j (\vec{a} \cdot \vec{a} ) - \vec{a}_j ( \vec{b} \cdot \vec{a} )) - \vec{a}_i ( \vec{b} \cdot \vec{a} )(\vec{b}_j (\vec{a} \cdot \vec{a} ) - \vec{a}_j ( \vec{b} \cdot \vec{a} )) \\ = \vec{b}_i \vec{b}_j (\vec{a}\cdot \vec{a})^2 - \vec{b}_i \vec{a}_j (\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{a}) - \vec{a}_i\vec{b}_j(\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{a})+\vec{a}_i\vec{a}_j(\vec{b}\cdot\vec{a})^2 \\ = \vec{b}_i \vec{b}_j (\vec{a}\cdot \vec{a})^2 + \vec{a}_i\vec{a}_j(\vec{b}\cdot\vec{a})^2 - (\vec{b}_i \vec{a}_j+\vec{a}_i\vec{b}_j)(\vec{a}\cdot\vec{a})( \vec{b} \cdot \vec{a} ) $$