Question
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big).$

Answer

Let:
$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{a}_1&\text{a}_2&\text{a}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
Now,
$\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big).\big[\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)\\-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)\big]$
$=\text{a}_1(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\text{a}_2(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\text{a}_3(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
$=\text{a}_1\text{b}_2\text{a}_3-\text{a}_1\text{b}_3\text{a}_2-\text{a}_2\text{b}_1\text{a}_3+\text{a}_2\text{b}_3\text{a}_1+\text{a}_3\text{b}_1\text{a}_2-\text{a}_3\text{b}_2\text{a}_1$
$=0$

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