$\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})$ is always equal to:
Answer & explanation
Correct answer: option 3
$\hat{i} \times(\vec{a} \times \hat{i})=(\hat{i} . \hat{i}) \vec{a}-(\vec{a} . \hat{i}) \vec{i}=\vec{a}-(\vec{a} . \hat{i}) \hat{i}$
Similarly, $\hat{j} \times(\vec{a} \times \hat{j})=\vec{a}-(\vec{a} . \hat{j}) \hat{j}$
and $\hat{k} \times(\vec{a} \times \hat{k})=\vec{a}-(\vec{a} . \hat{k}) \hat{k}$
$\Rightarrow \hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})$
$=3 \vec{a}-((\vec{a} . \hat{i}) \hat{i}+(\vec{a} . \hat{j}) \hat{j}+(\vec{a} . \hat{k}) \hat{k})$
$=2 \vec{a}$
Hence (3) is correct answer.