If \(\vec{u}=\hat{i}\times (\vec{a}\times \hat{i})+\hat{j} \times (\vec{a}\times \hat{j})+\hat{k}\times (\vec{a} \times \hat{k})\) then \(\vec{u}\) is equal to

\(2\vec{a}\)

\(\begin{aligned}\hat{i}\times (\vec{a}\times \hat{i})&=(\hat{i}\cdot \hat{i})\vec{a}-(\hat{i}\cdot \hat{a})\hat{i}\\ &=\vec{a}-(\hat{i}-\vec{a})\hat{i}=\vec{a}-x\hat{i}\\ \text{similarly, }\hat{j}\times (\vec{a}\times \hat{j}&=\vec{a}-y\hat{j}\\ \hat{k}\times (\vec{a}\times \hat{k})&=\vec{a}-z\hat{k}\\ \vec{u}=3\vec{a}-(x\hat{i}+y\hat{j}+z\hat{k})&=3\vec{a}-\vec{a}=2\vec{a}\end{aligned}\)