Question

For any vector $\vec{a}$
$\hat{i}\times (\vec{a}\times \hat{i})+\hat{j}\times (\vec{a}\times \hat{j})+\hat{k}\times (\vec{a}\times \hat{k})$

• A. $\hat{i}+\hat{j}+\hat{k}$
• B. $3\vec{a}$
• C. $\vec{a}$
• D. $2\vec{a}$

Answer 1

Answer: (D)

$\therefore \vec{a}\times \hat{i}=x(\hat{i}\times \hat{i})+y(\hat{j}\times \hat{i})+z(\hat{k}\times \hat{i})=-y\hat{k}+z\hat{j}$
Similarly, $\vec{a}\times \hat{j}=x(\hat{i}\times \hat{j})+y(\hat{j}\times \hat{j})+z(\hat{k}\times \hat{j})$
          $=x\hat{k}-z\hat{i}$
and $\vec{a}\times \hat{k}=x(\hat{i}\times \hat{k})+y(\hat{j}\times \hat{k})+z(\hat{k}\times \hat{k})$
          $=-x\hat{j}+y\hat{i}$
Now , $\hat{i}\times (\vec{a}\times \hat{i})+\hat{j}\times (\vec{a}\times \hat{j})+\hat{k}\times (\vec{a}\times \hat{k})$
     $=\hat{i}\times (-y\hat{k}\times z\hat{j})+\hat{j}\times (-z\hat{i}\times x\hat{k})+\hat{k}\times (-x\hat{j}\times y\hat{i})$
     $=y\hat{j}+z\hat{k}+z\hat{k}+x\hat{i}+x\hat{i}+y\hat{j}$
      $=2(x\hat{i}+y\hat{j}+z\hat{k})=2\vec{a}$