If a→ is any Vector, then i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^) is equal to

Let a→=a1i^+a2j^+a3k^Now, i^×(a→×i^)=(i^i^)a→−(i^⋅a→)i^=a→−a1i^Similarly, j^×(a→×j^)=a→−a2j^and k^×(a→×k^)=a→−a3k^∴i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^)=3a→−a1i^+a2j^+a3k^=3a→−a→=2a→

If a→ is any Vector, then i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^) is equal to

Let a→=a1i^+a2j^+a3k^Now, i^×(a→×i^)=(i^i^)a→−(i^⋅a→)i^=a→−a1i^Similarly, j^×(a→×j^)=a→−a2j^and k^×(a→×k^)=a→−a3k^∴i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^)=3a→−a1i^+a2j^+a3k^=3a→−a→=2a→

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If $\vec{a}$ is any Vector, then $\hat{i} \times (\vec{a} \times \hat{i}) + \hat{j} \times (\vec{a} \times \hat{j}) + \hat{k} \times (\vec{a} \times \hat{k})$ is equal to

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$\vec{a}$

$2 \vec{a}$

$3 \vec{a}$

answer is B.

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Detailed Solution

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$

Now, $\hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \hat{i}) \vec{a} - (\hat{i} \cdot \vec{a}) \hat{i} = \vec{a} - a_{1} \hat{i}$

Similarly, $\hat{j} \times (\vec{a} \times \hat{j}) = \vec{a} - a_{2} \hat{j}$

and $\hat{k} \times (\vec{a} \times \hat{k}) = \vec{a} - a_{3} \hat{k}$

$\begin{aligned} ∴ \hat{i} \times (\vec{a} \times \hat{i}) + \hat{j} \times (\vec{a} \times \hat{j}) + \hat{k} \times (\vec{a} \times \hat{k}) \\ = 3 \vec{a} - (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) = 3 \vec{a} - \vec{a} = 2 \vec{a} \end{aligned}$