Consider three vectors v→1,v→2 and v→3 such that v→1=v→2−v→3 where v→1=(a→×i^)×i^,v→2=(a→×j^)×j^ and v→3=(a→×k^)×k^. If a→ is non-zero vector, then

v→1=(a→×i^)×i^=(a→⋅i^)i^−a→v→2=(a→×j^)×j^=(a→⋅j^)j^−a→v→3=(a→×k^)×k^=(a→⋅k^)k^−a→v→1=v→2−v→3⇒ (a→⋅i^)i^−a→=(a→⋅j^)j^−a→−(a→⋅k^)k^+a→⇒ (a→⋅i^)i^−{(a→⋅i^)i^+(a→⋅j^)j^+(a→⋅k^)k^}=(a→⋅j^)j^−(a→⋅k^)k^⇒ 2(a→⋅j^)j^=0⇒ a→⋅j^=0

Consider three vectors v→1,v→2 and v→3 such that v→1=v→2−v→3 where v→1=(a→×i^)×i^,v→2=(a→×j^)×j^ and v→3=(a→×k^)×k^. If a→ is non-zero vector, then

v→1=(a→×i^)×i^=(a→⋅i^)i^−a→v→2=(a→×j^)×j^=(a→⋅j^)j^−a→v→3=(a→×k^)×k^=(a→⋅k^)k^−a→v→1=v→2−v→3⇒ (a→⋅i^)i^−a→=(a→⋅j^)j^−a→−(a→⋅k^)k^+a→⇒ (a→⋅i^)i^−{(a→⋅i^)i^+(a→⋅j^)j^+(a→⋅k^)k^}=(a→⋅j^)j^−(a→⋅k^)k^⇒ 2(a→⋅j^)j^=0⇒ a→⋅j^=0

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$Consider three vectors {\vec{v}}_{1}, {\vec{v}}_{2} and {\vec{v}}_{3} such that$ ${\vec{v}}_{1} = {\vec{v}}_{2} - {\vec{v}}_{3} where {\vec{v}}_{1} = (\vec{a} \times \hat{i}) \times \hat{i}, {\vec{v}}_{2} = (\vec{a} \times \hat{j}) \times \hat{j} and$
${\vec{v}}_{3} = (\vec{a} \times \hat{k}) \times \hat{k} . If \vec{a} is non-zero vector, then$

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$\vec{a} \cdot \hat{j} = 0$

$\vec{a} \cdot \hat{i} = 0$

$\vec{a} \cdot \hat{k} = 0$

${\vec{v}}_{1} \cdot {\vec{v}}_{2} = (\vec{a} \cdot \hat{j})^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} {\vec{v}}_{1} = (\vec{a} \times \hat{i}) \times \hat{i} = (\vec{a} \cdot \hat{i}) \hat{i} - \vec{a} \\ {\vec{v}}_{2} = (\vec{a} \times \hat{j}) \times \hat{j} = (\vec{a} \cdot \hat{j}) \hat{j} - \vec{a} \\ {\vec{v}}_{3} = (\vec{a} \times \hat{k}) \times \hat{k} = (\vec{a} \cdot \hat{k}) \hat{k} - \vec{a} \\ {\vec{v}}_{1} = {\vec{v}}_{2} - {\vec{v}}_{3} \\ \Rightarrow (\vec{a} \cdot \hat{i}) \hat{i} - \vec{a} = (\vec{a} \cdot \hat{j}) \hat{j} - \vec{a} - (\vec{a} \cdot \hat{k}) \hat{k} + \vec{a} \\ \Rightarrow (\vec{a} \cdot \hat{i}) \hat{i} - {(\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k}} = (\vec{a} \cdot \hat{j}) \hat{j} - (\vec{a} \cdot \hat{k}) \hat{k} \\ \Rightarrow 2 (\vec{a} \cdot \hat{j}) \hat{j} = 0 \\ \Rightarrow \vec{a} \cdot \hat{j} = 0 \end{array}$