If a→ satisfies a→×(i^+2j^+k^)=i^−k^ , then a→ is equal to

a→×(i^+2j^+k^)=i^−k^=(j^×(i^+2j^+k^))⇒(a→−j^)×(i^+2j^+k^)=0→⇒a→−j^=λ(i^+2j^+k^)⇒a→=λi^+(2λ+1)j^+λk^,λ∈R

If a→ satisfies a→×(i^+2j^+k^)=i^−k^ , then a→ is equal to

a→×(i^+2j^+k^)=i^−k^=(j^×(i^+2j^+k^))⇒(a→−j^)×(i^+2j^+k^)=0→⇒a→−j^=λ(i^+2j^+k^)⇒a→=λi^+(2λ+1)j^+λk^,λ∈R

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If $\vec{a}$ satisfies $\vec{a} \times (\hat{i} + 2 \hat{j} + \hat{k}) = \hat{i} - \hat{k}$ , then $\vec{a}$ is equal to

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$λ \hat{i} + (2 λ - 1) \hat{j} + λ \hat{k}, λ \in R$

$λ \hat{i} + (1 - 2 λ) \hat{j} + λ \hat{k}, λ \in R$

$λ \hat{i} + (2 λ + 1) \hat{j} + λ \hat{k}, λ \in R$

$λ \hat{i} - (1 + 2 λ) \hat{j} + λ \hat{k}, λ \in R$

answer is C.

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Detailed Solution

$\begin{array}{l} \vec{a} \times (\hat{i} + 2 \hat{j} + \hat{k}) = \hat{i} - \hat{k} = (\hat{j} \times (\hat{i} + 2 \hat{j} + \hat{k})) \\ \Rightarrow (\vec{a} - \hat{j}) \times (\hat{i} + 2 \hat{j} + \hat{k}) = \vec{0} \\ \Rightarrow \vec{a} - \hat{j} = λ (\hat{i} + 2 \hat{j} + \hat{k}) \\ \Rightarrow \vec{a} = λ \hat{i} + (2 λ + 1) \hat{j} + λ \hat{k}, λ \in R \end{array}$