If a→=(i^+j^+k^),a→⋅b→=1 and a→×b→=j^−k^, then b→ is

(a→×b→)×a→=(a→⋅a→)b→−(a→⋅b→)a→ (j^−k^)×(i^+j^+k^)=(3)2b→−(i^+j^+k^) ijk01-1111=3b→-(i^+j^+k^) 2i-j-k+(i^+j^+k^)=3b→⇒ 3b→=3i^ or b→=i^

If a→=(i^+j^+k^),a→⋅b→=1 and a→×b→=j^−k^, then b→ is

(a→×b→)×a→=(a→⋅a→)b→−(a→⋅b→)a→ (j^−k^)×(i^+j^+k^)=(3)2b→−(i^+j^+k^) ijk01-1111=3b→-(i^+j^+k^) 2i-j-k+(i^+j^+k^)=3b→⇒ 3b→=3i^ or b→=i^

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$If \vec{a} = (\hat{i} + \hat{j} + \hat{k}), \vec{a} \cdot \vec{b} = 1 and \vec{a} \times \vec{b} = \hat{j} - \hat{k}, then \vec{b} is$

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$\hat{i} - \hat{j} + \hat{k}$

$2 \hat{j} - \hat{k}$

$2 \hat{i}$

$\hat{i}$

answer is C.

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Detailed Solution

$\begin{array}{l} (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{a} \\ (\hat{j} - \hat{k}) \times (\hat{i} + \hat{j} + \hat{k}) = (\sqrt{3})^{2} \vec{b} - (\hat{i} + \hat{j} + \hat{k}) \end{array} |\begin{matrix} i & j & k \\ 0 & 1 & - 1 \\ 1 & 1 & 1 \end{matrix}| = 3 \vec{b} - (\hat{i} + \hat{j} + \hat{k}) 2 i - j - k + (\hat{i} + \hat{j} + \hat{k}) = 3 \vec{b} \Rightarrow$