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If $\vec{a}=(\hat{i}+\hat{j}+\hat{k}), \vec{a} \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{j}-\hat{k}$, then $\vec{b}$ is
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$\hat{i}$
$\because(\vec{a} \times \vec{b}) \times \vec{a}=(\vec{a} \cdot \vec{a}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{a}$ $\therefore \quad(\hat{j}-\hat{k}) \times(\hat{i}+\hat{j}+k)=(\sqrt{3})^{2}(\vec{b})-(\hat{i}+\hat{j}+k)$ $\Rightarrow 3 \hat{b}=3 \hat{i} \Rightarrow \hat{b}=\hat{i}$
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