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If $\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+\mathbf{j}, \mathbf{c}=\mathbf{i}$ and $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\lambda \mathbf{a}+\mu \mathbf{b}$, then $\lambda+\mu=$
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a. $\mathbf{c}=1$ and $\mathbf{b} \cdot \mathbf{c}=1$
Given that $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=(\mathbf{c} . \mathbf{a}) \mathbf{b}-(\mathbf{c} . \mathbf{b}) \mathbf{a}=\mu \mathbf{b}+\lambda \mathbf{a}$
where $\mu=\mathbf{c} \cdot \mathbf{a}=1, \lambda=-(\mathbf{c} . \mathbf{b})=-1$
$\Rightarrow \mu+\lambda=1-1=0$
Given that $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=(\mathbf{c} . \mathbf{a}) \mathbf{b}-(\mathbf{c} . \mathbf{b}) \mathbf{a}=\mu \mathbf{b}+\lambda \mathbf{a}$
where $\mu=\mathbf{c} \cdot \mathbf{a}=1, \lambda=-(\mathbf{c} . \mathbf{b})=-1$
$\Rightarrow \mu+\lambda=1-1=0$
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