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If the vectors $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}$ are mutually orthogonal, then $(\lambda, \mu)$ is equal to
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The correct answer is:
$(-3,2)$
Given, $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}$
$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are mutually perpendicular,
$\therefore \quad \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=0$
$\mathbf{a} \cdot \mathbf{c}=\lambda-1+2 \mu=0$ ...(i)
$\mathbf{b} \cdot \mathbf{c}=2 \lambda+4+\mu=0$ ...(ii)
Solving Eqs. (i) and (ii), we get
$\lambda=-3$ and $\mu=2$
$\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}$
$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are mutually perpendicular,
$\therefore \quad \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=0$
$\mathbf{a} \cdot \mathbf{c}=\lambda-1+2 \mu=0$ ...(i)
$\mathbf{b} \cdot \mathbf{c}=2 \lambda+4+\mu=0$ ...(ii)
Solving Eqs. (i) and (ii), we get
$\lambda=-3$ and $\mu=2$
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