If the vectors overline{a}=hat{i}-hat{j}+2 ha | vedclass

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(A) Since the vectors $\overline{a}$,$\overline{b}$,and $\overline{c}$ are mutually orthogonal,their dot products must be zero: $\overline{a} \cdot \o

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$(-3, 2)$

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(A) Since the vectors $\overline{a}$,$\overline{b}$,and $\overline{c}$ are mutually orthogonal,their dot products must be zero: $\overline{a} \cdot \overline{c} = 0$ and $\overline{b} \cdot \overline{c} = 0$. For $\overline{a} \cdot \overline{c} = 0$: $(\hat{i}-\hat{j}+2 \hat{k}) \cdot (\lambda \hat{i}+\hat{j}+\mu \hat{k}) = 0$ $\lambda - 1 + 2\mu = 0 \implies \lambda + 2\mu = 1 \quad ...(i)$ For $\overline{b} \cdot \overline{c} = 0$: $(2 \hat{i}+4 \hat{j}+\hat{k}) \cdot (\lambda \hat{i}+\hat{j}+\mu \hat{k}) = 0$ $2\lambda + 4 + \mu = 0 \implies 2\lambda + \mu = -4 \quad ...(ii)$ Multiplying equation $(ii)$ by $2$,we get $4\lambda + 2\mu = -8 \quad ...(iii)$. Subtracting $(i)$ from $(iii)$: $(4\lambda + 2\mu) - (\lambda + 2\mu) = -8 - 1$ $3\lambda = -9 \implies \lambda = -3$. Substituting $\lambda = -3$ into $(i)$: $-3 + 2\mu = 1 2\mu = 4 \mu = 2$. Thus,$(\lambda, \mu) = (-3, 2)$.

If the vectors $\overline{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\overline{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$ are mutually orthogonal,then $(\lambda, \mu) = $

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