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

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(D) Given vectors are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \bet

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$(-3, 2)$

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(D) Given vectors are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$.Since the vectors are mutually orthogonal,their dot products are zero: $\vec{a} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 0$.For $\vec{a} \cdot \vec{c} = 0$: $(\hat{i} - \hat{j} + 2\hat{k}) \cdot (\alpha\hat{i} + \hat{j} + \beta\hat{k}) = \alpha - 1 + 2\beta = 0 \implies \alpha + 2\beta = 1$ (Equation $1$).For $\vec{b} \cdot \vec{c} = 0$: $(2\hat{i} + 4\hat{j} + \hat{k}) \cdot (\alpha\hat{i} + \hat{j} + \beta\hat{k}) = 2\alpha + 4 + \beta = 0 \implies 2\alpha + \beta = -4$ (Equation $2$).From Equation $2$,$\beta = -4 - 2\alpha$. Substituting this into Equation $1$:$\alpha + 2(-4 - 2\alpha) = 1 \implies \alpha - 8 - 4\alpha = 1 \implies -3\alpha = 9 \implies \alpha = -3$.Substituting $\alpha = -3$ into $\beta = -4 - 2\alpha$: $\beta = -4 - 2(-3) = -4 + 6 = 2$.Thus,$(\alpha, \beta) = (-3, 2)$.

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

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