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

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(A) Two vectors are perpendicular if their dot product is zero. Since $\vec{a}, \vec{b},$ and $\vec{c}$ are mutually perpendicular,we have $\vec{a} \c

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$(-3, 2)$

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(A) Two vectors are perpendicular if their dot product is zero. Since $\vec{a}, \vec{b},$ and $\vec{c}$ are mutually perpendicular,we have $\vec{a} \cdot \vec{b} = 0$,$\vec{a} \cdot \vec{c} = 0$,and $\vec{b} \cdot \vec{c} = 0$. First,check $\vec{a} \cdot \vec{b} = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0$. This is consistent. Now,use $\vec{a} \cdot \vec{c} = 0$: $(1)(\lambda) + (-1)(1) + (2)(\mu) = 0 \implies \lambda - 1 + 2\mu = 0 \implies \lambda + 2\mu = 1$ (Equation $1$). Next,use $\vec{b} \cdot \vec{c} = 0$: $(2)(\lambda) + (4)(1) + (1)(\mu) = 0 \implies 2\lambda + 4 + \mu = 0 \implies 2\lambda + \mu = -4$ (Equation $2$). From Equation $2$,$\mu = -4 - 2\lambda$. Substitute into Equation $1$: $\lambda + 2(-4 - 2\lambda) = 1 \lambda - 8 - 4\lambda = 1 -3\lambda = 9 \lambda = -3$. Now find $\mu$: $\mu = -4 - 2(-3) = -4 + 6 = 2$. Thus,$(\lambda, \mu) = (-3, 2)$.

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

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