If vec{a} = 2 hat{i} - hat{j} + hat{k},vec{b} | vedclass

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(A) We are given the vectors: $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} -

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$-2$

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(A) We are given the vectors: $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$.First,we calculate the vector $\lambda \vec{b} + \vec{c}$:$\lambda \vec{b} + \vec{c} = \lambda (\hat{i} + \hat{j} - 2 \hat{k}) + (\hat{i} + 3 \hat{j} - \hat{k})$$= (\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}$.Since $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,their dot product must be zero:$\vec{a} \cdot (\lambda \vec{b} + \vec{c}) = 0$$(2 \hat{i} - \hat{j} + \hat{k}) \cdot ((\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}) = 0$$2(\lambda + 1) - 1(\lambda + 3) + 1(-2 \lambda - 1) = 0$$2 \lambda + 2 - \lambda - 3 - 2 \lambda - 1 = 0$$-\lambda - 2 = 0$$\lambda = -2$.

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ are given vectors. If $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,then $\lambda = . . . . . .$.

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