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

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

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$-1$

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(C) Given vectors are $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 4\hat{k}$,and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.Since $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$,their dot product must be zero:$(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$First,calculate $\vec{a} + \lambda\vec{b}$:$\vec{a} + \lambda\vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k}) = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + 4\lambda)\hat{k}$Now,take the dot product with $\vec{c} = \hat{i} + \hat{j} + \hat{k}$:$((1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + 4\lambda)\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$$(1 + \lambda)(1) + (2 - \lambda)(1) + (1 + 4\lambda)(1) = 0$$1 + \lambda + 2 - \lambda + 1 + 4\lambda = 0$$4 + 4\lambda = 0$$4\lambda = -4$$\lambda = -1$

If $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+4\hat{k}$,and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ are such that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

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