If bar{a}=2 hat{i}+2 hat{j}+3 hat{k},bar{b}=- | vedclass

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(C) Given vectors are $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$.First,we find the v

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$8$

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(C) Given vectors are $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$.First,we find the vector $\bar{a}+\lambda \bar{b}$:$\bar{a}+\lambda \bar{b} = (2 \hat{i}+2 \hat{j}+3 \hat{k}) + \lambda(-\hat{i}+2 \hat{j}+\hat{k})$$= (2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}$.Since $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,their dot product must be zero:$(\bar{a}+\lambda \bar{b}) \cdot \bar{c} = 0$$((2-\lambda) \hat{i} + (2+2\lambda) \hat{j} + (3+\lambda) \hat{k}) \cdot (3 \hat{i} + 1 \hat{j} + 0 \hat{k}) = 0$$(2-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0$$6 - 3\lambda + 2 + 2\lambda = 0$$8 - \lambda = 0$$\lambda = 8$.

If $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$ are vectors such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

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