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

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(A) Given vectors are $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}.$First,calculate the

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$8$

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

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

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