Let vec{r} be a vector in the plane of hat{i} | vedclass

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(C) Since $\vec{r}$ lies in the plane of $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} - \hat{k}$,we can write $\vec{r} =

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$\sqrt{35}$

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(C) Since $\vec{r}$ lies in the plane of $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} - \hat{k}$,we can write $\vec{r} = \lambda \vec{a} + \mu \vec{b} = \lambda(\hat{i} - 2\hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} - \hat{k})$.$\vec{r} = (\lambda + \mu)\hat{i} - (2\lambda + \mu)\hat{j} + (\lambda - \mu)\hat{k}$.Given $\vec{r} \cdot (\hat{i} + \hat{j}) = -2$,we have $(\lambda + \mu) - (2\lambda + \mu) = -2$,which simplifies to $-\lambda = -2$,so $\lambda = 2$.Now,the projection of $\vec{r}$ on $\hat{i} - \hat{j}$ is given by $\frac{|\vec{r} \cdot (\hat{i} - \hat{j})|}{\sqrt{1^2 + (-1)^2}} = 4\sqrt{2}$.$\vec{r} \cdot (\hat{i} - \hat{j}) = (\lambda + \mu) + (2\lambda + \mu) = 3\lambda + 2\mu$.Substituting $\lambda = 2$,we get $3(2) + 2\mu = 6 + 2\mu$.Thus,$\frac{|6 + 2\mu|}{\sqrt{2}} = 4\sqrt{2} \Rightarrow |6 + 2\mu| = 8$.Case $1$: $6 + 2\mu = 8 \Rightarrow 2\mu = 2 \Rightarrow \mu = 1$.Then $\vec{r} = (2+1)\hat{i} - (2(2)+1)\hat{j} + (2-1)\hat{k} = 3\hat{i} - 5\hat{j} + \hat{k}$.$|\vec{r}| = \sqrt{3^2 + (-5)^2 + 1^2} = \sqrt{9 + 25 + 1} = \sqrt{35}$.Case $2$: $6 + 2\mu = -8 \Rightarrow 2\mu = -14 \Rightarrow \mu = -7$.Then $\vec{r} = (2-7)\hat{i} - (2(2)-7)\hat{j} + (2+7)\hat{k} = -5\hat{i} + 3\hat{j} + 9\hat{k}$.$|\vec{r}| = \sqrt{(-5)^2 + 3^2 + 9^2} = \sqrt{25 + 9 + 81} = \sqrt{115}$.Since $\sqrt{35}$ is an option,the magnitude is $\sqrt{35}$.

Let $\vec{r}$ be a vector in the plane of $\hat{i} - 2\hat{j} + \hat{k}$ and $\hat{i} - \hat{j} - \hat{k}$ such that $\vec{r} \cdot (\hat{i} + \hat{j}) + 2 = 0$ and the length of the projection of $\vec{r}$ on $\hat{i} - \hat{j}$ is $4\sqrt{2}$. Then,the magnitude of vector $\vec{r}$ is:

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