Let vec{a}=2 hat{i}+hat{j}-hat{k} and vec{b}= | vedclass

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(C) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$.First,calculate the vector $\vec{v} = 3 \vec{a}-2 \vec{b}$:$\v

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$-4 \hat{i}+3 \hat{j}-7 \hat{k}$

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(C) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$.First,calculate the vector $\vec{v} = 3 \vec{a}-2 \vec{b}$:$\vec{v} = 3(2 \hat{i}+\hat{j}-\hat{k})-2(\hat{i}+3 \hat{j}-5 \hat{k}) = (6-2) \hat{i} + (3-6) \hat{j} + (-3+10) \hat{k} = 4 \hat{i}-3 \hat{j}+7 \hat{k}$.Since $\vec{r}$ is along the vector $\vec{v}$,we can write $\vec{r} = x \vec{v} = x(4 \hat{i}-3 \hat{j}+7 \hat{k})$ for some scalar $x$.Given $|\vec{r}| = \sqrt{74}$,we have $|x| \sqrt{4^2+(-3)^2+7^2} = \sqrt{74}$.$|x| \sqrt{16+9+49} = \sqrt{74} \Rightarrow |x| \sqrt{74} = \sqrt{74} \Rightarrow |x| = 1$.Since the direction of $\vec{r}$ is opposite to that of $\vec{v}$,we must have $x = -1$.Therefore,$\vec{r} = -1(4 \hat{i}-3 \hat{j}+7 \hat{k}) = -4 \hat{i}+3 \hat{j}-7 \hat{k}$.

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