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

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

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$\frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}$

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(A) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k}.$Let $\vec{v} = 2 \vec{a}-\vec{b}+3 \vec{c}.$Substituting the values,we get $\vec{v} = 2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k}).$Expanding this,$\vec{v} = (2\hat{i}+2\hat{j}+2\hat{k}) - (2\hat{i}-\hat{j}+3\hat{k}) + (3\hat{i}-6\hat{j}+3\hat{k}).$Grouping the components,$\vec{v} = (2-2+3)\hat{i} + (2+1-6)\hat{j} + (2-3+3)\hat{k} = 3\hat{i}-3\hat{j}+2\hat{k}.$The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{3^{2}+(-3)^{2}+2^{2}} = \sqrt{9+9+4} = \sqrt{22}.$The unit vector parallel to $\vec{v}$ is $\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}} = \frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}.$

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k},$ find a unit vector parallel to the vector $2 \vec{a}-\vec{b}+3 \vec{c}.$

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