If vec{f}=hat{i}+hat{j}+hat{k} and vec{g}=2 h | vedclass

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(B) Given vectors are $\vec{f} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{g} = 2\hat{i} - \hat{j} + 3\hat{k}$.The formula for the projection vector of $

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$\frac{2}{7}(2 \hat{i}-\hat{j}+3 \hat{k})$

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(B) Given vectors are $\vec{f} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{g} = 2\hat{i} - \hat{j} + 3\hat{k}$.The formula for the projection vector of $\vec{f}$ on $\vec{g}$ is given by $	ext{proj}_{\vec{g}} \vec{f} = \left( \frac{\vec{f} \cdot \vec{g}}{|\vec{g}|^2} ight) \vec{g}$.First,calculate the dot product $\vec{f} \cdot \vec{g} = (1)(2) + (1)(-1) + (1)(3) = 2 - 1 + 3 = 4$.Next,calculate the square of the magnitude of $\vec{g}$,which is $|\vec{g}|^2 = (2)^2 + (-1)^2 + (3)^2 = 4 + 1 + 9 = 14$.Now,substitute these values into the formula:$	ext{proj}_{\vec{g}} \vec{f} = \frac{4}{14} (2\hat{i} - \hat{j} + 3\hat{k}) = \frac{2}{7} (2\hat{i} - \hat{j} + 3\hat{k})$.Thus,the correct option is $B$.

If $\vec{f}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{g}=2 \hat{i}-\hat{j}+3 \hat{k}$,then the projection vector of $\vec{f}$ on $\vec{g}$ is

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