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(C) Given $|\bar{u}|=1, |\bar{v}|=2, |\bar{w}|=3$. Since the projection of $\bar{v}$ on $\bar{u}$ is equal to the projection of $\bar{w}$ on $\bar{u}$

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$\sqrt{14}$

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(C) Given $|\bar{u}|=1, |\bar{v}|=2, |\bar{w}|=3$. Since the projection of $\bar{v}$ on $\bar{u}$ is equal to the projection of $\bar{w}$ on $\bar{u}$,we have $\frac{\bar{v} \cdot \bar{u}}{|\bar{u}|} = \frac{\bar{w} \cdot \bar{u}}{|\bar{u}|}$. Since $|\bar{u}|=1$,this implies $\bar{v} \cdot \bar{u} = \bar{w} \cdot \bar{u}$,or $(\bar{v}-\bar{w}) \cdot \bar{u} = 0$. Also,$\bar{v} \perp \bar{w}$ implies $\bar{v} \cdot \bar{w} = 0$. We need to calculate $|\bar{u}-\bar{v}+\bar{w}|^2 = (\bar{u}-\bar{v}+\bar{w}) \cdot (\bar{u}-\bar{v}+\bar{w})$. $= |\bar{u}|^2 + |\bar{v}|^2 + |\bar{w}|^2 - 2(\bar{u} \cdot \bar{v}) + 2(\bar{u} \cdot \bar{w}) - 2(\bar{v} \cdot \bar{w})$. Since $\bar{u} \cdot \bar{v} = \bar{u} \cdot \bar{w}$ and $\bar{v} \cdot \bar{w} = 0$,the terms $-2(\bar{u} \cdot \bar{v}) + 2(\bar{u} \cdot \bar{w})$ cancel out. $= |\bar{u}|^2 + |\bar{v}|^2 + |\bar{w}|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$. Therefore,$|\bar{u}-\bar{v}+\bar{w}| = \sqrt{14}$.

Let $\bar{u}, \bar{v}, \bar{w}$ be vectors such that $|\bar{u}|=1, |\bar{v}|=2, |\bar{w}|=3$. If the projection of $\bar{v}$ on $\bar{u}$ is equal to that of $\bar{w}$ on $\bar{u}$,and the vectors $\bar{v}, \bar{w}$ are perpendicular to each other,then $|\bar{u}-\bar{v}+\bar{w}|=$

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