Let overline{u}, overline{v}, overline{w} be | vedclass

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(A) Given that the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$,we have: $\frac

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$\sqrt{14}$

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(A) Given that the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$,we have: $\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|} = \frac{\overline{w} \cdot \overline{u}}{|\overline{u}|}$. Since $|\overline{u}|=1$,this implies $\overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u}$,or $(\overline{v}-\overline{w}) \cdot \overline{u} = 0$. Also,$\overline{v} \perp \overline{w}$,so $\overline{v} \cdot \overline{w} = 0$. We need to find $|\overline{u}-\overline{v}+\overline{w}|$. Let $X = \overline{u}-\overline{v}+\overline{w}$. Then $|X|^2 = |\overline{u}-\overline{v}+\overline{w}|^2 = (\overline{u}-\overline{v}+\overline{w}) \cdot (\overline{u}-\overline{v}+\overline{w})$. Expanding this,we get: $|X|^2 = |\overline{u}|^2 + |\overline{v}|^2 + |\overline{w}|^2 - 2(\overline{u} \cdot \overline{v}) + 2(\overline{u} \cdot \overline{w}) - 2(\overline{v} \cdot \overline{w})$. Substituting the known values: $|X|^2 = 1^2 + 2^2 + 3^2 - 2(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) - 2(0)$. Since $\overline{u} \cdot \overline{v} = \overline{u} \cdot \overline{w}$,the term $(\overline{u} \cdot \overline{v} - \overline{u} \cdot \overline{w}) = 0$. Thus,$|X|^2 = 1 + 4 + 9 = 14$. Therefore,$|\overline{u}-\overline{v}+\overline{w}| = \sqrt{14}$.

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