Let $$\vec u,\,\vec v,\,\vec w$$   be such that $$\left| {\vec u} \right| = 1,\,\left| {\vec v} \right| = 2,\,\left| {\vec w} \right| = 3.$$      If the projection $${\vec v}$$ along $${\vec u}$$ is equal to that of $${\vec w}$$ along $${\vec u}$$ and $$\vec v,\,\vec w$$  are perpendicular to each other then $$\left| {\vec u - \vec v + \vec w} \right|$$   equals :

Solution :
Projection of $${\vec v}$$ along $$\vec u = \frac{{\vec v.\vec u}}{{\left| {\vec u} \right|}} = \frac{{\vec v.\vec u}}{2}$$
projection of $${\vec w}$$ along $$\vec u = \frac{{\vec w.\vec u}}{{\left| {\vec u} \right|}} = \frac{{\vec w.\vec u}}{2}$$
$$\eqalign{ & {\text{Given }}\frac{{\vec v.\vec u}}{2} = \frac{{\vec w.\vec u}}{2}.....(1) \cr & {\text{Also,}}\,\,{\text{ }}\vec v.\,\vec w = 0.....(2) \cr & {\text{Now }}\,\,{\text{ }}{\left| {\vec u - \vec v + \vec w} \right|^2} \cr & = {\left| {\vec u} \right|^2} + {\left| {\vec v} \right|^2} + {\left| {\vec w} \right|^2} - 2\vec u.\vec v - 2\vec v.\vec w + 2\vec u.\vec w \cr & = 1 + 4 + 9 + 0\,\,\,\,\,\,\left[ {{\text{ From equation (1) and (2)}}} \right] \cr & = 14 \cr & \therefore \,\left| {\vec u - \vec v + \vec w} \right| = \sqrt {14} \, \cr} $$