Answer
$${\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = \left\langle { - 1,1, - 2} \right\rangle {\text{ and sca}}{{\text{l}}_{\bf{v}}}{\bf{u}} = - \sqrt 6 $$
Work Step by Step
$$\eqalign{
& {\bf{u}} = \left\langle {3,3, - 3} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {1, - 1,2} \right\rangle \cr
& {\text{The orthogonal projection of }}{\bf{u}}{\text{ onto }}{\bf{v}}{\text{, denoted pro}}{{\text{j}}_{\bf{v}}}{\bf{u}}, \cr
& {\text{,where }}{\bf{v}} \ne 0,{\text{is}} \cr
& {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}}\left( {\frac{{\bf{v}}}{{\left| {\bf{v}} \right|}}} \right) = \left( {\frac{{{\bf{u}} \cdot {\bf{v}}}}{{{\bf{v}} \cdot {\bf{v}}}}} \right){\bf{v}} \cr
& {\text{The scalar component of }}{\bf{u}}{\text{ in the direction of }}{\bf{v}}{\text{ is}} \cr
& {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}} = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{v}} \right|}} = \frac{{\left\langle {3,3, - 3} \right\rangle \cdot \left\langle {1, - 1,2} \right\rangle }}{{\left| {\left\langle {1, - 1,2} \right\rangle } \right|}} = \frac{{3 - 3 - 6}}{{\sqrt {1 + 1 + 4} }} = \frac{{ - 6}}{{\sqrt 6 }} = - \sqrt 6 \cr
& \cr
& {\text{Calculating pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} \cr
& {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}}\left( {\frac{{\bf{v}}}{{\left| {\bf{v}} \right|}}} \right) = - \sqrt 6 \left( {\frac{{\left\langle {1, - 1,2} \right\rangle }}{{\sqrt 6 }}} \right) \cr
& {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = - \left\langle {1, - 1,2} \right\rangle \cr
& {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = \left\langle { - 1,1, - 2} \right\rangle \cr} $$
