Find the vector projection of $\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$ onto $\mathbf{w}=\mathbf{i}+\mathbf{j}+\mathbf{k}$. Decompose $\mathbf{v}$ into two vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$, where $\mathbf{v} _{1}$ is parallel to $\mathbf{w}$ and $\mathbf{v}_{2}$ is orthogonal to $\mathbf{w}$.

Solution We use the formula for the projection of $\mathbf{v}$ onto $\mathbf{w}.$ $$\begin{eqnarray*} \mathbf{v}_{1}& =&\hbox{proj}_{\mathbf{w}}\mathbf{v}=\frac{\mathbf{v}\,{\cdot}\, \mathbf{w}}{\Vert \mathbf{w}\Vert ^{2}}\mathbf{w} \underset{\underset{\underset{{\color{#0066A7}{\hbox{\(\Vert \mathbf{w}\Vert =\sqrt{1^{2}+1^{2} +1^{2}} =\sqrt{3}\)}}}} {\color{#0066A7}{\hbox{\(\mathbf{v}\,{\cdot}\, \mathbf{w=}2-1+1=2\)}}}} {\color{#0066A7}{\uparrow }}}{=}\frac{2}{(\sqrt{3})^{2}}\mathbf{w}=\frac{2}{3}( \mathbf{i}+\mathbf{j}+\mathbf{k})=\frac{2}{3}\,\mathbf{i}+\frac{2}{3}\, \mathbf{j}+\frac{2}{3}\,\mathbf{k} \\ \mathbf{v}_{2}& =&\mathbf{v}-\mathbf{v}_{1}=(2\mathbf{i}-\mathbf{j}+\mathbf{k} )-\left( \frac{2}{3}\,\mathbf{i}+\frac{2}{3}\,\mathbf{j}+\frac{2}{3}\, \mathbf{k}\right) =\frac{4}{3}\,\mathbf{i}-\frac{5}{3}\,\mathbf{j}+\frac{1}{3 }\,\mathbf{k} \end{eqnarray*}$$