Precalculus (6th Edition) Blitzer

The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)$ is $5\mathbf{i}-5\mathbf{j}$.
The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)$ can be obtained as, \begin{align} & \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)\mathbf{=}\frac{\left( \mathbf{v}+\mathbf{w} \right)\cdot \mathbf{u}}{{{\left| \mathbf{u} \right|}^{2}}}\mathbf{u} \\ & =\frac{\left[ \left( 3\mathbf{i}-2\mathbf{j} \right)+\left( -5\mathbf{j} \right) \right]\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{{{\left( \sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}} \right)}^{2}}}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =\frac{\left( 3\mathbf{i}-7\mathbf{j} \right)\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}}}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =\frac{3\cdot \left( -1 \right)+\left( -7 \right)\cdot 1}{2}\left( -\mathbf{i}+\mathbf{j} \right) \end{align} Solve ahead to get the result as, \begin{align} & \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)=\frac{-3-7}{2}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =-\frac{10}{2}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =-5\left( -\mathbf{i}+\mathbf{j} \right) \\ & =5\mathbf{i}-5\mathbf{j} \end{align} Hence, the vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)$ is $5\mathbf{i}-5\mathbf{j}$.