## Precalculus (6th Edition) Blitzer

The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=\frac{1}{2}\mathbf{i}+\frac{3}{2}\mathbf{j}$.
The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ can be obtained as below: \begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v=}\frac{\mathbf{v}\cdot \mathbf{w}}{{{\left| \mathbf{w} \right|}^{2}}}\mathbf{w} \\ & =\frac{\left( -\mathbf{i}+2\mathbf{j} \right)\cdot \left( 3\mathbf{i}-\mathbf{j} \right)}{{{\left( \sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}} \right)}^{2}}}\left( 3\mathbf{i}-\mathbf{j} \right) \\ & =\frac{-1\left( 3 \right)+2\left( -1 \right)}{{{\left( \sqrt{9+1} \right)}^{2}}}\left( 3\mathbf{i}-\mathbf{j} \right) \\ & =\frac{-3-2}{{{\left( \sqrt{10} \right)}^{2}}}\left( 3\mathbf{i}-\mathbf{j} \right) \end{align} Solve ahead to get the result as, \begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=-\frac{5}{10}\left( 3\mathbf{i}-\mathbf{j} \right) \\ & =-\frac{1}{2}\left( 3\mathbf{i}-\mathbf{j} \right) \\ & =-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \end{align} Now, we will obtain ${{\mathbf{v}}_{\mathbf{1}}}$ such that ${{\mathbf{v}}_{\mathbf{1}}}$ is parallel to $\mathbf{w}$ as, \begin{align} & {{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v} \\ & =-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \end{align} Now, we will obtain ${{\mathbf{v}}_{\mathbf{2}}}$ such that ${{\mathbf{v}}_{\mathbf{2}}}$ is orthogonal to $\mathbf{w}$ as, \begin{align} & {{\mathbf{v}}_{\mathbf{2}}}=\mathbf{v}-{{\mathbf{v}}_{\mathbf{1}}} \\ & =\left( -\mathbf{i}+2\mathbf{j} \right)-\left( -\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \right) \\ & =-\mathbf{i}+2\mathbf{j}+\frac{3}{2}\mathbf{i}-\frac{1}{2}\mathbf{j} \\ & =\frac{1}{2}\mathbf{i}+\frac{3}{2}\mathbf{j} \end{align} Hence, the projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=-\frac{3}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=\frac{1}{2}\mathbf{i}+\frac{3}{2}\mathbf{j}$.