## Precalculus (6th Edition) Blitzer

$\mathbf{v}=\mathbf{i}-2\mathbf{j}$
The vector $\mathbf{v}$ with initial point ${{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)$ and the terminal point ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)$ can be written as $\mathbf{v}=\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j}$ Here, $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -3,0 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,-2 \right)$. So, the vector $\mathbf{v}$ is \begin{align} & \mathbf{v}=\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j} \\ & =\left[ -2-\left( -3 \right) \right]\mathbf{i}+\left( -2-0 \right)\mathbf{j} \\ & =\mathbf{i}-2\mathbf{j} \end{align}