## Calculus (3rd Edition)

Assume that the terminal point of the vector $a=\langle 1,3\rangle$, is $(x,y)$, then we have $$\langle 1,3\rangle =(x,y) - (2,2)\Longrightarrow x=3, \quad y=5$$ and hence $a$ is based at $(2,2)$ and the terminal point at $(3,5)$. So, the equivalent vector to $a$ and based at the origin is $\langle 1,3\rangle$ as in the following figure.