Thomas' Calculus 13th Edition

${\bf u}+{\bf v}$ is obtained geometrically by 1. translating the vector ${\bf v}$ so that its initial point is at the terminal point of ${\bf u}$. (head-to-tail translation) 2. The resultant sum vector has the initial point of ${\bf u}$ and the terminal point of ${\bf v}$ (in its new position) (See image below) Algebraically, we write the vectors in component form, ${\bf u}=\langle u_{1}, u_{2}\rangle$ or $\langle u_{1}, u_{2}, u_{3}\rangle$ ${\bf v} =\langle v_{1}, v_{2}\rangle$ or $\langle v_{1}, v_{2},\ v_{3}\rangle$ The sum of the two vectors has the sums of the corresponding components as its components, ${\bf u}$+${\bf v} =\langle u_{1}+v_{1}, u_{2}+v_{2}\rangle$ or $\langle u_{1}+v_{1}, u_{2}+v_{2}, u_{3}+v_{3}\rangle$ To subtract, we add the negative vector of ${\bf v}$ ${\bf u-v} ={\bf u}+(-{\bf v}).$ Geometrically, $(-{\bf v})$ is the vector in which we invert the initial and terminal endpoints of $(-{\bf v})$. (see image) Algebraically,$-{\bf v}=\langle-v_{1}, -v_{2}\rangle$ or $\langle-v_{1}, -v_{2},\ -v_{3}\rangle$. The difference ${\bf u-v} =\langle u_{1}-v_{1}, u_{2}-v_{2}\rangle$ or $\langle u_{1}-v_{1}, u_{2}-v_{2}, u_{3}-v_{3}\rangle$