#### Answer

Explanations given below.

#### Work Step by Step

${\bf u}+{\bf v}$ 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 $\langle u_{1}-v_{1}, u_{2}-v_{2}, u_{3}-v_{3}\rangle$