# How do you find a) u+v, b) u-v, c) 2u-3v given u=<0,0>, v=<2,1>?

Apr 15, 2017

$\left(\begin{matrix}2 \\ 1\end{matrix}\right) , \left(\begin{matrix}- 2 \\ - 1\end{matrix}\right) , \left(\begin{matrix}- 6 \\ - 3\end{matrix}\right)$

#### Explanation:

$\text{Note the notation " <0,0>" is equivalent to } \left(\begin{matrix}0 \\ 0\end{matrix}\right)$

Add/subtract corresponding x and y components from each vector.

$\Rightarrow \underline{u} + \underline{v} = \left(\begin{matrix}0 \\ 0\end{matrix}\right) + \left(\begin{matrix}2 \\ 1\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times x} = \left(\begin{matrix}0 + 2 \\ 0 + 1\end{matrix}\right) = \left(\begin{matrix}2 \\ 1\end{matrix}\right)$

$\Rightarrow \underline{u} - \underline{v} = \left(\begin{matrix}0 \\ 0\end{matrix}\right) - \left(\begin{matrix}2 \\ 1\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times x} = \left(\begin{matrix}0 - 2 \\ 0 - 1\end{matrix}\right) = \left(\begin{matrix}- 2 \\ - 1\end{matrix}\right)$

multiply each x and y component by the scalar quantity.

$\Rightarrow 2 \underline{u} - 3 \underline{v} = 2 \left(\begin{matrix}0 \\ 0\end{matrix}\right) - 3 \left(\begin{matrix}2 \\ 1\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times \times x} = \left(\begin{matrix}2 \times 0 \\ 2 \times 0\end{matrix}\right) - \left(\begin{matrix}3 \times 2 \\ 3 \times 1\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times \times \times x} = \left(\begin{matrix}0 \\ 0\end{matrix}\right) - \left(\begin{matrix}6 \\ 3\end{matrix}\right) = \left(\begin{matrix}- 6 \\ - 3\end{matrix}\right)$