# How do you find a unit vector which is parallel to the vector which points from (4,13)(19,8)?

Mar 3, 2018

$\frac{\sqrt{10}}{10} \left(\begin{matrix}3 \\ - 1\end{matrix}\right) = \frac{\sqrt{10}}{10} \left(3 \hat{i} - \hat{j}\right)$

#### Explanation:

call the first coordinate $A$, and the second $B$

then

$\vec{O A} = \left(\begin{matrix}4 \\ 13\end{matrix}\right)$

$\vec{O B} = \left(\begin{matrix}19 \\ 8\end{matrix}\right)$

now

$\vec{A B} = \vec{A O} + \vec{O B}$

$\therefore \vec{A B} = - \vec{O A} + \vec{O B}$

$\vec{A B} = - \left(\begin{matrix}4 \\ 13\end{matrix}\right) + \left(\begin{matrix}19 \\ 8\end{matrix}\right)$

$\vec{A B} = \left(\begin{matrix}15 \\ - 5\end{matrix}\right)$

now

|vec(AB)|=sqrt(15^2+5^2#

$| \vec{A B} | = \sqrt{250} = 5 \sqrt{10}$

a unit vector parallel to$\text{ } \vec{A B}$

is given by

$\frac{1}{5 \sqrt{10}} \left(\begin{matrix}15 \\ - 5\end{matrix}\right)$

$\frac{\sqrt{10}}{50} \left(\begin{matrix}15 \\ - 5\end{matrix}\right)$

cancelling

$\frac{\sqrt{10}}{10} \left(\begin{matrix}3 \\ - 1\end{matrix}\right) = \frac{\sqrt{10}}{10} \left(3 \hat{i} - \hat{j}\right)$