# Given two points (4i, -2j, 3k) and (i, j, -3k), how do you find a unit vector parallel to the line segment AB?

Dec 20, 2016

$\hat{\vec{A B}} = \left(\frac{\sqrt{6}}{6}\right) \left(- \vec{i} + \vec{j} - 2 \vec{k}\right)$

#### Explanation:

assuming $\vec{O A} = \left(\begin{matrix}4 \\ - 2 \\ 3\end{matrix}\right) \text{ }$&$\text{ } \vec{O B} = \left(\begin{matrix}1 \\ 1 \\ - 3\end{matrix}\right)$

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

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

$\vec{A B} = - \left(\begin{matrix}4 \\ - 2 \\ 3\end{matrix}\right) + \left(\begin{matrix}1 \\ 1 \\ - 3\end{matrix}\right)$

$\vec{A B} = \left(\begin{matrix}- 4 + 1 \\ 2 + 1 \\ - 3 - 3\end{matrix}\right)$

$\vec{A B} = \left(\begin{matrix}- 3 \\ 3 \\ - 6\end{matrix}\right)$

unit vector parralel to $\text{ } \vec{A B}$

$\hat{\vec{A B}} = \frac{\vec{A B}}{|} \vec{A B} |$

$| \vec{A B} | = \sqrt{{3}^{2} + {3}^{2} + {6}^{2}} = 3 \sqrt{6}$

$\hat{\vec{A B}} = \left(\frac{1}{3 \sqrt{6}}\right) \left(\begin{matrix}- 3 \\ 3 \\ - 6\end{matrix}\right)$

$\hat{\vec{A B}} = \left(\frac{1}{3 \sqrt{6}}\right) \left(- 3 \vec{i} + 3 \vec{j} - 6 \vec{k}\right)$

$\hat{\vec{A B}} = \left(\frac{\sqrt{6}}{18}\right) \left(- 3 \vec{i} + 3 \vec{j} - 6 \vec{k}\right)$

cancelling a factor of $3$ we have

$\hat{\vec{A B}} = \left(\frac{\sqrt{6}}{6}\right) \left(- \vec{i} + \vec{j} - 2 \vec{k}\right)$