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Let $(\mathbb R^3,d)$ where $d$ is Euclidean distance be called Euclidean space.

For any two points $a,b \in \mathbb R^3$, let the set ${\{s:s=ta+(1-t)b:0 \leq t \leq 1}\}$ be called line segment from $a$ to $b$.

In this context, how are directed line segments defined?

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    The parametrization gives an orientation to the segment. The set $\{s:s=(1-t)a+tb:0\le t\le1\}$ describes the same set of points, but they’re “traced” in the opposite direction. – 2017-02-11

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One could define a directed line segment to be an ordered triple $$\vec{ab} = (a,b,\overline{ab}) $$ where $\overline{ab}$ is the set that you defined.

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    How would one define "equal direction" in this definition? – 2017-02-11
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    Two ordered triples $(x,y,z)$, $(x',y',z')$ are equal if and only if $x=x'$ and $y=y'$ and $z=z'$. So, for example any line segment $\overline{ab}$ has two directed line segments associated to it, namely $(a,b,\overline{ab})$ and $(b,a,\overline{ab})$; these are unequal because $a \ne b$. – 2017-02-11
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    Consider though two line segments, laying on each other, one being shorter than other. Intuitively, it′s possible for them to point in the same direction, however under your definition it′s not. – 2017-02-11
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    You bring up a different issue which, while interesting, is not part of your question as currently written. So I'll just say very briefly that this can easily be handled with the definition that I wrote. – 2017-02-11