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I have been given a definition in my geometry notes that two lines in $\Bbb R^3$ are parallel if

$1)$ They lie in the same plane

$2)$ They do not intersect.

How can I use this to show that they have the same direction vector. Suppose they lie in the plane $H = \{ v \in \Bbb R^3 | a \cdot v = b \}$ for some $b \in \Bbb R^3$. Then let the lines be named $L_1, L_2$. Not sure how to proceed. Hints appreciated.

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Changing coordinates, you may as well assume that $L_1$ and $L_2$ lie in the $xy$-plane and that neither passes through the origin. In this case the equation for $L_i$ is $a_i x + b_i y = 1$ for non-zero $a_i$ and $b_i$. Assuming that $L_1$ and $L_2$ are not the same line, you get a common solution to these equations unless $(a_1,b_1)$ is a scalar multiple of $(a_2,b_2)$. Since these are the normal vectors to $L_1$ and $L_2$, you get the result you want.

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    $\begingroup$ why equal to 1? $\endgroup$ – IntegrateThis Mar 17 '17 at 14:24
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    $\begingroup$ They don't go through the origin, so you'd have $a x + b y = c$, with $c\neq 0$. You don't change the solution set by switching to $(a/c) x + (b / c) y = 1$. $\endgroup$ – Louis Mar 17 '17 at 14:27
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    $\begingroup$ Great. Only one more thing what if one of them does go through the origin. $\endgroup$ – IntegrateThis Mar 17 '17 at 14:30
  • $\begingroup$ Same thing is true. The point here is that being parallel doesn't depend on where the origin is or orientation, so you can make an affine coordinate change so that nothing goes through the origin or is axis-aligned. If you don't want to do that, you can just replace "1" with "0" when a line goes through the origin, and the same argument applies. $\endgroup$ – Louis Mar 17 '17 at 14:32

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