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Line $L_2$ is $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$

Find the parametric equation of a line $L_4$ which is different from, yet parallel to, the line $L_2$ given above.

Where I am at so far:

All I know is that two lines in three dimensions are parallel if the direction vectors of both lines are scalar multiples of each other. So I know $L_4$'s direction vector is $(1, -2, -1)$. But that's all I got.

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  • $\begingroup$ To complete your parametric definition of $L_4$ you need one point on the line $L_4$. But how can you choose that point so that $L_4$ will not be the same line as $L_2$? $\endgroup$ – GEdgar Aug 5 '19 at 10:41
  • $\begingroup$ @GEdgar I'm not sure. I've thought about it but nothing is coming to mind. $\endgroup$ – CubbyKushi Aug 5 '19 at 10:42
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    $\begingroup$ You have to understand what $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$ means. $\endgroup$ – GEdgar Aug 5 '19 at 10:43
  • $\begingroup$ It's the vector equation of a line. $\endgroup$ – CubbyKushi Aug 5 '19 at 10:44
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Convince yourself that $(0,0,0)$ is not a point on $L_2$.

Then

$$L_4: \quad ( x, y, z ) = (0, 0, 0) + t(1, -2, -1)$$

will do the job.

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    $\begingroup$ So all I need to do is just find any old point that doesn't lie on $L_2$ and use that for the vector equation of $L_4$? Because that makes sense. $\endgroup$ – CubbyKushi Aug 5 '19 at 11:00
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In the equation $$( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$$ the $(1, 1, -1)$ is a point on the line (for $t=0$) and the $(1, -2, -1)$ is a vector, that gives the direction of the line. Therefore parallel lines are $$( x, y, z ) = (\alpha, \beta, \gamma) + t(1, -2, -1)$$ and $\alpha, \beta, \gamma\in \mathbb{R}$.

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