# Find the parametric equation of a line different from, yet parallel to $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$

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.

• 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 $L_2$? Aug 5, 2019 at 10:41
• @GEdgar I'm not sure. I've thought about it but nothing is coming to mind.
– SFR
Aug 5, 2019 at 10:42
• You have to understand what $( x, y, z ) = (1, 1, -1) + t(1, -2, -1)$ means. Aug 5, 2019 at 10:43
• It's the vector equation of a line.
– SFR
Aug 5, 2019 at 10:44

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.

• 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.
– SFR
Aug 5, 2019 at 11:00

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 $$\alpha, \beta, \gamma\in \mathbb{R}$$.