I'm doing some practice problems and have the following question which I am able to produce a 'correct' result for however I believe my reasoning is off.

Question: Show that $L_1$ and $L_2$ are equations of the same line.

$L_1: (x,y,z)=(3,1,-7)+\lambda(2,2,-1), \qquad L_2: 2(6-x)=2(4-y)=2(2z+17)$


Rewrite $L_2$ as $2x+2y+8z=-48$.

I can then say $(2,2,-1)$ is parallel to $L_1$ and $(2,2,8)$ is perpendicular to $L_2$. Equate the dot product to zero, showing the lines are parallel, then input $(3,1,-7)$ into my rewritten $L_2$ to show the point on $L_1$ also falls on $L_2$, hence they are the same line.

Is this an incorrect approach? My issue is that my rewritten $L_2$ appears to be an equation for a plane which $L_2$ lies completely within, and then I'm selecting the only vector which is perpendicular to the plane, even though it would be one of infinite vectors perpendicular to the line. Or would any plane that $L_2$ lies completely within have a normal vector perpendicular to $L_2$ which covers my concern?


Yes, your "reformulation" of $L_2$ is a plane, hence it is not correct.


show that $L_2$ indeed is a line.

After which, notice that $2$ points on a line fully determine the line. Hence find two points on $L_1$ and verify that it satisfies $L_2$.

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  • $\begingroup$ Thanks, would I essentially be trying to write $L_2$ in vector form to show parallel components are the same (as first step)? I'm just missing something when converting between forms which is where my issue arose $\endgroup$ – user497663 Oct 31 '17 at 5:24
  • $\begingroup$ $L_2$ is the intersection of two non-parallel planes. The two non-parallel planes are $6-x=4-y$ and $6-x=2z+17$. Hence it is a line. We do not need to write it out explicitly isn't it. We just have to find two points on $L_1$ and substitute it to $L_2$. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 5:28
  • $\begingroup$ Understood, thanks again $\endgroup$ – user497663 Oct 31 '17 at 5:29

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