# What is wrong with this technique for proving skew lines?

We have the following set of lines: $$L_1: \frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}$$ $$L_2:\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{-7}$$

This leads to the following parametric equations: $$L_1:x=t+2,\space y=-2t+3,\space z=-3t+1$$ $$L_2: x=s+3,\space y=3s-4,\space z=-7s+2$$ The $x$ line looked pretty simple, so I did this: $$t+2=s+3$$$$t=s+1$$

Then I simply substituted this into the Y equations $$-2(s+1)+3=3s-4$$ which yields $$s=5,\space t=6$$ That, in turn, gives me: $$x_1=8,x_2=8 \space z_1=-17,z_2=-38$$ forcing me to conclude the lines are skew.

The lines are not skew - there is an intersection at point $(4,-1,-5)$.

Where is the flaw in my analysis?

I suspect taking $t=s+1$ is insufficiently bounded to use in this system? If so, how do we prove $x$ equations is sufficient in an $n$-dimensional system of lines to still apply to the system?

EDIT: $5=5s$ actually does not mean that $s=5$.

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Your error is in going from

$$-2(s+1)+3=3s-4$$

to

$$s=5,\space t=6$$

I don't know how you got that, but it should be $s=1$, $t=2$. Maybe you dropped a term or misplaced a sign while solving the equation.

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Yes, you'll get $5s = 5$, and then $s = 1$. I think he forgot the $5$ on the LHS. – M. Vinay Jun 11 '14 at 6:21

You didn't solve for $s$ and $t$ correctly. $s=5$ doesn't satisfy your equation $-2(s+1)+3 = 3s-4$.

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