Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $s=5$.

share|improve this question

2 Answers 2

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.

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.