Simple but Stuck: How do I find the point of intersection of two lines in Vector Calculus? Given symmetric equations and deriving parametric equations.

Find the point of intersection of the lines L1: $\frac{x}{2} =\frac{y+1}{3} = \frac{z-1}{2}$ and L2: $\frac{x}{4} = \frac{y}{5} =\frac{z}{5}$

The parametric forms should be $x_1 = 2t$, $y_1 = -1+3t$, $z_1=1+2t$ and $x_2=4t$, $y_2=5t$, $z_2 = 5t$

The final solution is $(4, 5, 5)$. It almost looks like someone just grabbed the coefficients from the parametric form of the second line to make the point. Working backwards, how do I get this answer?

In need of the process.
Thank you very much.


Let the parameter of the second line be $s$ and solve $2t=4s, -1+3t=5s, 1+2t=5s$. If they are not consistent, there is no intersection.

  • $\begingroup$ Forgive me if this is a seemingly uninspired question, but do I solve for s or do I solve for t? If I plug-in t=2s, I end up with s = 1 in each equation. Though, I'm still at a loss on how to get to (4, 5, 5). It's late, and I'm probably missing a logical step that is right under my nose. $\endgroup$ – user7140 Feb 16 '11 at 6:18
  • 2
    $\begingroup$ @Vector-Calculus Student: As you get s=1, plug that value into the equations for x, y, and z (L2) that you parametrized for s. That will give you the point at which the lines intersect (4, 5, 5). $\endgroup$ – InterestedGuest Feb 16 '11 at 6:29
  • $\begingroup$ I knew it -- something right under my nose. Thank you Ross and InterestedQuest. Though, why s and L2? Could I not solve for t and plug those values into L1, but I would get a different (presumably inaccurate) result? $\endgroup$ – user7140 Feb 16 '11 at 6:36
  • 2
    $\begingroup$ @Vector-Calculus Student: You could, indeed, and the result would have to be the same, since they intersect at that point. In fact, given Ross' equations, let's solve for $t$. Given $2t=4s$, and, consequently, $s=.5t$, plug that value into the other two equations. You will get $t=2$ in both equations (assuming correct arithmetic). Then, plug this value of $t$ into your original L1 equation -- you will get precisely the point (4, 5, 5). $\endgroup$ – InterestedGuest Feb 16 '11 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy