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.

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: x/2 = (y+1)/3 = (z-1)/2 and L2: x/4 = y/5 = 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.

share|improve this question

1 Answer 1

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.

share|improve this answer
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. –  user7140 Feb 16 '11 at 6:18
@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). –  InterestedGuest Feb 16 '11 at 6:29
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? –  user7140 Feb 16 '11 at 6:36
@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). –  InterestedGuest Feb 16 '11 at 6:43

Your Answer


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