3
$\begingroup$

This one's probably easy, but I'm dreadfully stuck and can't seem to figure out a decent method.

I have the following lines:

$$a: \vec{x}(\lambda)= \left( \begin{array}{ccc} 4 \\ -2 \\ -2 \end{array} \right) + \lambda\left( \begin{array}{ccc} 1 \\ -1 \\ -1 \end{array} \right) $$ $$b: \vec{x}(\mu)= \left( \begin{array}{ccc} -1 \\ 1 \\ -3 \end{array} \right) + \mu\left( \begin{array}{ccc} 1 \\ 0 \\ 2 \end{array} \right) $$ $$c: \vec{x}(\nu)= \left( \begin{array}{ccc} 1 \\ 0 \\ 5 \end{array} \right) + \nu\left( \begin{array}{ccc} 0 \\ -2 \\ 1 \end{array} \right) $$ $$d: \vec{x}(\tau)= \left( \begin{array}{ccc} 3 \\ -2 \\ 0 \end{array} \right) + \tau\left( \begin{array}{ccc} -1 \\ 1 \\ 1 \end{array} \right) $$ I have to find the line that intersects all four of these lines. How do I go about doing this?

Cheers!

$\endgroup$
4
  • $\begingroup$ Hmm, is it true in general that for any four lines in $\mathbb{R}$, there's a line that intersects all four of them? My guess is that it's true. Since any two lines are contained in a pair of (not necessarily distinct) parallel planes, you could get two pairs of parallel planes ... not sure if this could be used to prove the existence of the intersecting line in any way? $\endgroup$ Nov 20, 2014 at 23:55
  • $\begingroup$ @ZubinMukerjee It is decidedly not true if the of the lines are parallel but not coplanar. $\endgroup$
    – Arthur
    Nov 21, 2014 at 0:00
  • $\begingroup$ @Arthur Do you mean skew? I thought parallel implied coplanar. $\endgroup$ Nov 21, 2014 at 0:05
  • 1
    $\begingroup$ @Arthur I think I understand; three of the lines are pairwise parallel but there's no plane that contains all three. $\endgroup$ Nov 21, 2014 at 0:07

1 Answer 1

2
$\begingroup$

Note that $a$ and $d$ are parallel, so whatever line it is, it needs to lie in the plane containing $a$ and $d$. Considering it also needs to intersect $b$ and $c$, try figuring out which two points those two lines intersect the $ad$-plane.

Edit: Here's a full answer.

$ad$-plane: The $ad$-plane's normal vector is orthogonal to $(-1, 1, 1)$ as well as $\vec{a}(0) - \vec{d}(0) = (1, 0, -2)$. We therefore have a normal vector given by $$ (-1, 1, 1)\times (1, 0, -2) \\ = (1\cdot(-2) - 1\cdot 0, 1\cdot 1 - (-1)\cdot (-2), (-1)\cdot 0 - 1\cdot 1)\\ = (-2, -1, -1) $$ I elect to choose the negative of this vector, for estethic reasons.

Inserting $\vec{d}(0)$ into the general equation for a plane, we have: $$ 2\cdot 3 + 1\cdot (-2) + 1 \cdot 0 = 4 $$ and therefore the $ad$-plane is given by $2x + y + z = 4$.

$b$-intersection: The $\mu$ for the point where the $b$-line intersects the $ad$-plane is given by $$ 2(-1 + \mu) + 1 -3 + 2\mu = 4\\ 4\mu = 8\\ \mu = 2 $$ so the intersection point is $B = \vec{b}(2) = (1, 1, 1)$.

$c$-intersection: The $\nu$ for the point where the $c$-line intersects the $ad$-plane is given by $$ 2\cdot 1 -2\nu + 5 + \nu = 4\\ -\nu = -3\\ \nu = 3 $$ so the intersection point is $C = \vec{c}(3) = (1, -6, 8)$.

The line: We need the line $\vec{l}(\gamma)$ that goes from $B$ to $C$. It is given by $$ \vec{l}'(\gamma) = B + \gamma(C - B)\\ = (1, 1, 1) + \gamma(0, -7, 7) $$ which I would like to rewrite to: $$ \vec{l}(\gamma) = (1, 1, 1) + \gamma(0, -1, 1) $$ For reference, the four intersection points are:

  • $al$: $\lambda = -3, \gamma = 0, (1, 1, 1)$
  • $bl$: $\mu = 2, \gamma = 0, (1, 1, 1)$
  • $cl$: $\nu = 3, \gamma = 7, (1, -6, 8)$
  • $dl$: $\tau = 2, \gamma = 1, (1, 0, 2)$
$\endgroup$
6
  • $\begingroup$ What am I supposed to do when I find the points of intersection of b and c with the plane through a and d? How will I know if these points are also contained in lines a and d, and not just their plane? Just by checking for each line? Seems like a lot of work! $\endgroup$
    – Ius Klesar
    Nov 21, 2014 at 0:06
  • 1
    $\begingroup$ Well, confirming that the line goes through the points is easy. By construction it goes through $b$ and $c$, and it's coplanar yet hopefully nonparallel to $a$ and $d$, so minimal calculation on the verification. But yes, getting to the line equation takes some work. $\endgroup$
    – Arthur
    Nov 21, 2014 at 0:15
  • $\begingroup$ Yet hopefully nonparallel to a and d? So pretty much the only verification needed is checking whether the obtained line is parallel to a (or d)? $\endgroup$
    – Ius Klesar
    Nov 21, 2014 at 0:25
  • $\begingroup$ Hey Arthur, I'm not getting the right answer here. I've tried your method, found two points, but the line through these points is not the correct answer. The directional vector of my line should be (0,1,-1), and I'm not getting this out of any of my attempts. $\endgroup$
    – Ius Klesar
    Nov 21, 2014 at 16:28
  • 1
    $\begingroup$ @Luke i now added a full solution to my answer. $\endgroup$
    – Arthur
    Nov 21, 2014 at 23:43

Your Answer

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

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