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Is it always possible to find a line perpendicular to two skew lines in space? And how can we visualise the proof geometrically? And if anyone could present the proof that it is always possible to exist a line perpendicular to both skew lines, please elaborate.

Gerry Myerson
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Jyotishraj Thoudam
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    Given two skew lines, you can always rotate space so both are horizontal, and then it's pretty clear that there's a vertical line that goes through both of them. – Gerry Myerson Aug 07 '16 at 11:12
  • How do I rotate space? – Jyotishraj Thoudam Aug 07 '16 at 11:19
  • OK, don't rotate space, just rotate the two lines until one of them is horizontal, then rotate the second line around the first one until it, too, is horizontal. – Gerry Myerson Aug 07 '16 at 12:24
  • But wouldn't that change the equation in the first place? – Jyotishraj Thoudam Aug 07 '16 at 15:02
  • What equation? Jyo, you asked, "how can we visualize the proof geometrically?" I have given you geometry, not equations. You imagine the two lines fixed in their relationship to each other, but free to move in space. You visualize moving them until they are both horizontal, at which point it's easy to see that there's a vertical line skewering the skew lines. No equations – geometry! – Gerry Myerson Aug 07 '16 at 22:48
  • David I tried. But you could supplement with another concept. @Gerry I am almost convinced. Are you saying that we should rotate the line so that their skewness is invariant? – Jyotishraj Thoudam Aug 08 '16 at 02:30
  • I was mistaken in my earlier comment (which I have deleted); the other questions of which I was thinking were concerned with the shortest distance between skew lines, which occurs along a mutually perpendicular line, but none of them was particularly concerned with the geometric visualization of that line. I think this question deserves its own answers. – David K Aug 08 '16 at 02:36
  • I don't know what you mean by "their skewness is invariant". Imagine space to be a lump of plastic, and the two lines to be wires embedded therein. Rotate the plastic so one line is horizontal, then rotate using that horizontal line as an axis until the other line is also horizontal. – Gerry Myerson Aug 08 '16 at 02:39

5 Answers5

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Here's an attempt at a purely geometric approach.

Label the two lines $\ell_1$ and $\ell_2$. Select any point $A$ on line $\ell_1$. Construct line $\ell_3$ through $A$ parallel to $\ell_2$. Then the plane $\pi_1$ containing the lines $\ell_1$ and $\ell_3$ is parallel to the line $\ell_2$.

Now find the perpendicular projection $\ell_2'$ of the line $\ell_2$ on the plane $\pi_1$. (One way to do this is to pick any two distinct points $M$ and $N$ on $\ell_2$, find the points $M'$ and $N'$ in plane $\pi_1$ closest to $M$ and $N$, respectively, and construct the line $\ell_2'$ through $M'$ and $N'$.) Let $P$ be the point where the lines $\ell_1$ and $\ell_2'$ intersect.

By similar methods, find the plane $\pi_2$ through line $\ell_2$ parallel to line $\ell_1$, find the projection $\ell_1'$ of the line $\ell_1$ on the plane $\pi_2$, and let $Q$ be the intersection point of the lines $\ell_1'$ and $\ell_2$.

Now planes $\pi_1$ and $\pi_2$ are parallel, lines $\ell_1$ and $\ell_1'$ are perpendicular projections of each other on planes $\pi_1$ and $\pi_2$, and lines $\ell_2$ and $\ell_2'$ are perpendicular projections of each other on planes $\pi_2$ and $\pi_1$, respectively.

In particular, $P$ (the intersection of lines $\ell_1$ and $\ell_2'$) is the perpendicular projection of $Q$ (the intersection of lines $\ell_1'$ and $\ell_2$) on plane $\pi_1$, and likewise $Q$ is the perpendicular projection of $P$ on plane $\pi_2$. The line $PQ$ is perpendicular to both planes $\pi_1$ and $\pi_2$ and to the lines $\ell_1$ and $\ell_2$ in those planes; that is, $PQ$ is the line that was to be found.

David K
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  • $l_1$ and $l_3$ cannot be in the same plane can it? That would defy the skewness – Jyotishraj Thoudam Aug 08 '16 at 03:07
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    The skew lines are $\ell_1$ and $\ell_2$. The line $\ell_3$ is a new, third line. Even though we have two lines that are skew, that does not imply that every other line in space must be skew to either of them. – David K Aug 08 '16 at 03:30
  • I think I got some part. Will update my understanding – Jyotishraj Thoudam Aug 08 '16 at 05:40
  • I had a similar question as the OP, and I'm trying to follow your answer... How is $l'_2$ constructed? In parentheses you only get to $l_4$ (which never gets used again), but never actually say to construct $l'_2$ itself. I don't mean this as a demeaning criticism, but I'm having a hard time following your answer. – Fine Man Sep 14 '16 at 17:28
  • @SirJony I think what happened is that originally I started writing this answer using the symbol $\ell_4$ for the line parallel to $\ell_2$, and later decided it made more sense to call that line $\ell_2'$ since it was a projection of $\ell_2$. I must have neglected to change one of the occurrences of the symbol $\ell_4$ to $\ell_2'$. I have edited the answer to use the symbol $\ell_2'$ consistently; the construction "One way to do this" gives $\ell_2'$ (not some other line named $\ell_4$). – David K Sep 14 '16 at 18:09
  • @SirJony Thanks for pointing out the strange use of the symbol $\ell_4$, by the way; it was a mistake that could have remained there for a long time if you had not noticed it. – David K Sep 14 '16 at 18:13
  • @DavidK : I assumed that was the case. :) – Fine Man Sep 14 '16 at 20:41
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Consider two skew lines: $$l_1=\mathbf{a}_1+p\mathbf{v}_1$$ $$l_2=\mathbf{a}_2+q\mathbf{v}_2$$ where $p,q\in\mathbb{R}$. The cross product of the direction vectors, $\mathbf{v}_3=\mathbf{v}_1\times\mathbf{v}_2$, is by definition perpendicular to both $l_1$ and $l_2$. Now find a point each on $l_1$ and $l_2$ such that the line between them is parallel to $\mathbf{v}_3$, or $\alpha,\beta,\gamma$ such that $$\mathbf{a}_1+\alpha\mathbf{v}_1+\gamma\mathbf{v}_3=\mathbf{a}_2+\beta \mathbf{v}_2$$ Since this is a system of three linear equations with three unknowns, it has a unique solution. The line perpendicular to both skew lines is then $\mathbf{a}_1+\alpha\mathbf{v}_1+r\mathbf{v}_3$ (or $\mathbf{a}_2+\beta\mathbf{v}_2+r\mathbf{v}_3$), $r\in\mathbb{R}$.

Parcly Taxel
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Think about how you would find the shortest distance between the two lines. You know that the shortest distance from a line to a point not on the line is along a segment that’s perpendicular to the line. By symmetry, this means that the shortest distance between two skew lines must be along a segment that’s perpendicular to both of them.

amd
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  • wouldn't we see the other line as a projection instead? – Jyotishraj Thoudam Aug 08 '16 at 02:32
  • @jyotishrajthoudam There’s no need to involve projections. This is straightforward geometry: you drop a perpendicular to the line $\ell_2$ from a point $P_1$ on $\ell_1$ that intersects $\ell_2$ at $P_2$ and measure the length of $\overline{P_1P_2}$. Where that’s minimal, $\overline{P_1P_2}$ must also be perpendicular to $\ell_1$. – amd Aug 08 '16 at 03:57
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If you like to use visualization and manipulatives, hold a pair of pencils as your two skew lines. Pick any random orientation for each pencil. Visualize a line segment that represents the closest approach between the pencils.

Is the closest approach path perpendicular to both pencils? Keeping the two pencils in a fixed juxtaposition to each other, rotate the ensemble so that both pencils are level to the ground. This special case of skew lines demonstrates that the skew lines lie in two parallel planes, and the closest approach runs perpendicular to these planes.

Since angles are preserved under 3D rotation, this same relationship where the closest approach runs perpendicular to both skew lines must apply to all pairs of skew lines, since they are just 3D rotations of the special case.

pbierre
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You can do better than to just arrive at the separation distance of two skew lines at closest approach. With some software computing, you can solve for the specific line segment bridging between them (located by its endpoints):

This approach takes advantage of extrusion.

We can obtain the 3D run direction of the bridging line segment LS by taking the normalized cross-product of the run-dirs of the two lines:

dir-LS <-- (dir-L1 X dir-L2) norm (mutual orthogonal direction to d1 and d2)

Using dir-LS, extrude L1 into a plane PL1. All you need to do is generate 2 points on L1, L1p1 and L1p2, and then obtain a 3rd point p3 = L1p1 + dir-LS. These 3 points spawn the plane PL1 extruding L1 in direction LS-dir:

PL1 <-- extrude( L1, LS-dir )

The point where L2 pierces through PL1 gives endPt2 of the bridge LS:

LS.endPt2 <-- intersectionOf ( L2, PL1 )

By symmetry, we can compute the location of LS.endPt1:

PL2 <-- extrude( L2, LS-dir )

The point where L1 pierces through PL2 gives endPt1 of the bridge LS:

LS.endPt1 <-- intersectionOf ( L1, PL2 )

With both endpoints, we've solved for the bridging line segment of the 2 skew lines. (The length of LS gives the separation distance.)

pbierre
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