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.
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$\begingroup$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.
$\endgroup$ 7 $\begingroup$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 $\mathbf{a}_2+\beta\mathbf{v}_2+r\mathbf{v}_3$), $r\in\mathbb{R}$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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.)
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