Question

Prove the following is the minimum distance between two line in 3D.

\[
l_1(t)=\mathbf{x}t+\mathbf{y}\\
l_2(t)=\mathbf{a}t+\mathbf{b}\\
\vec{l_1l_2}=l_2(u)-l_1(v)\text{ for some }u,v\in \mathbb{R}\\
\mathbf{n}=x\times a\\
\|\mathbf{proj}_\mathbf{n}\,\vec{l_1l_2}\|=\text{minimum distance}
\]

Answer 1

x and a are the direction vectors for l1 and l2 and so x cross a should give the line that is perpendicular to both, ie the *direction* of the shortest distance between them.

be careful, however, as the question is a tad ambiguous.

the min distance will be the projection - via the dot product- of a line joining any point on l1 to any point on l2 (eg using the u and v parameters) BUT it has to be dotted onto the **unit** normal, which is not necessarily the primitive normal you get from x cross a. IOW:
\[\hat n = \frac{\vec x \times \vec a}{|\vec x||\vec a|}\]
\[\min \ dist = \vec l_{12} \ \bullet \hat n\]
not sure how your teacher expects you to present it but that is the wordy proof.

Answer 2

Why \(\|\mathbf{n}\|=1\) is necessary?
\[
\|\mathbf{proj}_\mathbf{n}\,\vec{l_1l_2}\|
=\frac{\mathbf{n}\cdot\vec{l_1l_2}}{\|\mathbf{n}\|^2}\|\mathbf{n}\|
=\frac{\|\mathbf{n}\|\|\vec{l_1l_2}\|\cos(\theta)}{\|\mathbf{n}\|^2}
=\|\vec{l_1l_2}\|\cos(\theta)\\

\|\mathbf{proj}_\mathbf{\hat{n}}\,\vec{l_1l_2}\|
=\frac{\mathbf{\hat{n}}\cdot\vec{l_1l_2}}{\|\mathbf{\hat{n}}\|^2}\|\mathbf{\hat{n}}\|=\|\vec{l_1l_2}\|\cos(\theta)
\]

Answer 3

My question is that why the projection must yield the minimum distance. Why the component of \(\vec{l_1l_2}\) in the direction of \(\mathbf{n}\) must be the same for all \(\vec{l_1l_2}\) and corresponds to the minimum distance?

Answer 4

*

Answer 5

The given works for skewed lines (lines that don't lie in one plane), but consider two parallel lines that lie in same plane for simplicity :

Answer 6

We know that the shortest distance between \(l_1\) and \(l_2\) is the perpendicular distance : suppose AB is the shortest distance and perpendicular to both the lines