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I have two skew lines in $\mathbb{R}^N$ ($N > 2$) defined as $\vec{x} = \vec{x}_A + \vec{d}_A t$ and $\vec{x} = \vec{x}_B + \vec{d}_B s$ ($t, s \in \mathbb{R}$). Now, I'd like to calculate the shortest distance between those lines. In 3D, this seems to be rather simple since the cross product $[\vec{d}_A \times \vec{d}_B]$ is a vector. However, in $\mathbb{R}^N$, there is an infinite number of vectors that are perpendicular to $\vec{d}_A$ and $\vec{d}_B$ and that lie on a subset $H^{\perp}$ of dimension $N - 2$.

My question is: How can one calculate the minimal distance without generalizing the cross product to $N$ dimensions?

1 Answers 1

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You could compute the minimum of

$$ d(s,t)=\Vert(\vec x_A+\vec d_At)-(\vec x_B+\vec d_Bs)\Vert=\Vert(\vec x_A-\vec x_B)+\vec d_At-\vec d_Bs\Vert $$

using basic analysis. In more detail: the above gives you a function $\mathbb R^2\rightarrow \mathbb R$. Compute its gradient, and look for zeros.

Hint: Even easier, use $d(s,t)^2$.

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    Thanks for your answer! I though of minimizing $d(s, t)^2$ too. This would give me two points $P_A$ and $P_B$ that lie on either line, and the vector connecting the points $\vec{v}_{AB}$ would give the direction. However, I still have a problem with the uniqueness of $\vec{v}_{AB}$. It's true that $|\vec{v}_{AB}|$ is the minimal distant I am after, but how many vectors in $H^{\perp}$ would have the same distance?.. Is my logic correct? – 2017-02-20
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    I'm not sure what you mean with "vectors in $H^\bot$ having a _distance_". Only the vector $\vec v_{AB}$ gives you the shortest connection between the lines. Of course, there are other vectors perpendicular to $\vec d_A$ and $\vec d_B$ at the same time, but if they are not a solution to your minimization, they cannot point from one line to the other while being also short as $\vec v_{AB}$. – 2017-02-20
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    Yes, this was exactly what I meant. In this sense, $\vec{v}_{AB}$ must be unique since it's a) perpendicular to $\vec{d}_A$ and $\vec{d}_B$ and b) points from one line to the other. Other vectors from $H^{\perp}$ satisfy only a). Thanks for clarifying. – 2017-02-20