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I've got the lines' points and vectors $p,q$. My idea was to find a subspace (plane) with the basis of $p,q$ - perpendicular to the lines' axis. Then find the intersecting point $P$ of the lines' projections onto that plane. Then project $P$ onto both lines, get two points $a,b$ and calculate their distance. The problem is that the lines don't go through the origin, which means they're not subspaces and I can't project onto them. Any thoughts?

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First I find the orthogonal complement to the line vectors. Since the lines are skew and we're in $\mathbb{R}^3$, I get the lines' axis. Now I just orthogonally project difference vector of any points on the lines onto the axis. The projected vector's norm is the answer. Any formal explanation why this works appreciated.

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The lines are affine subspaces, let $P+\lambda p$ and $Q+\mu q$.

The direction of the common perpendicular is $r=p\times q$. If you look in any direction perpendicular to $r$, you will see the projections of the two lines as two parallels and their distance is just the orthogonal projection of $PQ$ onto $r$,

$$d=\frac{PQ\cdot(p\times q)}{\|p\times q\|}.$$


You can obtain the same result by minimizing the squared distance vector,

$$d^2=(PQ+\lambda p-\mu q)^2.$$

To minimize, you cancel the derivatives wrt to the parameters,

$$\frac{\partial d^2}{\partial\lambda}=(PQ+\lambda p-\mu q)p=PQ\cdot p+\lambda p^2-\mu p\cdot q=0,\\ \frac{\partial d^2}{\partial\mu}=(PQ+\lambda p-\mu q)q=PQ\cdot q+\lambda p\cdot q-\mu p^2=0$$ and solve for $\lambda,\mu$.

The determinant of this system is $p^2q^2-(p\cdot q)^2$ which you should recognize to be $(p\times q)^2$.

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Using the definition of a norm, the distance between two vectors $d(A,B)=||A-B||$.

Source: http://www.math.vanderbilt.edu/~msapir/msapir/feb5.html

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  • $\begingroup$ i need to find these vectors first... $\endgroup$ – k5f Mar 8 '14 at 17:33
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We are in the affine euclidean space $\mathbb{R}^n$ and we consider two non-parallel lines $D_1=\{P+\lambda u|\lambda\in \mathbb{R}\}$ and $D_2=\{Q+\mu v|\mu\in \mathbb{R}\}$. Then $d(D_1,D_2)=RS$ where $R=P+\lambda u\in D_1,S=Q+\mu v\in D_2$ and $\vec{SR}=P-Q+\lambda u-\mu v$ is perpendicular to $D_1,D_2$.

We obtain the system in the unknowns $\lambda,\mu$: $(P-Q).u+\lambda||u||^2-\mu v.u=0,(P-Q).v+\lambda u.v-\mu ||v||^2=0$. The determinant is $-||u||^2||v||^2+(u.v)^2<0$ and there is a unique solution. Then we obtain $R,S,d(D_1,D_2)$.

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If you have already figured out how to orthogonally project each of the lines onto your chosen subspace, then you can take one of the lines (call it $L$) and a point on that line, and take any point $Q$ on $L.$ Project $Q$ onto the subspace; that gives you a projected point $Q',$ and the vector $v = Q - Q'$ is orthogonal to the subspace. Then $P' = P + v$ is the orthogonal projection of $P$ onto $L.$

Do that once for each line, and then you have your points $a$ and $b.$

On the other hand, the first part of Yves Daoust's answer is equivalent to this (after some algebraic simplification) and constructs a lot fewer intermediate objects on the way to its result, so that's what I'd do.

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