How can I find the straight line incident on 4 given straight lines, all embedded in 3D space?
I'm usually comfortable with linear algebra; my understanding of projective geometry is limited.
(My ultimate goal is to find a probability distribution for the straight line incident on N given straight lines, where N-1 of them are uncertain/fuzzy. This may be an entirely different question; this question is just to get me started.)
(I will be offline for about 20 hours.)
On
I don't think the problem is well defined. If all 4 given lines go through a point O, any other line going through that point is incident on all 4. Similarly, if the 4 lines are co-planar, I can find an infinite number of lines incident on all 4. Now if you have for example 4 parallel lines incident on a plane, and the intersections form a non degenerate quadrilateral, there is no single line that intersects all of them
On
I am assuming you would consider $\mathbb{RP}^3$ and not just $\mathbb{R}^3$ to make things more homogeneous. And I believe, assuming your four lines are in general position (i.e. meaning no two lines intersect in $\mathbb{RP}^3$). Then, generically, it seems like there might be a pair of lines that intersect four lines lines in the projective three-space in general position. However, there also might be one or no solution to this problem.
There are various approaches, some directly geometric, some a bit more sophisticated involving spaces of lines (the Plucker quadric, which is a type of Grassman manifold). I think in your case, it is more instructive to learn a bit about the Plucker quadric. One can do this using the conceptual construction of wedge product and exterior algebra. Consider $\mathbb{R}^4$ and as usual $\mathbb{RP}^3 = \big\{[u] \, : \, u \in \mathbb{R}^4 \big\}$ with $[u] = \big\{\lambda \, u \, \, : \,\, \lambda \in \mathbb{R}\setminus\{0\} \, \big\}$. Projective lines in the projective three space $\mathbb{RP}^3$ correspond to the projectivisation of two-dimensional vector subspaces in $\mathbb{R}^4$. If I fix a two-dimensional subspace in $\mathbb{R}^4$ I can define it as the span of two linearly independent vectors $u,v \in \mathbb{R}^4$ lying on it and I can form the wedge product $q = u \wedge v \, \in \, \wedge^{2} \,\mathbb{R}^4 = \mathbb{R}^4 \wedge \mathbb{R}^4$. Moreover, if $u_1, v_1$ are two different vectors spanning the same subspace as $u, v$, then there is $\lambda \in \mathbb{R}$ such that $u_1\wedge v_1 = \lambda (u \wedge v)$. This prompts us to projectivize the wedge product: $$\mathbb{P}(\wedge^{2} \,\mathbb{R}^4 ) = \big\{ [\omega] \,\, : \,\, \omega \in \wedge^2 \, \mathbb{R}^4\big\} $$ $$[\omega] = \big\{\lambda \, \omega \in \wedge^{2} \,\mathbb{R}^4 \,\, : \,\, \lambda \in \mathbb{R}\setminus\{0\}\big\}$$ and the elements that describe the lines in $\mathbb{RP}^3$, which is the same as the two dimensional subspaces of $\mathbb{R}^4$, have the form $[u \wedge v]$. However, not all elements of $\mathbb{P}(\wedge^{2} \,\mathbb{R}^4 ) $ are of this form. How to distinguish the ones that define lines from the one that do not, i.e. when is an element of $\wedge^{2} \,\mathbb{R}^4$ of the form $u \wedge v$? As it turns out, $\omega \in \wedge^{2} \,\mathbb{R}^4$ has the form $\omega = u \wedge v$ if and only if $\omega \wedge \omega = 0$, where $\omega \wedge \omega \in \wedge^{4} \,\mathbb{R}^4$. Since there exists an isomorphism (non-canonical) $\phi : \wedge^{4} \,\mathbb{R}^4 \to \mathbb{R}$, we can define the complex bilinear dot product (nondegenerate bilinear form) $(\omega \cdot \sigma) = \phi(\omega \wedge \sigma)$ for $\omega, \sigma \in \wedge^2 \, \mathbb{R}^4$. Then, this dot product has two very important properties:
1. $(\omega \cdot \omega) = \phi(\omega \wedge \omega) = 0$ if and only if $\omega \wedge \omega = 0$ if and only if $\omega=u \wedge v$;
2. If two lines in $\mathbb{RP}^3$ intersect at a common point, then their corresponding two two-dimensional subspaces of $\mathbb{R}^4$ intersect at a common one dimensional subspace, spanned by a vector $v_0$, so the two lines will be represented by two wedge products $q_1 = u_1 \wedge v_0$ and $q_2 = u_2 \wedge v_0$ and so $(q_1 \cdot q_2) = \phi(q_1 \wedge q_2) = \phi(0) = 0$. The converse is also true, if two wedge products $q_1 = u_1 \wedge v_1$ and $q_2 = u_2 \wedge v_2$ are such that $(q_1 \cdot q_2) = \phi(q_1 \wedge q_2) = \phi(0) = 0$, then there is at least one vector $v_0 \in \mathbb{R}^4$ such that $q_1 = \lambda_1 (w_1 \wedge v_0)$ and $q_2 = \lambda _2 (w_2 \wedge v_0)$.
With this information at hand, define the quadric $$\hat{Q} = \big\{\omega \in \wedge^2 \, \mathbb{R}^4 \,\, : \,\, (\omega \cdot \omega) = 0 \big\}.$$ Then its projectivization $$Q =\big\{\, [\omega] \in \mathbb{P}(\wedge^2 \, \mathbb{R}^4) \,\, : \,\, \omega \in \hat{Q} \, \big\} = \big\{\, [\omega] \in \mathbb{P}(\wedge^2 \, \mathbb{R}^4) \,\, : \,\, (\omega \cdot \omega) = 0 \, \big\}.$$ Thus $\mathbb{G}(1,3) \cong Q$. If you have a plane $H$ in $\mathbb{RP}^3$ and a point on it $p \in H$ one can pick two different lines $l_1$ and $l_2$ lying on $H$ and passing through $p$. Then $H$ is spanned by these two lines. Let $l_1 = [u_1 \wedge v_0]$ and $l_2 = [u_2 \wedge v_0]$, where $p = [v_0]$. Denote $\hat{l}_1 = u_1 \wedge v_0$ and $\hat{l}_2 = u_2 \wedge v_0$. Then any other line on $H$ passing through $p$ should look like $$l = [\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2] = [\lambda_1 u_1 \wedge v_0 + \lambda_2 u_2 \wedge v_0] = [(\lambda_1 u_1 + \lambda_2 u_2) \wedge v_0]$$ for any $\lambda_1, \lambda_2 \in \mathbb{R}$. Thus the set $\Sigma_{p,H}$ of all lines lying on $H$ and passing through $p$ is the projectivisation of a two-dimensional subspace of $\wedge^2 \, \mathbb{R}^4$ so it is a one-dimensional line on $\mathbb{P}(\wedge^2 \, \mathbb{R}^4)$ composed of elements $l = [\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2] \in \Sigma_{p,H}$ with the property that $$\big((\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2)\cdot (\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2) \big) = 0$$ which means that $l \in Q \cong \mathbb{G}(1,3)$ and hence $\Sigma_{p,H} \subset Q \cong \mathbb{G}(1,3)$. Observe that $\dim_{\mathbb{R}} \big(\wedge^2 \, \mathbb{R}^4\big) = 6$, so $\wedge^2 \, \mathbb{R}^4 \cong \mathbb{R}^6$ and so $\mathbb{P}(\wedge^2 \, \mathbb{R}^4) \cong \mathbb{RP}^5$ and thus $Q = \mathbb{G}(1,3)$ is a four dimensional projective quadric in $\mathbb{RP}^5$. The signature of the quadratic form $(\sigma \cdot \omega)$ is $+,+,+,-,-,-$ so technically, $Q = \mathbb{G}(1,3)$ lies in the projectivizaiton of $\mathbb{R}^{3,3}$.
The converse. Let $\Sigma$ be a line lying on the conic $Q$. Then there are two different points $[l_1]$ and $[l_2]$ lying on $\Sigma$, where $l_i = u_i\wedge v_i$ for $i=1,2$. So these two points span $\Sigma$, i.e. $[l] \in \Sigma$ if and only if $[l] = [\lambda \, l_1 + \lambda_2 \, l_2] \, \in \, \Sigma \, \subset \, Q$ for any complex numbers $\lambda_1, \lambda_2 \in \mathbb{R}$ not simultaneously zero. Then since $[l] \in \Sigma$ and $\Sigma \subset Q$, $$0 = (l \cdot l) = \big((\lambda \, l_1 + \lambda_2 \, l_2) \cdot(\lambda \, l_1 + \lambda_2 \, l_2)\big) = \lambda_1^2 (l_1 \cdot l_1) + 2\lambda_1 \lambda_2 (l_1 \cdot l_2) + \lambda_2^2 (l_2 \cdot l_2) = 2\lambda_1 \lambda_2 (l_1 \cdot l_2)$$ because $(l_1\cdot l_1) = (l_2 \cdot l_2) = 0$ as $[l_1], [l_2]$ are points lying on the quadric $Q$. Hence $(l_1 \cdot l_2) = 0$. But this means that $\phi(l_1 \wedge l_2)=0$ i.e. $u_1 \wedge v_1 \wedge u_2 \wedge v_2 = 0 \in \wedge^4 \, \mathbb{R}^4$. The latter statement means that the four vectors $u_1, v_1, u_2, v_2 \in \mathbb{R}^4$ cannot span the whole space and are linearly dependent. Hence there is a vector $v_0 \in \mathbb{R}^4$ such that $$\text{span}\{u_1, v_1\} \cap \text{span}\{u_2, v_2\} = \text{span}\{v_0\}$$ Hence $l_1 = [u_1 \wedge v_0]$ and $l_2 = [u_2 \wedge v_0]$ and $[\lambda_1 \, l_1 + \lambda_2\, l_2] = [\,(\lambda_1 v_1 + \lambda_2 u_2) \wedge v_0]$ represents the two-space $\text{span}\{\lambda_1 v_1 + \lambda_2 u_2, v_0\}$ in $\mathbb{R}^4$ which means that all the lines from $\Sigma$ pass through the same point $p = [v_0] \in \mathbb{RP}^3$ all of them are contained in the plane $$H = \mathbb{P}\big(\text{span}\{u_1, u_2, v_0\}\big) \, \subset \, \mathbb{RP}^3$$
Now, after this introduction to the construction and the structure of the Plucker quadric, let us identify $\wedge^2 \, \mathbb{R}^4$ together with the bilinear form $\phi(\omega \wedge \sigma)$ with $\mathbb{R}^6$ with a bilinear form $(\omega \cdot \sigma)$ of signature $+,+,+,-,-,-$. This is possible as discussed above. Now, if we pick four lines $l_1, \, l_2, l_3$ and $l_4$ from $\mathbb{RP}^3$ in general position (no two intersect), then this is equivalent to selecting the corresponding four points $l_1, \, l_2, l_3$ and $l_4$ on the four dimensional projective quadric $Q \, \subset \, \mathbb{RP}^5$. These four points $l_1, \, l_2, l_3$ and $l_4$ span a three dimensional projective subspace $L$ of $\mathbb{RP}^5$ so that $L \cap Q$ is, in general, a two dimensional projective surface (a two dimensional quadric) lying inside $Q$. Then the polar space $L^*$ of $L$ with respect to the quadric $Q$ is a one dimensional projective line $L^*$ which could intersects the four dimensional quadric $Q$ at two points lying on $Q$. The intersection is nonempty (generlic) depending on the general position of the original four lines, which determines the signature of the spaces $L$ and $L^*$. Based on this signature, one can determine how many solutions the problem could have. The claim is that any of the points $l \in L^* \cap Q$ is a line in $\mathbb{RP}^3$ that intersects the four lines $l_1, l_2, l_3$ and $l_4$.
To understand this from another point of view, look at the pre-projective space $\mathbb{R}^6$. Then the four lines $l_1, \, l_2, l_3$ and $l_4$ of $\mathbb{RP}^3$ give rise to four one dimensional vector subspaces $\hat{l}_1, \, \hat{l}_2, \hat{l}_3$ and $\hat{l}_4$ of $\mathbb{R}^6$. The span $\hat{L}$ of $\hat{l}_1, \, \hat{l}_2, \hat{l}_3$ and $\hat{l}_4$ is a four dimensional vector subspace of $\mathbb{R}^6$. Then it's orthogonal complement $\hat{L}^*$ in $\mathbb{R}^6$ with respect to the bilinear form $(\omega \cdot \sigma)$ is the two dimensional vector subspace $$\hat{L}^* = \{ \, \sigma \in \mathbb{R}^6 \,\, | \,\, (\sigma \cdot \omega) = 0 \, \text{ for all } \omega \in \hat{L} \, \}$$ After projectivization, this construction of orthogonality becomes polarity with respect to $Q$. Recall that the five dimensional cone $$\hat{Q} = \{ \, \omega \in \mathbb{R}^6 \,\, | \,\, (\omega \cdot \omega) = 0 \, \}$$ is what becomes the four dimensional quadric $Q$ after projectivization. Moreover, $l \in Q \cap L^*$ is equivalent to $\hat{l} \in \hat{Q} \cap \hat{L}^*$ so for that reason $(\hat{l} \cdot \hat{l}_j) = 0$ for all $j=1,2,3,4$. Here I have abused notation, thinking of $\hat{l}$ and $\hat{l}_j$ as nonzero vectors, spanning the actual one dimensional subspaces $\hat{l}$ and $\hat{l}_j$ of $\mathbb{R}^6$. By construction, two lines $l$ and $l_j$ from $\mathbb{RP}^3$ intersect if and only if their representatives $l$ and $l_j$ from the Plucker quadric $Q$ satisfy $(\hat{l} \cdot \hat{l}_j) = 0$.
In your case, later you could consider introducing a probability measure on $Q$, which will allow you to write down probabilities of line incidences.
The other post got too long and the LaTeX starts to lag.
One can also reason geometrically, directly in $\mathbb{R}^3$ and $\mathbb{RP}^3$. I am going to assume that $\mathbb{RP}^3$ = $\mathbb{R}^3 \, \cup P_{\infty}$ where $P_{\infty}$ is the plane at infinity. Let $l_1, \, l_2, \, l_3$ and $l_4$ be the four lines such that no two intersect (or are parallel). Take three of them, say $l_1, \, l_2, \, l_3$. Then there is a unique quadratic surface $H$ (aka quadric) in $\mathbb{R}^3$ (and in fact in $\mathbb{RP}^3$) such that all three lines $l_1, \, l_2, \, l_3$ lie on that quadric $H$. In fact, $H$ is a doubly ruled surfaces, i.e. there is one family of non intersecting lines (call them $l_{\alpha} \, : \, \alpha$) and another family of non intersecting lines (call them $m_{\beta} \, : \, \beta$). Any line $l_{\alpha}$ from the first family intersects any line $m_{\beta}$ from the second family. The three lines $l_1, \, l_2, \, l_3$ are members of the first family $l_{\alpha}$ (they determine it). Think of $H$ as a one sheeted hyperboloid (up to projective transformation) or a hyperbolic paraboloid if you wish (wiki these and you will see what I am talking about).
Now, if you know the three lines in coordinates, you will be able to write explicitly the quadratic equation for $H$.
Assume you have a line $m$ that intersects all three lines $l_1, \, l_2, \, l_3$. Since $m$ intersects $l_1$ and $l_2$, where $l_1 \cap l_2 = \varnothing$, then $m$ intersects $H$ at two different points. Recall that a line and a qadratic surface can intersect at no more than two points, unless the line lies completely on the surface. However, $m$ intersects also line $l_3$ which is disjoint from the other two lines, so $m$ has a third point of intersection with $H$ so $m$ must lie on $H$. Consequently, $m$ must belong to one of the two families of lines on $H$ and since all lines from $l_{\alpha}$ are disjoint, $m$ is a member of the other family $m_{\beta}$.
Finally, one concludes that if there is a line $m$ that intersects all four lines $l_1, \, l_2, \, l_3$ and $l_4$ it must be a line lying on $H$ so the forth line $l_4$ must have a common point with $H$. The converse is also true, if $l_4$ intersects $H$ at a point $P$, then one can take the unique line $m$ from family $m_{\beta} $ that passes through point $P$. Then $m$ also would intersect the other three lines $l_1, \, l_2, \, l_3$.
Thus, the solution to your problem boils down to construct the equation of the quadric $H$, generated by the three lines $l_1, \, l_2, \, l_3$, and to check how many intersection points has $l_4$ with $H$.
1. If there are two (which is max if all lines are in general position, i.e. if $l_4$ is not a member of $l_{\alpha}$, in which case there would be infinitely many solutions), then there are two lines $m_1$ and $m_2$ that intersect all four lines $l_1, \, l_2, \, l_3$ and $l_4$.
2. If $l_4$ is tangent to $H$, then there is exactly one line $m$ that intersects all four lines $l_1, \, l_2, \, l_3$ and $l_4$.
3. If $l_4$ doesn't intersect $H$ at all, then there is no solution.