Identify the hyperboloid of one sheet that contains three skew lines on its surface

Question

I claim that three skew lines define a unique hyperboloid of one sheet that contains all of the three lines on its surface.

Suppose you are given three lines in parametric form in $3D$, described as follows

$r_i(t) = P_i + t\ d_i ,\ t \in \mathbb{R},\ i = 1, 2, 3 $

where $P_i$ is a point on the $i$-th line and $d_i$ is the direction vector of the $i$-th line.

Find the equation of the hyperboloid of one sheet that contains all three lines on its surface.

My attempt:

My attempt at this problem is contained in my solution that follows.

My question:

Is it true that three skew-lines define a unique hyperboloid of one sheet that contains them on its surface ? Any hints, remarks, and alternative solutions are appreciated.

Answer 1

I quote "Geometry and the imagination" pp 14-15, Hilbert, David, 1862-1943, author; Cohn-Vossen, S. (Stephan), 1902-1936, author; Nemenyi, P., translator

... the general hyperboloid of one sheet contains two families of straight lines, since a dilation always transforms straight lines into straight lines. Again the lines are arranged in such a way that every line of one family has a point in common with every line of the other family and any two lines of the same family are skew.

This gives rise to the following construction of the hyperboloid of one sheet (see Fig. 21). We start with any three straight lines of one family. Since no two of them have a plane in common, every point $P$ on one of them is on one and only one straight line $p$ meeting the other two given lines, namely the intersection of the plane containing $P$ and the second line with the plane containing $P$ and the third line. $p$ has three points in common with the hyperboloid. But no straight line can intersect a quadric in more than two points. Consequently $p$ must be one of the lines on the hyperboloid. If the point $P$ traverses the first line, the corresponding line $p$ will take on the positions of all the straight lines of that family of the hyperboloid to which the first line does not belong. If we choose any three straight lines of this family, we can get the other family by the same procedure, and of course this will include the three lines with which we started originally. The construction shows that every pair of straight lines of the same family must be skew, provided it is at all possible to find three non-coplanar straight lines in one of the families. For, if we could find three straight lines with which to carry out the construction, while $p$ and $p'$ were to meet at a point $Q$, (Fig. 21), then the original lines would all have to lie in the plane $PP'Q$, in contradiction to one of the assumptions. On the other hand, it is clear that our surface would be not a hyperboloid but a plane, if three lines of the same family always turned out to be coplanar.

Thus three skew straight lines always define a hyperboloid of one sheet, except in the case where they are all parallel to one plane (but not to each other). In this case they determine a new type of second-order surface, called the hyperbolic paraboloid, which does not include any surface of revolution as a special case. ...

Answer 2

The equation of the hyperboloid to be identified is of the form

$ (r - r_0)^T Q (r - r_0) = 1 \hspace{15pt}(1) $

where $r$ is the position coordinate vector of a point on the hyperboloid, $r = [x, y, z]^T $, and $r_0 = [x_0, y_0, z_0] $ is the center of the hyperboloid, and symmetric $3 \times 3 $ matrix $Q$ has two positive eigenvalues and one negative eigenvalue.

Setting $r = r_i(t) = P_i + t d_i $ then we have

$ ( P_i - r_0 + t d_i )^T Q (P_i - r_0 + t d_i) = 1 \hspace{5pt} \forall \ t \in \mathbb{R},\ i=1,2,3 \hspace{15pt} (2) $

Since $P_i$ is on the line, and the line is contained in the hyperboloid, then $P_i$ satisfies its equation, i.e.

$ (P_i - r_0)^T Q (P_i - r_0) = 1 \hspace{15pt} (3) $

This reduces the equation $(2)$ to

$ t^2 d_i^T Q d_i + 2 t d_i^T Q (P_i - r_0) = 0 \hspace{15pt} (4)$

Dividing by $t$

$ t d_i^T Q d_i + 2 d_i^T Q (P_i - r_0) = 0 \hspace{15pt} (5)$

Since this is an identity that holds for any $t$ then we must have

$ d_i^T Q (P_i - r_0) = 0 \hspace{15pt} (6)$

and

$ d_i^T Q d_i = 0 \hspace{15pt} (7)$

Now matrix $Q$ is symmetric, and is therefore determined by $6$ unknowns,

$ Q = \begin{bmatrix} a && d && e \\ d && b && f \\ e && f && c \end{bmatrix} \hspace{15pt} (8) $

From equation $(7), (8)$ we can write $3$ linear homogeneous equations in the variables $a,b,c,d,e,f$. The solution will give result in matrix $Q$ being expressed in terms of three basis matrices $Q_1, Q_2, Q_3$ that are determined by solving the $3 \times 6$ homogenous system.

i.e.

$ Q = \alpha Q_1 + \beta Q_2 + \gamma Q_3 $

where $Q_1, Q_2, Q_3 $ are now known, and $\alpha, \beta, \gamma $ are yet to be determined.

Now, from equation (6),

$ d_i^T Q P_i = d_i^T Q r_0 \hspace{15pt} (9)$

Since the three given lines are skew, then matrix $[d_1, d_2, d_3]$ spans $\mathbf{R}^3$, therefore, we can express $ Q r_0$ as

$ Q r_0 = V u $

where $ V = [d_1, d_2, d_3] $

Let $ \Omega = [\alpha, \beta, \gamma]^T$. Now equation (9) becomes

$ [ d_i^T Q_1 P_i , d_i^T Q_2 P_i, d_i^T Q_3 P_i ]^T \Omega = d_i^T V u , i = 1,2,3 \hspace{15pt} (10) $

Iterating equation $(10)$ for $i=1,2,3$ gives the matrix-vector equation

$ F \Omega = G u $

where $F_{ij} = d_i^T Q_j P_i $ and $G = V^T V $

From this, it follows that

$ u = G^{-1} F \Omega $

Hence,

$ Q r_0 = V G^{-1} F \Omega \hspace{15pt}(11)$

Now back to equation (3),

$ (P_i - r_0)^T Q (P_i - r_0) = 1, i = 1, 2, 3 \hspace{15pt} (3) $

Subtracting pairs of equations corresponding to $i=1,2$ and $i=1,3$ gives

$ P_1^T Q P_1 - P_2^T Q P_2 - 2 (P_1 - P_2)^T Q r_0 = 0 \hspace{15pt} (12) $

and

$ P_1^T Q P_1 - P_3^T Q P_3 - 2 (P_1 - P_3)^T Q r_0 = 0 \hspace{15pt} (13) $

Using $(11)$ in $(12), (13)$:

$ P_1^T Q P_1 - P_2^T Q P_2 - 2 (P_1 - P_2)^T V G^{-1} F \Omega = 0 \hspace{15pt} (14) $

and

$ P_1^T Q P_1 - P_3^T Q P_3 - 2 (P_1 - P_3)^T V G^{-1} F \Omega = 0 \hspace{15pt} (15) $

Note $ P_i^T Q P_i = P_i^T [\alpha Q_1 + \beta Q_2 + \gamma Q_3] P_i $

Thus, equations $(14)$ and $(15)$ are two homogeneous linear equations in $ \Omega = [\alpha, \beta, \gamma]^T $

Solving them using Gauss-Jordan elimination results in a solution of the form

$ \Omega = \Omega_0 t $

where $\Omega_0 = [\alpha_0, \beta_0, \gamma_0] $ is a constant vector, and $ t \in \mathbb{R} $ is yet to be found.

Now we have

$ Q = t Q_0$ , where $ Q_0 = \alpha_0 Q_1 + \beta_0 Q_2 + \gamma_0 Q_3$

Now we can find $r_0$ from equation $(11)$,

$ t Q_0 r_0 = V G^{-1} F \Omega_0 t \hspace{15pt}(11')$

Hence

$ r_0 = Q_0^{-1} V G^{-1} F \Omega_0 $

Finally, the value of scalar $t$ can found from equation $(3)$

$ t = \dfrac{1}{ (P_i - r_0)^T Q_0 (P_i - r_0) } $

Now the hyperboloid that contains the three lines is fully specified.

Answer 3

I am not sure I understood the question right. There is no unique way to span three skewed straight lines... the possibilities are many.

... Is it true that three skew-lines define a unique hyperboloid of one sheet that contains them on its surface ?

A hyperboloid of 1 sheet is formed by sweeping a skew straight generator S around a straight axis of symmetry A guided along a generator/director circle G whose radius is the minimum distance between the axis and skew line. The generator does not rotate around the axis. There can be three or many more generators depending on the fineness or coarseness of the mesh chosen. Two examples:

If the guided curve G is a straight line and the skew line S rotates uniformly, a right ot left helicoid surface is swept out.

If the guided curve G is a circle and the skew line S rotates uniformly at double the angular speed of the axis a right ot left Moebius Band is swept out.

The swept surfaces have negative Gauss curvature and the shown skewed lines are asymptotic straight generators ( zero normal curvature ) of the warped/skewed surface.

Depending on the director curve and relative speed of skew line rotation, a differential equation of a variety of such warped surfaces can be generated. The association of three lines to form an inclusive surface with no director is not unique.

Let me have your comment about my understanding your question.

The graphics label should be Helicoid, but not Catenoid, shall correct the error later.