11 Jan 72 Alien Newell

Size: px

Start display at page:

Download "11 Jan 72 Alien Newell"

Maryann Pearson
5 years ago
Views:

1 11 Jan 72 Alien Newell Comments on Alexandra Forsythe's "What Points are Equidistant from Two Skew Lines," The Mathematics Teacher, 62, 2 February 1969, Before the analysis is over a rather substantial amount of algebra has been * waded through. This does not appear to detract very much from the total feeling of understanding, since it is sandwiched in between layers of more direct per ception (looking at simple cases in front, reflecting on the solution at the end). However, it raises the, issue of whether the insightful steps really by passed would have happened if the problem had been done in the straightforward way. The Analytic Geometric Straightforward Way: Introduce general representations for the various entities expressed in the problem, write the indicated expression for the answer, and crank. Side advice: If the algebraic thicket becomes too thickety, crank harder and have faith that all terms will cancel. Let us try to apply this to the problem: (1) Problem: Find the locus of points, P, such that: Distance(P, Line.l) = Distance(P, Line.2) (2) Represent the entities: P is a general point, hence (x y z) A line in space has several representations. The most explicit (why do I believe this?) is the parametric representation: x=a + b t, y = a * b t, z = a +bt xx y y z z To distinguish the two lines we will use 1's and 2 f s: (3) Subproblem: al, bl, x x a2, b2, x x Find an expression for Distance(P, Line) This follows directly from the formulation in (1). I even know it is the right think to do because it occurs twice in the formulation -- so it is really essential.

2 - 2 - (4) The definition of the distance of a point from a line is the shortest distance from the point to any on the line: d a Min( Distance(P, P 1 ), where P 1 is on the Line) P 1 But I know the distance formula between two points: Distance = Sqrt( (x - x 1 ) + (y - y 1 ) + (z - z 1 ) ) And I know enough to substitute in the formula: d = Min( Sqrt( (x - a -b t) 2 + (y - a - b t) 2 + (z - a - b t) 2 ) ) xx y y z z It appears that the real reason for liking the parametric form was two-fold: that it permitted direct substitution for x f, y 1 and z 1 ; and that it reduced the variability to a single variable, so that the minimization problem is immediately simplified. This latter virtue need not be known by the problem solver it is provided automatically by the analytic representation. All the problem solver has to know is that parametric representations of lines are good things. (5) Is this where this method stops for AF's students, since they cannot solve the minimization problem? That is, it requires a "higher" mechanical technique to crank it through. (6) Well let's go ahead anyway: d(d)» 1 1 (-2b (x - a - b t) - 2b (y - a - b t) -2b (z - a - b t)) dt F x x x y y y y ^ z z ' Getting: (b/ + byz + bz 2)tmln = bx (x - ax) + by (y - ay) (7) A most important part of analytic geometry, of the cranking variety that tends to use general formulas, is to know what degress of freedom are available in a parametric representation.

3 - 3 - The six parameters in the general parametric equation of a straight line only contain four free parameters. One relation may hold between the b f s One relation may hold between the a's Subproblem: To determine the degrees of freedom, if don't know it (which I didn't). i This is something you know about in general, I think. (8) There Is now a powerful temptation to simplify the formula for t. by cashing the two free relations. This is the equivalent of picking a useful coordinate system. The main additional power from the cranking, method is that, if you will wait (putting up with ever more noxious cranking) you can cash the relations later when it is clear what they should be. However, we can't wait: 222 Set: b + b + b =1 which removes a divisor forever x y z Then get: ab + a b + a b = 0 which gets rid of dependence on the a's xxyyzz t.=bx + mm x (9) Now we can get an explicit formula for the distance ^ '' d2 =(x-a) 2 +(y-a) 2 +(z-a) 2 -(bx + b + b z) 2 x y z x y z Or, as an explicit polynomial in (x y z): d - (1 - b )x + (1 - b *)y^ + (1 - b )z *» z - 2t> b yz y z - 2a x x + a 2 X 2b b xz x z 2a y y + a 2 y 2b b xy x y 2a z z + a 2 Z

4 - 4 - The above requires an ever so mild amount of huffing and puffing, but it 2 all comes through. Also, I had to know to use d rather than d, but that really makes no difference either except to the appearance of complexity. (10) At last we are ready to write the equation: dl = d2... Or: 2 2 dl = d2 (x - al ) 2 + (y - al ) 2 + (z - al ) 2 x y z - (bl x + bl y + bl z) 2 = x y z (z - a2x) 2 + (y - a2y ) 2 + (z - a2^) 2 - (b2xx + b2yy + b2 z z) 2 All the terms are, in general, different on the two sides, so no. structural simplification can occur. We can however, push the expressions back and forth to obtain familiar forms. Just responding to the formal urge to 2 2 exploit all those X - Y -> (X + Y) (X - Y) factorizations that are latent above, yields the following. Since we now have many differences and sums of similar terms, we introduce the following systematic abbreviations: Da = al - a2, Da = al - a2,... Db = bl - b2 xxxyyy zzz Sa = al + a2, Sa = al + a2,... Sb = bl + b2 xx x y y y z z z (Db x + Db y + Db z) (Sb x + Sb y + Sb z) x y z x y z +2Da x + 2Da y + 2Da z x y z -,Sa Da - Sa Da - Sa Da = 0 'xx y y z z (11) We have lucked out (something that is supposed to happen with the cranking method). For we can now start to interpret our answer: It is clearly a quadratic equation. It is a hyperbolic surface, since the quadratic part consists of the product of two linear equations -- which are indeed the equations of the asymptotes.

5 - 5 - (12) To proceed further we need again to find out and use degrees of freedom we have in the parameters which correspond to localizing the two lines in special coordinate systems. First, how much freedom do we have? There are 12 parameters in the equation (either the system: al, al,... bl, a2,... b2 or the equivalent x y z x z system: Db, Db,... Sb, Da,... Sa.) Only 8 of these are independent, J xyzxz and we actually have the four relations specified (in terms of the a's and b's). These put some restrictions on the complete independence of the parameter determination, as does the structure of the parameters themselves (e.g., All three Da = 0 implies the two lines intersect, no matter what the other are). Thus, we must be a little careful. (A) We can look at what we care about: apparently only the distance between the two lines and the angle the line make with each other (measured, say, on their projection onto a plane parallel both of them). Thus there seems to be only two relations that count and all the rest (8-2=6) are free to be specified. (B) We can look at what we might specify. We can fix one line completely, taking 4 parameters. We might want to take the line perpendicular to both as one of the coordinates. That takes _^"' two more, giving us six again. (C) To be pure about the method, one should now compute as sub- problems the two quantities of interest -- the distance between the lines and the angle (or some trigonometric indicator thereof) and substitute these into the general expression for the surface, thus explicitly revealing the structure. That is more huffery-puffery than I am prepared for.

6 - 6 - (13) Let's specialize: (A) Let the lines lie in planes parallel to the xy-plane: bl a b2 =.0, hence Db =. Sb =0 z z z z. And the side conditions become: We get: b 2 + b 2 = l, ab + a b =0 forl and 2 x y c x y y (Db x + Db y) (Sb x + Sb y) v x y x y7 +2Da x + 2Da y + 2Da z x y z -Sa Da - Sa Da - Sa Da =0 x x y y z z Observe that we could have obtained this simplification just be wanting to get rid of the z-terms in the product. (B) Let the two lines "cross" at the z-axis; that is, each line goes through the z-axis: One passes through (0 0 zl), the other through (0 0 z2) Now these points are the al and a2 points respectively. I guess this should be shown, in terms of the specialization of the a and b coefficients. But it is so: Thus: We get: The b-constraint makes the b f s into direction cosines. The a-constraint (cross-product of a and b) makes the a-point to be the ore nearest the origin (0 0 0), since it makes the a-vector orthogonal to the b-vector. al = a2 = 0, Da = Sa = 0 X X X X al = a2 = 0, Da = Sa =0 y y y y. zl = al, z2 = a2 (just to remain with the original notation) z z a b + a b =0 satisfied for both 1 and 2 xx y y (Db x + Db y)(sb x + Sb y) + 2Da z - Sa Da =0 x y x y z z z This is the simplest form of formal simplification on the a's, subject to the general constraints. However, we might not have been led to it if we wanted to treat all terms symmetrically.

7 - 7 - (C) Locate the two lines symmetrically above and below the xy-plane: Thus, we get: al = +k, a2 = -k, Da = 2k, Sa = 0 z z z ' z (Db x + Db y) (Sb x + Sb y) + 4kz = 0 x y x yj Except for the factor of 2 we could have obtained this just as easily by deciding to take Da as one parameter and eliminate the z constant term by'setting Sa =0. z (D) There now exist just two free parameters, one of which expresses the separation and the other of which locates the two lines (viewed as a fixed wedge) with respect to the coordinate system. We can fix the coordinate system by taking it symmetrically with respect to the lines, namely: bl = b2 = b, giving Db = 0, Sb = 2b xx x x bl = -b2 = b 1, giving Db = 2b', Sb =0 Thus, we get: bb'xy = -kz This is the simple solution we were after. This last step could 2 2 have been obtained formally by eliminating the x and y coefficients and finding out that it required one to be done by D = 0 and the other by S = 0 to avoid making the two lines identical. The other simple solution comes from eliminating the xy-term.

8 - 8 - (14) Now we are through. To rephrase the question: Is this more complex than the way in the paper, which also ended up with some algebraic manipulation? (15) To ask a second question: Could one design a program so that it would find a solution by this straightforward method. Here the answer seems to be yes. The path we have been following and the choices we have been making, including those of representation (which are critical in any such assessment), all seem simple enough to be capturable by machines of the current sort.

9 (an ellipse? a circle? a hyperbolic surface, etc.)* (This was the path AN started down and then chickened out.) (7) Try for a differential calculation, by starting with some point that satisfies the condition and then compute where a nearby point would be, if a change is forced in some coordinate system. (8) There must be a number of deformational approaches. E.g., start with a special solution of two lines parallel and then start to twist the lines into skew. This same approach might have been tried instead of the one I did, in which we start with the two lines intersecting in the plane and then try to lift one of them gradually. (9) I don't see any way to start with a degenerate figure, say a point and a line, and then grow the point into a line. (10) Likewise, I don't see any useful way to start with a line and a solution assumed and then find the second line (now some sort of curve, since we chose the "solution" to be simple in some way). The true solution is to be found by deforming the second line back into a straight line and watching the effect on the hypothesized solution. ****** (11) Is there any way of generating all possible approaches? None of these methods looks unique to the particular problem, but applies to a whole class of locus problems. How would one classify all these methods? What are the essential ideas: special cases; vary a part of the problem, embed the problem in a space of problems so that approximating sequences can be built,...

Section 13.5: Equations of Lines and Planes. 1 Objectives. 2 Assignments. 3 Lecture Notes

Section 13.5: Equations of Lines and Planes 1 Objectives 1. Find vector, symmetric, or parametric equations for a line in space given two points on the line, given a point on the line and a vector parallel