Calculus: Analytic Geometry and Calculus, with Vectors

108 Vectors and geometry in three dimensions

16 Prove that a line which is not completely contained in a quadric surface

can intersect the quadric surface at most twice. Solution: Suppose the coordi-

nate system is so chosen that the given line has the equations y = z = 0, and

let the equation of the quadric surface be (2.68). A point (x,0,0) then lies on

the surface if and only if 11x2 + Gx + J = 0. If there is an x for which this

equation is not satisfied, then at least one of !1, G, I must be different from 0 and

there are at most two values of x for which the equation is satisfied.

17 The purpose of this long problem is to develop ideas about the transversals

of three given skew (no two lying in the same plane) lines PIP2, P3P4, P5P6. A

line L is called a transversal of the given lines if it intersects the lines P1P2, P3P4i

PbP6 at points Q, Q1, Qs as in Figure 2.693. For each k = 1, 2, ., 6 the coordi-

nates (xk,yk,Zk) of Pk are given numbers and we want information about the

coordinates of Q, Q1, Q2. The latter coordinates are determined with the aid of

numbers X, X1, X2 for which

(1) P1Q = XP1P2, P3Q1 = a1P3Q4, P5Q2 = XZP5P6

Our first step is to select a number X (or to think of X as being "fixed") and ask

whether a transversal through Q exists. If X is so chosen that the plane 7r1 con-

taining Q, P3, P4 does not intersect the line P,P66 or intersects the line P5P6 at a

point Q2 for which the line QQ2 is parallel to the line P3P4, then no transversal

through Q exists. Henceforth, we suppose that X is not so unhappily chosen.

A transversal through Q is then obtained by drawing the line L through Q and

the Point Qz where al intersects the line QsQ6. The value of X2 can be calculated

in terms of X from the equation

xs + X 2(X6 - x5) Ys + X2(Y6 - ys) Z5 + X2(Z6 - Zs) 1

(2) xa 1a za (^1) = 0
X4 y4 Z4 1
xl + X(xs - x1) Y1 + M(y2 - y1) zl + X(z2 - 71) 1
which says that Q2 is the point on the line PiP6 which lies in the plane a1 contain-
ingPa, P4, Q. Similarly, the value of X1 can be calculated in terms of X from the
equation
(3)
xa + X1(x4 - xa) ya + X1('4 - ya) Za + X1(24 - za)
X5 Y6 z5
xb Y6 Z6
x1 + X(xt - x1) Y1 + X(ya - yl) a1 + X(z2 - zl)
=0