Calculus: Analytic Geometry and Calculus, with Vectors

62 Vectors and geometry in three dimensions

for the length of the vector P1P2, that is, for the distance between P1 and

P2. This formula will be derived more carefully in the next section. A

sphere is defined to be the set of points P in E3 which lie at a fixed distance

r (called the radius) from a fixed point PO (called the center). It follows

from this definition and the distance formula that the equation

(2.26) (x - xo)2 + (y - yo)2 + (z - zo)2 = r2

is the equation of the sphere with center at Po(xo,yo,ze) and radius r.

A paraboloid should be something resembling a parabola, because the

Greek suffix "oid" means "like" or "resembling." A paraboloid (or

circular paraboloid) is defined to be the set of points P(x,y,z) in E3

equidistant from a fixed point F which is called the focus and a fixed plane

r which is called the directrix and which does not contain the focus. In

Figure 2.27

order to obtain the equation of a paraboloid

in an attractive form, we let 1/2k denote the

distance from F to r so that 1/2k = p and

k = 1/2p, where p is the length of the per-

pendicular from F to r. Then we put the z

axis through F perpendicular to r and put

the origin midway between F and r as in

Figure 2.27. The paraboloid is then the set

of points P(x,y,z) for which IFPI = I DPI,

where D is the projection of P on the plane r

and has coordinates (x, y, -1/4k). The

present situation is very similar to that in

Section 1.4, where the equation of a parabola

was worked out. A point P(x,y,z) lies on the paraboloid if and only if

IFPI2= I DPI1 and hence, as use of the distance formula shows, if and only if

x2+y2+(z-

1)2

= Cz+-4k J2.

Simplifying this gives the more attractive equation

(2.28) z = k(x2 + y2).

This is the equation of the paraboloid shown in Figure 2.27.

Problems 2.29

(^1) Plot the points 4(1,1,0), B(1,0,1), and C(0,1,1).
(^2) In a new figure, repeat the construction of Problem 1 and insert the
horizontal and vertical line segments running from 11, B, and C to the coordinate
axes.
3 In a new figure, repeat all of the construction of Problem 2. Then insert
the point D(1,1,1) and the line segments joining D to 1, to B, and to C. Remark: