Calculus: Analytic Geometry and Calculus, with Vectors

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2.6 Vector products and changes of coordinates in E3 103

So far as we know, some quadric surfaces may be rather complicated
things, and it is of interest to know what the intersection Sl of a quadric
surface and a plane -7r may be. Such a set Si is a quadric section. A little
thought can save us a lot of trouble. We can introduce a coordinate
system in such a way that the plane a is the plane having the equation
z = 0. The equation of the quadric surface must have the form (2.68).
Putting z = 0 in (2.68) shows that the quadric section must be the set
of points P(x,y) in the xy plane whose coordinates satisfy the equation


(2.682) 4X2 + Bye + Dxy + Gx -}- Hy + J = 0.

Quadric surfaces and quadric sections will be studied later. Meanwhile,
we make some remarks that may be at least partially understood. The
family of quadric surfaces includes spheres, circular cylinders, circular
cones, ellipsoids, various kinds of paraboloids and hyperboloids, and,
in addition, assorted degenerate things such as empty sets, lines, planes,
and pairs of planes. The equations z2 + 1 = 0, x2 + y2 = 0, z2 = 0,
and z2 - 1 = 0 do have the form (2.68). The family of quadric sections
includes circles, parabolas, ellipses, hyperbolas, and, in addition, such
degenerate things as the empty set, points, lines, pairs of lines, and the
whole plane.


Problems 2.69


1 Supposing that u and v are vectors in E3 having scalar components u,,
u2i us and v, V2, v3, tell how u X v can be expressed as a determinant. Ans.:
(2.62).
2 Calculate w = u X v and check your answer by showing that^0
and 0 when

(a) u=21-3j+ 4k,
(b) u=21-3j+-4k,
(c) u=i+j,
(d) u=i,

v=2i+3j+4k
v=2i+3j+4k
v= i+ j+ k
v= i+ j

3 Two unit vectors u, and u2 have their tails at the same point P on a
sphere which we consider to be the surface of an idealized earth. Suppose that
P is neither the north pole nor the south pole of the earth, that u, points east,
and that u2 points north. Find the direction of ui X u2. Ans.: UP-
4 Show that the vectors i' and j' defined by


i' = 1(i - 2j + 2k)
J1 = $(2i - j - 2k)
k'= ai -I- bj+ck

constitute a two-dimensional orthonormal system and then so determine a, b,
c that the three vectors constitute a right-handed three-dimensional orthonormal
system. Ins.: k' = 11(2i + 2j + k).
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