Calculus: Analytic Geometry and Calculus, with Vectors

2.6 Vector products and changes of coordinates in E, 99

neatly by writing

i j kI

uxv= 1 -2 31=5i+7j+3k.

12 -1 -11

We should all know enough to be able to guard against computational
errors by observing that our answer is perpendicular to u because
5 - 14 + 9 = 0 and is perpendicular to v because 10- 7 - 3 = 0.
To exhibit an application of vector products to a problem ingeometry,
we suppose that we are given two orthonormal vectors i' and j' in E3
(this means that i' and j' are unit vectors and are orthogonalor per-

pendicular) and are required to find a third vector k' such that the three
vectors i', j', k' constitute a right-handed orthonormal system. The
answer is given by the formula k' = i' X j'. This is so because

(2.621) it X j' = ji'I ji'l sin on = n,

where n is the unit thumb vector, and this is exactly what k' must be.
This problem can be put in a different form. Suppose we are given
a right-handed rectangular x, y, z coordinate system endowed with the
usual orthonormal vectors i, j, k. Suppose further we know the scalar
components all, a12, a13 and a21, all, a23 of two orthonormal vectors i'
and j' so that the coefficients in the first two of the equations

1' = a11i + aid + alak

(2.63) j' = a21i + a22j +a23k

k' = a31i+ a32j + aaak

are known. The problem is to determine the numbers as,, all, ass so
that the three vectors i', j', k' will form a right-handed orthonormal
system. These numbers are determined from the formula

asli+aa2j+aaak=k' =i' Xj' =

i j k

all a12 a13

a21 a22 a23

and the problem is solved or almost solved. To write more formulas is
somewhat anticlimactic, but we can do it. Equating coefficients of
i, j, and k gives

(2.631)

as, = al2a23 - alsa22

all = a1sa21 - a11a23

ass = a11a22 - al2a21

and then the problem is surely solved.
Everyone should read the remainder of this section, but teachers who
want to confine attention to other topics may inform their charges that