Mathematics of Physics and Engineering

Cross Product 21

Now, by simple algebra,

a^2 + b^2 + c^2 - (U1V3 - U3V1)^2 + (u 3 v 2 - u 2 v 3 )^2 + {uiv 2 - V1U2)^2

= (w^2 + u\ + ul){v^2 + v% + vl) - (ui«i + u 2 v 2 + u 3 v 3 )^2

= \u\^2 \v\^2 -(u.Wy llu||^2 IMI^22 (1-cos'^2 6)

Nil^2 Il-u||^2 sin^2 <

(1.2.23)

Applying Property (CI), we get a^2 +b^2 +c^2 = \w\^2. Using (1.2.22), w\ja^2 =

1. Hence, w 2 = ± a and by (1.2.21), wi = ± b and W3 = =F c. To determine
   the signs, consider the special case u = i and v — j. Then u\ = l,u 2 — 0
   and v\ = 0,^2 = 1 and c= 1-1-0-0 = 1. On the other hand, the
   determinant in Property (C4) for this choice of u and v is

det

1 0 0

0 10

±b ±a^fc

=pc,

depending on whether W3 = — 1 or w 3 = 1. Since the determinant must
be positive, we must take w 2 = —a, in order to make W3 = c = 1 in this
case. This implies wi = —b. Therefore, w is uniquely determined and
has components u>i = —b, w 2 = —a, w 3 = c. In other words, there exists a
unique vector with the properties (Cl), (C2), and (C4), and its components
are given by (1.2.16).
Conversely, let to be a vector with components given by (1.2.16). Then
direct computations show that w has the properties (C2) and (C4). After
that, we repeat the calculations in (1.2.23) to establish Property (Cl). The
details of this argument are the subject of Problem 1.3 on page 410.
Theorem 1.2.2 is proved. •

Remark 1.3 Formula (1.2.16) can be represented symbolically by

u x v = det

i 3 K

«1 U 2 U3

Vl v 2 V3

(1.2.24)

and expanding the determinant by co-factors of the first row. Together with
properties of the determinant, this representation implies Property (C3) of
the cross product. Also, when combined with (Cl), formula (1.2.24) can
be used to compute the angle between two vectors with known components.
Still, given the extra complexity of evaluating the determinant, the inner