(^20) Vector Operations
and the two matrices, A in (C4) and A' in (1.2.15) are related by A' =
BABT. Hence, detA = det.4' > 0 and (C4) holds.
EXERCISE 1.2.10? Verify that A' = BABT. Hint: see Exercise 8.1.4 on page
453 in Appendix. Pay attention to the basis in which each matrix is written.
The following theorem shows that the properties (Cl), (C2), and (C4)
define a unique vector w = u x v.
Theorem 1.2.2 For every two non-zero, non-parallel vectors u,v in
M^3 , there is a unique vector w — u x v satisfying (Cl), (C2), (C4)- If
(ux,U2,uz) and (vi,V2,vs) are the components of u and v in a cartesian
right-handed system i, j, k, then the components wi,u>2,u)3 ofuxv are
wi = U2V3 - U3V2, w 2 = U3V1 - uiv 3 , w 3 = uiv 2 - U2V1. (1.2.16)
Conversely, the vector with components defined by (1.2.16) has Properties
(Cl), (C2), and (C4).
Proof. Let w be a vector so that w • u — 0 and w • v = 0, that is, w
is orthogonal to both u and v. By the geometry of R^3 , there is such a
vector. Choose a w with magnitude ||tu|| = ||tt|| ||v|| sin#, satisfying (Cl).
By (C2),
uiwi + U2W2 + U3W3 = 0, (1.2.17)
wi^i+u 2 w 2 + W3W3 = 0. (1.2.18)
Multiply (1.2.17) by V3 and (1.2.18) by U3 and subtract to get
-A-a b -A.
(U1V3 - U 3 Vi)u>i = (U 3 V 2 - U2V 3 )W2. (1.2.19)
Similarly, multiply (1.2.17) by vi and (1.2.18) by ui and subtract to get
c ~a
(v 2 ui - viu 2 )w 2 = (U3V1 - uiv 3 )u>3. (1.2.20)
Abbreviating, let a = U1V3 — U3V\,b = U3V 2 — U2V3 and c = u\v 2 — v\u 2.
Then (1.2.19) and (1.2.20) yield
w\ = (b/a)w 2 ; W3 = {-c/a)w 2. (1.2.21)
Hence, ||u>||^2 = (b/a)^2 w% + w\ + (c/a)^2 u>2 and
HUJII^2 = (1 + (6^2 + c^2 )/a^2 )w^22 = (a^2 +b^2 + c^2 )(^/a^2 ). (1.2.22)
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