Mathematics for Enzyme Reaction Kinetics and Reactor Performance. F. Xavier Malcata

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hence, Eq. (3.111) can be rewritten as

      (3.113)

meaning that the vector product defines the vector area, Sn, of the portion of plane bounded by vectors u and v . The definition conveyed by Eq. (3.111) implies that the vector product is nil for two collinear vectors, because the sine of the angle formed thereby is nil; hence, the vector product being nil does not necessarily imply that at least one of the factors is a nil vector itself.

      The vector product is not commutative; in fact,

(3.114)

stemming from Eq. (3.111), where −n appears because the vector system is now left handed; Eq. (3.114) may thus be rewritten as

      (3.115)

      due to commutativity of the product of scalars, so one eventually finds

(3.116)

      – which means that the vector product is actually anticommutative.

Consider now vectors u, v, and w as depicted in Fig. 3.4a. Equation (3.59) still holds, relating ‖ u ‖ to L_[0A], as well as Eq. (3.111) pertaining to u × v, while one has that

(3.117)

      as per Fig. 3.4b – where [BD], with length L_[BD], denotes a straight segment opposed to ∠ u, v and obtained after projection of v onto u^⊥. Consequently,

(3.118)

based on Eqs. (3.59), (3.112), and (3.117) – and illustrated in Fig. 3.4f. By the same token,

(3.119)

      as per Fig. 3.4c, where [EF] denotes a straight segment opposed to ∠ u, w and obtained via projection of w onto u^⊥; therefore,

(3.120)

stemming from Eqs. (3.59), (3.112), and (3.119) – and apparent in Fig. 3.4g. Finally,

(3.121)

      as per Fig. 3.4d, where [CG] denotes a straight segment opposed to ∠ u, v + w and generated through projection of v + w onto u^⊥; hence,

(3.122)

in agreement with Fig. 3.4h and based on Eqs. (3.59), (3.112), and (3.121) – as represented in Fig. 3.4h. One may thus add the areas of the parallelograms in Figs. 3.4f and 3.4g to get

(3.123)

as emphasized in Fig. 3.4e via plain juxtaposition of the said parallelograms, with S_[BCKJ] = S_[0AHF]. However, triangles [0BJ] and [ACK] are identical as per Fig. 3.4i, owing to the geometrical features of parallelograms – i.e. [0B] and [AC] are parallel and share the same length, and the same applies to [BJ] and [CK]; hence, [0J] and [AK] must also be parallel, and have identical length. One accordingly concludes that

(3.124)

while Eqs. (3.118) and (3.120) allow transformation of Eq. (3.123) to

(3.125)

Recalling Eq.

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