Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR ALGEBRAnot coplanar. Moreover, if a, b and c are mutually orthogonal unit vectors thena ′ = a, b ′ = b and c ′ = c, so that the two systems of vectors are identical.◮Construct the reciprocal vectors of a =2i, b = j + k, c = i + k.First we evaluate the triple scalar product:a · (b × c) =2i · [(j + k) × (i + k)]=2i · (i + j − k) = 2.Now we find the reciprocal vectors:a ′ = 1 (j + k) × (i + k) = 1 (i + j − k),2 2b ′ = 1 (i + k) × 2i = j,2c ′ = 1 (2i) × (j + k) = −j + k.2It is easily verified that these reciprocal vectors satisfy their defining properties (7.47),(7.48). ◭We may also use the concept of reciprocal vectors to define the components of avector a with respect to basis vectors e 1 , e 2 , e 3 that are not mutually orthogonal.If the basis vectors are of unit length and mutually orthogonal, such as theCartesian basis vectors i, j, k, then (see the text preceeding (7.21)) the vector acanbewrittenintheforma =(a · i)i +(a · j)j +(a · k)k.If the basis is not orthonormal, however, then this is no longer true. Nevertheless,we may write the components of a with respect to a non-orthonormal basise 1 , e 2 , e 3 in terms of its reciprocal basis vectors e ′ 1 , e′ 2 , e′ 3 , which are defined as in(7.49)–(7.51). If we leta = a 1 e 1 + a 2 e 2 + a 3 e 3 ,then the scalar product a · e ′ 1 is given bya · e ′ 1 = a 1 e 1 · e ′ 1 + a 2 e 2 · e ′ 1 + a 3 e 3 · e ′ 1 = a 1 ,where we have used the relations (7.48). Similarly, a 2 = a·e ′ 2 and a 3 = a·e ′ 3 ;sonowa =(a · e ′ 1)e 1 +(a · e ′ 2)e 2 +(a · e ′ 3)e 3 . (7.52)7.10 Exercises7.1 Which of the following statements about general vectors a, b and c are true?(a) c · (a × b) =(b × a) · c.(b) a × (b × c) =(a × b) × c.(c) a × (b × c) =(a · c)b − (a · b)c.(d) d = λa + µb implies (a × b) · d =0.(e) a × c = b × c implies c · a − c · b = c|a − b|.(f) (a × b) × (c × b) =b[b · (c × a)].234