Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

VECTOR ALGEBRA


not coplanar. Moreover, ifa,bandcare mutually orthogonal unit vectors then


a′=a,b′=bandc′=c, so that the two systems of vectors are identical.


Construct the reciprocal vectors ofa=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′=^12 (j+k)×(i+k)=^12 (i+j−k),
b′=^12 (i+k)× 2 i=j,
c′=^12 (2i)×(j+k)=−j+k.

It 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 a

vectorawith respect to basis vectorse 1 ,e 2 ,e 3 that are not mutually orthogonal.


If the basis vectors are of unit length and mutually orthogonal, such as the


Cartesian basis vectorsi,j,k, then (see the text preceeding (7.21)) the vectora


canbewrittenintheform


a=(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 ofawith respect to a non-orthonormal basis


e 1 ,e 2 ,e 3 in terms of its reciprocal basis vectorse′ 1 ,e′ 2 ,e′ 3 , which are defined as in


(7.49)–(7.51). If we let


a=a 1 e 1 +a 2 e 2 +a 3 e 3 ,

then the scalar producta·e′ 1 is given by


a·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 anda 3 =a·e′ 3 ;sonow


a=(a·e′ 1 )e 1 +(a·e′ 2 )e 2 +(a·e′ 3 )e 3. (7.52)

7.10 Exercises

7.1 Which of the following statements about general vectorsa,bandcare 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+μbimplies (a×b)·d=0.
(e) a×c=b×cimpliesc·a−c·b=c|a−b|.
(f) (a×b)×(c×b)=b[b·(c×a)].
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