Section III, 1919 [201] Trans. R.S.C. 



Use of A nalogy in Vector A nalysis. 

 Professor Alfred Baker, M.A., LL.D. 



(Read May Meeting, 1919.) 



It is a commonplace to say that no system of geometry is the best 

 for the solution of all problems. Problems which are quite simple 

 under one system are sufficiently difficult under another. Thus 

 demonstrations which flow most readily from the very nature of, 

 say, geometry of position or reciprocal polars, may be very laborious 

 when attacked by Cartesian analysis. The extensive use made of 

 Cartesian forms shows that they constitute undoubtedly the system 

 of most general applicability. Next in general usefulness comes 

 vector analysis. In working with vector forms it is interesting to 

 observe the complete parallelism which often exists between them 

 and Cartesians: they appear as two languages expressing the same 

 thoughts, often employing the same idioms. If difficulties present 

 themselves when working under one system, it is frequently sufficient 

 to observe how the difficulties have been overcome in the other. 



I proceed to illustrate this: 



Suppose we are finding in the case of a conicoid the locus of the 

 intersection of three tangent planes mutually at right angles: 



Let them be 57r0a= 1, Siv(i>^= 1, Sir(j)-y= 1, where a, /3, 7 are semi- 

 diameters. Then 0a, 4>^, <^7, being perpendiculars to the planes, are 

 perpendicular to each other; so that S4)a<i>^ = Q = S(j)l3(f)y = S(t)y(j)a. 



T.T , iSia , jSja , kSka 

 j\OW 00= — - + -^^ + 



a- b- c'- 



. „. Sla „. Sja „ Ska .^. 



. . Oî0a = — , Sj(f)a = — ^^ , Sk(f)a = — (1) 



a- b^ c- 



and a'{Sicf>ay-+b%Sjcj>ay-^c%Skcl>ay= ^-^^ + ^^' + ^-^^ 



a^ b^ â 



= Sa(pa = 1 (2) 



Let i = .r0a+ 3/0/3+207 

 . * . 6'î0a = .T(0a)-, etc., and 



. Si(t>a .,^Si<t)0 Si(t)y 



Z = 0a \- 0/3 \- 07 



(0a)2 (0^)^ (07)2 



Sec. Ill, Sig. 14 



