[baker] vector analysis 203 



The complete parallelism between the above and the ordinary 

 method in Cartesian geometry will be perceived in noting that the 

 results in (3) correspond to lni-\-l'm'-{-l"m" = 0, etc.; and the results 

 in (4) correspond to r--\-r--\-l"^-= 1, etc. 



In the following treatment of the same proposition the analogy 

 with Cartesians is even more complete : 



Let ON = pa be the perpendicular from the origin on any plane, 



and 0P = p any vector to the plane. Ta= 1. Then 



a = —iSia —jSja — kSka, and 1 = (5/a)-+ (6)'a)^+ (Ska)'-^. 



p = — iSip —jSjp — kSkp. ' 



p = pa-\-NP 



Operating with a, and taking scalars 



SipSia-\- SjpSja-\- SkpSka = p, 



which is the equation of the plane. 



Comparing this with the equation of the tangent plane at the 

 extremity of the semi-diameter /3, Sp(f)0= 1, or 



SipSi^ SjpSj^ SkpSk^ _ 

 a- 0- c- 



we get as the equation of the tangent plane 



SipSia-\-SjpSja-\-SkpSka='^a"(Siay-{-b"{Sja)--\-c~{Skay. 



Similarly for the two other tangent planes. Squaring and adding, 

 remembering the relations between Sia, Sja, etc., and between Sia, 

 Sia, etc., we obtain the equation of the locus, (Sip)"-\-(Sjp)^-\-(Skp)''^ 



If the question be the determination of the axes in the central 

 section Sap = of the conicoid Sp(t)p=l, the finding of the maximum 

 and minimum values of p, when subject to the above conditions, is not 

 apparent. Converting, however, p into the form 



P= —iSip—jSjp — kSkp, 

 the equations become 



= SiaSip-\- Sj aSj p-\- SkaSkp, 



r^^= -p2= (^sipy-\-{Sjpy-{-(Skpy, 



