50 G. PLARR ON THE SOLUTION OF THE EQUATION Vpodp=0. 
§ 2. We introduce the six auxiliary vectors », o, 7, ,, 0,,7, by the follow- 
ing definitions : 
n= 2aSawa 9, = 2aSan’*a 
(12) o=2aSPay o, = 2aSho’y 
T= 2aSyaB T, = 2aSyoa'B 

The first three give : 
San =—Sawa, SBn=—S8aB, Syn =—Syay 
(13) Sac =—SBay, SBa=—Syaa, Syo=—SaaB 
Sar =—Syaf, SBr =—Sawy, Syr = —SBaa, 
and as we have generally, 
@(w) = —ZaSaa(w) , 
so we get by the preceding table : 
wa = aSan + BSyr + ySBo 
(14) aB = BSBn + ySar + aSyo 
ay = ySyn + aSBr + BSac , 
these three expressions may also be deduced, each from the preceding, by the 
circular permutation of aByaB, &c., into ByaBy, &c., and into yaBya, &e. 
With these expressions of ga, &c., we calculate 2P, 3Q, 2 P’, &e. 
By (9) and by (14) we have, namely : 
2 P= San(wa) = DSalwaSan+ wBSyr+aySBo}. 
But by the table of values (13), this becomes 
2 P=2| —S’an—SyoSyr—SBrSBo]. 
Applying the general formula : 
S00 =— SalSaf , 
and remarking that the two last sums in the expression P are one and the 
same, as by circular permutation each is composed of the identical same terms, 
we get 
(15) 2P=n7?+2Sor. 
Then by (10) and (14) we have : 
3 Q=2Saa"(wa) = ola’ aSan+a°BSy7t +a°ySBo | , 
and forming with the expressions of 4, o;, 7, a table of values similar to 
(13), we get : 
3 Q=2[—Say,San—SyoiSyt—S7,SBo! 
