58 G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0 . 
We might now be induced to consider the quaternion 
ne +266 , 
but as it enters here only by its scalar we may also consider the other 
quaternion (not precisely a conjugate to the first), namely 
g=nk +20¢. 
We have then the conjugate 
Kg=xn+2¢¢ . 
From this : 
qKq = (n+ 260) («'n + 266) 
S'a— V'g=7K? +4007 + 2[ KCC + O&'n |, 
and as the vectors of the last two products are equal, but opposed in sign, we 
have 
ne CC + Ook =2WyK' CL . 
Then we draw from the preceding 
mK? +40C?—S'g= — V?g—48nk' CC . 
On the other hand, by developing Ey we get 
ft / 9 19 9 7 3 ‘9 fi 
3 ii =1'k >+ 400 —S’q+2[ CK +176] : 
Substituting for the first three terms the prepared value we get 
i 7 / J 7 , 
al =TV9t 280K? +7°C?—2Syk' CE] . 
Now 
—2Snk' CO = —2Sy[ «SCO —OS0 K+ CSK'C], 
and the two first terms in the 2d member give 
[ —2Snk’ Sli + 2SynlSiK' ]=2S VE Vee. 
Replacing also (x? + 17°C? by 
Se C— VK C+ Sn —V'nG, 
we see that the second term in the value of a4 becomes : 
2 [ Sk C—28kK' Sn + 8’nG 
—VK'L+ QV LVL —V'G JS? 
which is 
2 S*(S— 0) — V%( C—9G) = 2T"(K'S—b,) 
Replacing now g, =nx’ +2¢¢, and as under the sign V we have: 
Vq=V(nk —220), 
we get finally : 
