[baker] vector analysis 205 



Evidently this equation is obtained from the usual Cartesian form by 

 writing — ^îp for x, —Sjp for y, and —Skp for z. 



If we wish to find the centre of the conicoid, suppose it at the term 

 of Ô, and write p+5 for p. Then that the term involving the first 

 power of p may vanish we must have 



aiSiÔ + bjSjô + ckSkÔ -\-f(jSkô + kSjÔ) -{-g{k Siô + iSkô) 

 + h {iSjô -\-jSiô) — ui —vj—wk = 0. 

 To solve this for ô, let d=xi-{-yj-\-zk, and 



— aix — bjy — ckz-'rfi — zj —yk)-\-g{— xk — zi) -\-h{—yi— xj) 

 — ui — vj —wk = 0. 

 Whence 



ax-^hy-\-gz-\-ii = 

 hx-{-by-{-fz-\-v = 

 gx-{-fy-\-cz-\-w = 0, 

 the ordinary forms for the centre in Cartesian coordinates. Thence 8 

 is found, and the surface is referred to its centre as origin. 



In reducing the general equation of the second degree to the various 

 forms which permit us to recognize the character of the surface, we 

 may first find the locus of the bisections of parallel chords having, say, 

 direction a. Let p = x±xa; then the locus required is 



STrii{aSia-\-hSja-\-gSka)-j-j{hSia-\-bSja-]-fSka) 

 -i-k{gSia-\-fSja-\-cSka)} —uSia—vSja—zvSka = 0. 



That this plane may be perpendicular to a, i.e., to —iSia—jSja 

 — kSka, we must have 



aSia-[-hSja-\-gSka hSia-\-bSja-\-fSka _ gSia-\-fSja-\-cSka _ ^ 



; — ; — —A, say, 



Sia Sja Ska 



Eliminating Sia, Sja, Ska, we obtain the usual discriminating 

 cubic in X, with real roots. Supposing one of these roots to give the 

 plane of {ij),f, g and w will disappear, and the equation becomes 

 a{Sipy-^b{Sjp)--^c(Skpy-{-2hSipSjp — 2uSip — 2vSjp-{-d = 0. 

 Next turn the axes i,j through an angle 6, so becoming i', j' , — for 

 this purpose operating with cos d — k sin 6. Then 

 i= (cos d — k sin d)i' = cos 6 . i'+sin 6 . j' 

 or Sip = cos dSi'p-\-sm OSj'p 



7 = (cos 6 — k sin d)j' = cos d . j' — sin 6 . i' 

 or 6)'p= —sin dSi' p-{-cos dSj'p 



Making these substitutions, a value can be found for d that will 

 make the term SipSjp vanish, and the equation takes the form 



a{Sipy-\-b{Sjpy~-\-c{Skp)~-2uSip-2vSjp-\-d = 



