929] 
APPLICABLE UPON A GIVEN SKEW SURFACE. 
233 
and therefore 
<xa! + ¡3/3' + <yy’ = 0; 
also 
x' 2 + y' 2 + z 2 = 1 ; 
and therefore 
xx" + y'y" + z'z" = 0 ; 
x, y', z are the cosine-inclinations of the tangent at the point x, y, z. 
I remark also that, writing 
p _1 = V x"' 1 + y" 2 + z" 2 , 
so that p is the radius of absolute curvature, we have px", py", pz" for the cosine- 
inclinations of the binomial (or perpendicular to the osculating plane). 
If from the point (x, y, z) we draw a perpendicular to the consecutive 
generating line through the point (x + dx, y+dy, z+dz), this will also be perpen 
dicular to the generating line through the point (x, y, z), and we thus find 
olx + (3'y' + y'z' = 0 
as the condition that the point (x, y, z) may be, as it is assumed to be, a point 
on the line of striction. 
For the proof hereof, take for the moment x„ y lt z, the coordinates of P, and 
A, Yi for the cosine-inclinations of X; then, considering the line PQ U this passes 
through P and cuts the line ft; taking its equation to be 
X-x _ 
Y~y 
Z-z 
this meets the line 
A 
B 
c 
X-x, 
Y-yx 
Z — z. 
and we thence find 
a i 
A 
7i 
A, 
B, 
G 
x, — X, 
yx-y, 
z x — z 
a i, 
ßx, 
7i 
Writing x lt y 1 , z 1 = x + x ds, y + y' ds, z + z'ds, ds divides out, and we have 
A (ft/ - y,y') + B (YiP - a,z) + G (a,y' - ft/) = 0; 
the line in question cuts ft and L; that is, we have 
A a, + Pft + G i y 1 = 0, 
Aa + P/3 + Cry = 0 ; 
or, eliminating A, B, G, we find 
(#Yi - ft y) (A^' - yi*/) + (y«i - Yi a ) yyi x ' ~ «i-0 + 0*A - a i A> ( a iV' ~ A«0 = 0, 
that is, 
fax' + ft//' + Yi /) (««i + /8ft + YYi) “ ( ax ' + /%' + l z> ) ( a i 2 + A 2 +1\) = 0. 
C. XIII. 
30