VECTOR ANALYSIS. 71 



That is, < al is a planar dyadic, which may be expressed by the 

 equation |$-al| = 0. 



(See No. 140.) Let $ = x; + M f+^; 



the equation becomes 



\[X-ai]i+[[jL-aj]j+[v-ak]k\ =0, 

 or, [X ai]x [/UL 



or, a 3 (i.\+j./uL + k.v 

 This may be written 



a* 

 Now if the dyadic <3? is given in any form, the scalars 



$ 8 , {S-'h, *| 



are easily determined. We have therefore a cubic equation in a, for 

 which we can find at least one and perhaps three roots. That is, we 

 can find at least one value of a, and perhaps three, which will satisfy 

 the equation |$-al|=0. 



By substitution of such a value, <i al becomes a planar dyadic, the 

 planes of which may be easily determined.! Let a be a vector 

 normal to the plane of the consequents. Then 



{$-aI}.a = 0, 



$. a = da. 

 If $ is a tonic, we may obtain three equations of this kind, say 



3?.a = aa, l?./3 = &/3, 3 > .y = cy, 

 in which a, /3, y are not complanar. Hence (by No. 108), 



$ = aaa' + 6/3/3' + Cyy, 

 where a', ft', y are the reciprocals pf a, /3, y. 



In any case, we may suppose a to have the same sign as |$|, since 

 the cubic equation must have such a root. Let a (as before) be 

 normal to the plane of the consequents of the planar < al, and a 

 normal to the plane of the antecedents, the lengths of a and a being 

 such that a. a' = 14 Let /3 be any vector normal to a, and such that 

 3>./3 is not parallel to /3. (The case in which <!>. is always parallel 

 to /3, if /3 is perpendicular to a', is evidently that of a tonic, and needs 

 no farther discussion.) {<!> aI}./3 and therefore <3?./3 will be per- 

 pendicular to a. The same will be true of 3? 2 ./3. Now (by No. 140) 



[$.a]. 

 that is, aa. 



* [See note on p. 90.] 



t In particular cases, $ - al may reduce to a linear dyadic, or to zero. These, how- 

 ever, will present no difficulties to the student. 



J For the case in which the two planes are perpendicular to each other, see No. 157. 



