476 THE THEORY OF SCREWS. [428- 



graphically, it follows that XX and YY are both generators. We there 

 fore find that after the two vectors all the right generators remain as before, 

 and so do also two left generators, i.e. the result of the two vectors could 

 have been attained by a single vector homonyrnous therewith. 



429. Geometrical proof of the Law of Permutability of Hetero- 

 nymous Vectors. 



Let AB and A B be a pair of right generators (fig. 49), and CD and 

 C U a pair of left generators. Let the right vector convey P to Q, and 



Fig. 49. 



then let the left vector carry Q to the final position P . We shall now show 

 that P would have been equally reached if P had gone first to R, so that 

 intervene PR = QP , and that then R was conveyed by the vector, RB B, 

 through a distance equal to PQ. 



Draw through P the transversal PRCC . Take R, so that 



but, because this relation holds, 



PQ, RP , CD, G D 

 must all lie on the same hyperboloid. 



Therefore RP must intersect AL and AM, and therefore, also, 



(PQAA ) = (RP BB }. 

 Hence, finally, we have for the intervenes 



PQ = RP and PR = QP . 



430. Determination of the Two Heteronymous Vectors equi 

 valent to any given Motor. 



If a right vector, a, ft, &amp;lt;y, B, be followed by a left vector, a, ft , 7 , 8 , then 



