8 



I. KINEMATICS OF A POINT. LAWS OF MOTION. 



The coordinates of the 

 the vector AB and of 0. 



axis may be found from those of 

 If we choose for origin, OAB 



for the plane of 

 (Fig. 3) and let the co- 

 ordinates of A be x, y, the 

 projections of AB, s x , s y9 

 we have for the area of 

 the triangle GAB 



-xy-s x s,} 



Accordingly we have for 

 m, the moment about 

 of a vector whose pro- 

 jections are s X9 s tj and 

 whose initial point has 

 the coordinates x, y, 



m = xs y ys x . 



To find the moment of the resultant of two vectors drawn from 

 the same initial point, whose plane contains 0, their projections being 

 sj, Sy', sj, s x ", Sy", s z ", we have 



m = x(s v '+s v ")-y(s x '+s x ") 



Fig. 3. 



ys x = 



m 



thus the moment of the resultant is equal to the sum of the mo- 

 ments. If the plane of OAB is not one of the coordinate -planes, 

 we may project the triangle OAB upon the three coordinate -planes, 

 and obtain three moments m x , m yj m z . If the direction cosines of the 

 axis of m are cos a, cos /3, cos y, we have by the rule for the projection 

 of areas, 



m x = m cos a, m y = m cos /?, m z = m cos 7, 



m' 



m z 



Therefore the moment m has three coordinates, m x , m y , m z , and may 

 itself be considered a vector m. Since the coordinates of the pro- 

 jections of A and AB on the YZ plane are y, 2, s y , S 2 , we have by 

 the preceding formula 



12) 



= ys, - es. 



xs z , m z = xSy 



In the language of Hamilton m is the vector product of the vector 

 OA into the vector AB. We have evidently 



