9, 10] COMPOSITION. 13 



vector is equivalent to vectors represented by OM, MP, and the 

 magnitudes of these are respectively R cos 6 and R sin 6, where 

 R is the magnitude of the vector to be resolved, and is the 

 angle between its direction and OA. 



The vector represented by MP is the resolved part of the 

 vector represented by OP at right angles to the line OA. 



10. Composition of any number of vectors. I. Consider 

 first the case where all the vectors are 

 parallel to a plane, and take it to be 

 the plane of so, y. Let OP 1} OP 2 , 

 ...OP n be lines representing the 

 vectors, (supposed to be n in num 

 ber,) in magnitude, direction, and 

 sense, and let 6 lt 6 2 ,...0 n be the 

 angles the lines OP ly OP 2 , ... OP n 

 make with Ox, i.e. the angles traced 

 out by a revolving line turning about from Ox towards Oy. Let 

 r 1} r 2 , ... r n denote the magnitudes of the vectors. 



Then the vector represented by OP! may be replaced by vectors 

 r x cos 6 parallel to Ox, and r^ sin X parallel to Oy, and similarly 

 for the others. 



All the resolved parts parallel to Ox are equivalent to a single 

 vector X parallel to Ox given by 



X = n cos #! + r 2 cos 2 -f . . . + r n cos O n = 2 (r cos 6). 



All the resolved parts parallel to Oy are equivalent to a single 

 vector Y parallel to Oy given by 



Y = i\ sin 0j + r 2 sin 2 + . . . -f r n sin 6 n = 2 (r sin 0). 



The vector whose resolved parts parallel to Ox and Oy are .Z 

 and Y is the resultant of all the vectors. Let the magnitude of 

 this vector be R, and let its direction and sense be those of a line 

 going out from and making an angle -fy with Ox. 



Then we have R cos -^ = X, and R sin i/r = F. 



These two equations determine the magnitude R and the 

 angle -fy, viz. : R is the numerical value of V (-3T 2 + F 2 ), and ty is 

 that one among the angles whose tangents are Y/X for which the 

 sine has the same sign as Y and the cosine has the same sign 

 as X. 



