GRAPHICAL ALGEBRA. 85 



156. Projections of a Vector; Scalar Product and Vector-Product. Let a 

 direction represented by the angle a be given everywhere in the field. We shall 

 form the projection A t of a given vector (F, <p) on this direction. This projection 

 will have the positive or the negative sign according as it points in or against the 

 direction represented by the given angle a. 



The projection is given by 

 (a) A , = F cos (<p a) 



We solve with respect to F, and to <p a 



r = 7 -v <p a = arc cos ^r 



cos (<p a) F 



We tabulate F as function of the variables A , and <p a (first of tables K) . In 

 the same manner we should have tabulated <p a as function of F and A t . But as we 

 deal here only with the general principles, and not with the tables for practical 

 use, we shall give here and in several cases below only one table. The field of the 

 projection A x can then be found in two operations. By graphical subtraction we 

 form the field of the angle <p a. This field is placed upon that which represents 

 the scalar value F of the given vector. Using the first table K, we derive from 

 the curves <p a = const, and F = const., the field of the scalar A lt proceeding 

 as we have exemplified several times already for graphical operations with two 

 variables. In this, as well as in several of the following tables, each tabulated 

 number corresponds to different sets of arguments. The arguments on the left side 

 and above belong together, and so do the arguments on the right side and below. 

 In order to avoid mistakes it may be favorable for practical use to have two tables 

 containing the same tabulated numbers, but each only with one set of arguments. 



We can now form the projection of (F, <p) on the positive normal to that direction 

 which is given by the angle a. For this projection we have 



(6) A 2 = Fsin (<p a) 



We solve this equation with respect to F 



(60 F = . f 2 , 



v sin (<p a) 



and tabulate F as function of A and <p a (second table K). Thus, in order to find 

 the field of this projection A 2 , we first form the same auxiliary field <p a as in the 

 preceding case, place this field upon that which represents the intensity F of the 

 given vector, and draw the curves A 2 = const, by use of the second table K. 



By the two tables K we can thus solve a vector F into orthogonal components 

 A x and A 2 . We thus have the way open to bring coordinate-methods into applica- 

 tion when this should be desirable. 



From expressions of the form (a) and (6) there is only one step to expressions 

 of the form 



AB cos (/3- a) and AB sin (p-a) 



i. e., to the formation of the complete scalar product or the complete vector-product 



