VECTOR ANALYSIS. 21 



Again, if a be directed toward the east, and ft lie in the same 

 horizontal plane and on the north side of a, ax ft will be directed 

 upward. 



15. It is evident from the preceding definitions that 



a.ft = ft.a, and axft=ftxa. 



16. Moreover, [na].ft = a.[nft] = n[a.ft], 

 and [na]xft = ax[nft] = n[axft]. 



The brackets may therefore be omitted in such expressions. 



17. From the definitions of No. 11 it appears that 



i.i=j.j = k.k=l, 

 ij =j.i = i.k = k.i=j. k = k.j == 0, 

 Q, kxk=0, 



jxi=k, kxj=i, ixk=j. 



18. If we resolve ft into two components ft' and ft", of which the 

 first is parallel and the second perpendicular to a, we shall have 



a.ft = a.ft' and axft = axft". 



19. a.[/3+y] = a./3 + a.y and ax[ft + y] = axft + aXy. 



To prove this, let <r = /3+y, and resolve each of the vectors ft, y, <r 

 into two components, one parallel and the other perpendicular to a. 

 Let these be ft', ft", y, y", <r', a-". Then the equations to be proved 

 will reduce by the last section to 



a.<r' = a.ft'-\-a.y and aX(r" = axft rr + aXy". 



Now since & = ft+y we may form a triangle in space, the sides of 

 which shall be ft, y, and <r. Projecting this on a plane perpendicular 

 to a, we obtain a triangle having the sides ft", y", and <r", which 

 affords the relation <r" = ft r ' + y". If we pass planes perpendicular 

 to a through the vertices of the first triangle, they will give on a 

 line parallel to a segments equal to ft', y', &'. Thus we obtain the 

 relation a-' = ft'+y. Therefore a.(r' = a.ft'+a.y, since all the cosines 

 involved in these products are equal to unity. Moreover, if a is a 

 unit vector, we shall evidently have aX(r n = axft" + aXy", since the 

 effect of the skew multiplication by a upon vectors in a plane 

 perpendicular to a is simply to rotate them all 90 in that plane. 

 But any case may be reduced to this by dividing both sides of the 

 equation to be proved by the magnitude of a. The propositions 

 are therefore proved. 



20. Hence, 



