OF A FUNCTION OF SEVEEAL VARIABLES. 255 



a + \a') Dx + (2y + \b + \V) Dy 



Therefore 2sc + \a + \a' = 0, | 



2y + \b+\b' = 0, [ ............... (3). 



2z + \c + \c = 0, J 



Multiply equations (3) by a, 6, c, respectively and add ; then 

 we have, by (1), 



2 +\ (a 2 + 5 2 + c 2 ) +\ 2 (aa + bb'+cc) = ......... (4). 



Similarly, 



c')=() ....... ..(5). 



Equations (4) and (5) determine \ and X 2 , and then by (3) 

 we find x, y, z. Also multiplying (3) by x, y, z, respectively 

 and adding, we have 



X, + \ = 0, 



which finds <. This is the solution of the following question 

 in Geometry of Three Dimensions : " In the line of inter- 

 section of two given planes to find the nearest point to the 

 origin of co-ordinates." From the nature of the question it 

 is evident there must be a minimum value of <. 



2. Determine the greatest quadrilateral which can be 

 formed with the four given sides a, /3, 7, 8, taken in this 

 order. 



Let x denote the angle between a and ft, y the angle between 

 7 and 8. The area of the figure is ^ (cc/3sin;c+ 78 sin^), 

 therefore we may put 



<j) (x, y} = a/3 sin # + 78 sin?/ ............ (1). 



If we draw a diagonal of the figure from the intersection 

 of /3 and 7 to the intersection of a and 8, we have from the 

 two different values which can be found for the length of this 

 diagonal, a 2 + /3 2 2a/3 cos x = 7* + S 2 278 cos y. 



Thus a 2 + /S 2 - 2a/3 cos x - 7" - S 2 + 2y8 cos y = ...... (2). 



