THE DYADICS IN THREE DIMENSIONS. 397 



to obtain this product we replace [AB] by its complementary form [a/3]. 

 Then 



[AB]-y= Mt]. 



The result is the point in which a, j3, y intersect. Since a and j3 are 

 planes passing through [AB], this is the point in which [AB] intersects 7. 

 This method of obtaining the product is simple in conception but 

 not analytically convenient. We shall therefore give a set of reduc- 

 tion formulas by which the same results can be obtained in more 

 useful form. The proof of these formulas is not essential for our 

 purposes. Hence we give them without proof. ^° The roman letters 

 represent points, the greek letters planes. 



[ABC-DEF] = {ABCF)[DE] - {ABCE)[DF] + {ABCD)[EF] 



= (ADEF)[BC] - iBDEF)[AC] + {CDEF)[AB]. (14) 



[ABC -DE] = iABCE)D - {ABCD)E 



= (ABDE)C - {ACDE)B + iBCDE)A. (15) 



[a- ABC] = {aC)[AB] - iaB)[AC] + {aA)[BC]. (16)" 



[a-AB] = iaB)A - {aA)B. (17) 



Similar formulas can be obtained by replacing points by planes and 

 planes by points. The following formula 



{ABCD)E = {ABCE)D - {ABDE)C + (ACDE)B - {BCBE)A (18) 



and the one that results by replacing points by planes are also some- 

 times found useful. 



5. The complex. Let A, B, C, D be four points not in a plane. 

 Any two points P], P2 of space can be expressed as linear functions 

 of A, B, C, D. 



Pi = \u-i + iiiB + a,C + piD, 

 P2 = X2.4 + M2jB + OiC + p.B. 

 Hence 



[P1P2] = [(Xi^ + Mi-B -f (TiC + PiB){\2A + M25 + aoC + p,D)] 



= (X1M2 - X2Mi)W5] -\- (X,(ro - X2<Ti)[.4C] + [X1P2 - \2Pi)[AD] 

 + Gui^2 - ^i2al)[BC] + (miP2 - i^2Pi)[BD] + ((x^po - <xopi)[CD]. 



This shows that any line is a linear function of the six edges of a tetra- 

 hedron. The sum of any number of lines is then obviously a linear 



10 See Linear distance and angle or any of the works mentioned in note 6. 



