THE DYADICS IN THREE DIMENSIONS. 435 



In a somewhat similar way from a line \BC] we get a dyadic 



CB - BC. 

 This is the product 



A[a-BC] = A{{aC)B - {aB)C] 



= AaCB - AaBC = CB - BC. 



This shows that if 



[BC] = [DE] 

 then 



BC - CB = DE - ED. 



In this discussion [BC] and [DE] may be complex lines or simple lines. 

 If [BC] is a complex, the dyadic BC — CB gives the plane-point polar 

 transformation with respect to the complex. 



Similarly, from a plane [BCD] we get two dyadics 



• A[a-BCD] = B[CD] - C[BD] + D[BC] 

 and 



[BCD-a]A = [CD]B - [BD]C + [BC]D. 



The first of these as an operator on lines gives the point in which the 

 line cuts the plane. The second as an operator on planes gives the 

 line in which [BCD] cuts the plane. 



Those same dyadics are also obtained by multiplying the plane 

 [BCD] with the idemfactor pq. 



In the same way by considering a point as the product of three 

 planes, two dyadics can be determined. 



19. Double products of dyadic with themselves. The 

 double product of a one-three dyadic 



B0 = B'I3' 



with itself is a two-two dyadic 



[BB'] [0] 



which gives the transformation of lines determined by the trans- 

 formation X' — b(^X) of points. For, if 6/3 transforms A^, 1' into 

 X', Y', then [BB'] [^j3'] transforms the line [XY] into a linear function 

 of [X'Y'] and [Y'X'], that is, into the line [X'Y']. 

 Similarly, the triple product 



^[BB'B"][^^'^"] 



