a = aiki + r/ok-j ± Va^ + a.>" kij. 

 The complement of a singular vector is 



A = a* = a'kij4 = (likn — chku =f Vor + «2-kij. 



This 2-vector A must be itself a singular plane vector; for we have 

 seen that the complement of any (5)-plane is a (7)-line and of any 

 (7)-plane a (5)-line, and vice versa. The inner product of A by itself 

 is obviously zero,^^ for, 



A-A = — oi- — ar + (ar + «•/') = 0. 



Conversely if we consider any 2-vector 



A = .-li..ki2 + An)^U + ^24k24, 



A-A = .li2- — An^ — AoJ = 0, 



such that 



its complement is a singular line, and it is itself a singular 2-vector. 

 The standard form may be taken as 



A = ± ylAu^ + yl24'ki2 + .■Il4ki4 + ^l24k24. 



The outer product of a singular vector by its complement, whether a 

 1-vector or a 2-vector, vanishes, as may be seen by multiplying out. 

 Thus the singular vector and its complement lie in the same plane, 

 that is, an element of the cone and the tangent plane through that 

 element are mutually complementary. 



When we have to consider the inner product of any singular vector 

 with any other vector, singular or not, the geometrical method de- 

 pendent on projection often fails to be applicable; for it is impossible 

 to project a vector upon a singular vector. We may in such cases 

 employ the analytical method, which is universally applicable, or 

 replace the inner product with an outer product by a method intro- 

 duced in a following section (§ 32). 



We have seen that an element of the cone is complementar}^ to the 

 tangent plane to the cone through that element, that is, the element 

 is perpendicular to the plane. Hence the element is perpendicular to 

 every line in the plane (including itself). 



26 A singular vector, or vector of zero magnitude, has, like any other vector, 

 a real geometrical existence and is not to be confused with a zero vector, that 



IS, a non-existent vector. 



