设AC=a,AB=b,A1A=c,.则:AB1=AB+AA1 =b-c.CB=AB-AC =b-a.
C1B=C1C+C1B1=A1A+CB= c+b-a.
由此:AB1*C1B=(b-c)*(c+b-a) 注意到:c垂直于a,且c垂直于b,故c*a=0,c*b=0.
又知|a|= |b| =(根号2)|c| ,从而a*b=|a|^2 *cos60度=(1/2)|a|^2
故::AB1*C1B=(b-c)*(c+b-a)=b*b-b*a-c*c= |b|^2 -(1/2)|a|^2 -|c|^2
=|b|^2 -(1/2)|a|^2 -(1/2)|a|^2= 0.
则AB1与C1B所成角的大小是90度.