题目

设alpha =(dfrac (1)(2),0,0,dfrac (1)(2)) ,=E-(alpha )^Talpha =E+2(alpha )^Talpha ,则AB=( )alpha =(dfrac (1)(2),0,0,dfrac (1)(2)) ,=E-(alpha )^Talpha =E+2(alpha )^Talpha ,

设则AB=(      )

题目解答

答案

答案为：D

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