设向
|
因为|
|=1,|a
|=1,所以b
|
-ta
|2=b
2 +t2a
2+2tb
•a b
=t2+2t(cos55°cos25°+sin55°sin25°)+1
=t2+2tcos30°+1=t2+
t+13
所以当t=
时,|3 2
-ta
|的最小值为b 1 2
故选B
设向
|
因为|
|=1,|a
|=1,所以b
|
-ta
|2=b
2 +t2a
2+2tb
•a b
=t2+2t(cos55°cos25°+sin55°sin25°)+1
=t2+2tcos30°+1=t2+
t+13
所以当t=
时,|3 2
-ta
|的最小值为b 1 2
故选B