问题
选择题
设向量
|
答案
∵
⊥a
,∴b
•a
=0,又|b
|=1,|a
|=2,b
由题意可得|
|=|-c
-a
|=|b
+a
|=b
=(
+a
)2b
=
2+2a
•a
+b
2b
=1+4
,5
故选D
设向量
|
∵
⊥a
,∴b
•a
=0,又|b
|=1,|a
|=2,b
由题意可得|
|=|-c
-a
|=|b
+a
|=b
=(
+a
)2b
=
2+2a
•a
+b
2b
=1+4
,5
故选D