问题
填空题
已知向量
|
答案
|
-a
|2=(b
-a
)2=b
2-2a
•a
+b
2b
=2-2cosαcos(α+
)+sinαsin(α+π 3
)π 3
=2-2cos[α-(α+
)]π 3
2-2cosπ 3
=1
∴|
-a
|=1b
故答案为:1
已知向量
|
|
-a
|2=(b
-a
)2=b
2-2a
•a
+b
2b
=2-2cosαcos(α+
)+sinαsin(α+π 3
)π 3
=2-2cos[α-(α+
)]π 3
2-2cosπ 3
=1
∴|
-a
|=1b
故答案为:1