题目内容
如图,在长方体ABCD-A1B1C1D1中,设AD=AA1=1,AB=2,则|
-
|=
,
•
=
CC1 |
BD1| |
5 |
5 |
CC1 |
CA1| |
1
1
.分析:由题意可得,
-
=
-(
+
)=-
,结合已知可求而
•
=
•(
+
+
)=
•
+
•
+
•
,结合长方体的性质及向量的数量积的性质即可求解
CC1 |
BD1 |
CC1 |
BB1 |
B 1D1 |
B 1D1 |
CC1 |
CA1 |
CC1 |
CB |
BA |
AA1 |
CC1 |
CB |
CC1 |
BA |
CC1 |
AA1 |
解答:解:由题意可得,|
|=
=
-
=
-(
+
)
=-
∴|
-
|=|
|=
∵
•
=
•(
+
+
)
=
•
+
•
+
•
=0+0+
2=1
故答案为:
,1
B1D1 |
1+4 |
5 |
CC1 |
BD1 |
CC1 |
BB1 |
B 1D1 |
=-
B 1D1 |
∴|
CC1 |
BD1| |
B1D1 |
5 |
∵
CC1 |
CA1 |
CC1 |
CB |
BA |
AA1 |
=
CC1 |
CB |
CC1 |
BA |
CC1 |
AA1 |
=0+0+
AA1 |
故答案为:
5 |
点评:本题主要考查了向量的数量积的性质的简单应用,属于基础 试题
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