题目内容
| AB |
| BD |
| DC |
| AB |
| BD |
| BD |
| DC |
| AB |
| BD |
| BD |
| DC |
| AB |
| DC |
| AC |
考点:平面向量数量积的运算
专题:平面向量及应用
分析:由于
•
=
•
=0,可得AB⊥BD,BD⊥DC.由于|
|+|
|+|
|=4,|
|•|
|+|
|•|
|=4,解得|
|+|
|=2=|
|.由于
与
方向相同,可得
•
=|
||
|.又
=
+
+
,代入(
+
)•
=(
+
)•(
+
+
)展开即可得出.
| AB |
| BD |
| BD |
| DC |
| AB |
| BD |
| DC |
| AB |
| BD |
| BD |
| DC |
| AB |
| DC |
| BD |
| AB |
| DC |
| AB |
| DC |
| AB |
| DC |
| AC |
| AB |
| BD |
| DC |
| AB |
| DC |
| AC |
| AB |
| DC |
| AB |
| BD |
| DC |
解答:
解:∵
•
=
•
=0,∴AB⊥BD,BD⊥DC.
∵|
|+|
|+|
|=4,|
|•|
|+|
|•|
|=4,
∴|
|+|
|=2=|
|.
∵
与
方向相同,∴
•
=|
||
|.
∵
=
+
+
,
∴(
+
)•
=(
+
)•(
+
+
)
=
2+
•
+
•
+
•
+
•
+
2
=
2+2
•
+
2
=(|
|+|
|)2
=22=4.
故答案为:4.
| AB |
| BD |
| BD |
| DC |
∵|
| AB |
| BD |
| DC |
| AB |
| BD |
| BD |
| DC |
∴|
| AB |
| DC |
| BD |
∵
| AB |
| DC |
| AB |
| DC |
| AB |
| DC |
∵
| AC |
| AB |
| BD |
| DC |
∴(
| AB |
| DC |
| AC |
| AB |
| DC |
| AB |
| BD |
| DC |
=
| AB |
| AB |
| BD |
| AB |
| DC |
| DC |
| AB |
| DC |
| BD |
| DC |
=
| AB |
| AB |
| DC |
| DC |
=(|
| AB |
| DC |
=22=4.
故答案为:4.
点评:本题考查了向量垂直与数量积的关系、向量的共线、数量积运算性质、向量的运算,考查了推理能力和计算能力,属于难题.
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