题目内容
如图,在平行六面体ABCD-A1B1C1D1中,AB=5,AD=3,AA1=4,∠DAB=90°,∠BAA1=∠DAA1=60°,E是CC1的中点,设
=
,
=
,
=
.
(1)用
,
,
表示
;
(2)求AE的长?

| AB |
| a |
| AD |
| b |
| AA1 |
| c |
(1)用
| a |
| b |
| c |
| AE |
(2)求AE的长?
(1)根据向量的三角形法则得到
=
+
+
=
+
+
(2)∵|
|2=(
+
+
)2
=
2+
2+
2+2
•
+
•
+
•
=25+9+4+0+(20+12)•cos60°
=54
∴|
|=3
,
即AE的长为3
.
| AE |
| AB |
| BC |
| CE |
| a |
| b |
| 1 |
| 2 |
| c |
(2)∵|
| AE |
| a |
| b |
| 1 |
| 2 |
| c |
=
| a |
| b |
| 1 |
| 4 |
| c |
| a |
| b |
| a |
| c |
| b |
| c |
=25+9+4+0+(20+12)•cos60°
=54
∴|
| AE |
| 6 |
即AE的长为3
| 6 |
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