题目内容

已知平行六面体ABCD-A′B′C′D′中,AB=4,AD=3,AA′=5,∠BAD=90°,∠BAA′=∠DAA′=60°,
(1)求AC′的长;(如图所示)
(2)求
AC/
AC
的夹角的余弦值.
(1)可得
AC′
=
AC
+
CC′
=
AB
+
AD
+
AA′

|
AC′
|2
=|
AB
+
AD
+
AA′
|2
=
AB
2
+
AD
2
+
AA′
2

+2(
AB
AD
+
AB
AA′
+
AD
AA′

=42+32+52+2(4×3×0+4×
1
2
+3×5×
1
2
)=85
故AC′的长等于|
AC′
|
=
85

(2)由(1)可知
AC′
=
AB
+
AD
+
AA′
|
AC′
|
=
85

AC
AC
=(
AB
+
AD
+
AA′
)•(
AB
+
AD

=
AB
2
+2
AB
AD
+
AD
2
+
AA′
AB
+
AA′
AD

=42+2×4×3×0+32+5×4×
1
2
+5×3×
1
2
=
85
2

|
AC
|
=
(
AB
+
AD
)2
=
AB
2
+2
AB
AD
+
AD
2
=
42+0+32
=5
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