题目内容

如图,平行六面体ABCD-A1B1C1D1中,以顶点A为端点的三条棱长都为1,且两夹角为60°.
(1)求AC1的长;
(2)求BD1与AC夹角的余弦值.
AB
=
a
AD
=
b
AA1
=
c
,则两两夹角为60°,且模均为1.
(1)
AC1
=
AC
+
CC1
=
AB
+
AD
+
AA1
=
a
+
b
+
c

∴|
AC1
|2=(
a
+
b
+
c
2=|
a
|2+|
b
|2+|
c
|2+2
a
b
+2
b
c
+2
a
c

=3+6×1×1×
1
2
=6,
∴|
AC1
|=
6
,即AC1的长为
6

(2)
BD1
=
BD
+
DD1
=
AD
-
AB
+
AA1
=
b
-
a
+
c

BD1
AC
=(
b
-
a
+
c
)•(
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