(1) AC 1的长为
(2) AC与BD 1夹角的余弦值为
记
=a,
=b,
=c,
则|a|=|b|=|c|=1,〈a,b〉=〈b,c〉=〈c,a〉=60°,
∴a·b=b·c=c·a=
.
(1)|
| 2=(a+b+c) 2
=a 2+b 2+c 2+2(a·b+b·c+c·a)
=1+1+1+2×(
+
+
)=6,
∴|
|=
,即AC 1的长为
.
(2)
=b+c-a,
=a+b,
∴|
|=
,|
|=
,
·
=(b+c-a)·(a+b)
=b 2-a 2+a·c+b·c=1.
∴cos〈
,
〉=
=
.
∴AC与BD 1夹角的余弦值为
.