=a2+b2+c2
+2(a²b+b²c+c²a)
=1+1+1+2³ 1 212+12 =6, ∴|AC1|=6, 即AC1的长为6.
(2)BD1=b+c-a,AC=a+b, ∴|BD1|=2,|AC|=3,BD1²AC =(b+c-a)²(a+b)
=b2-a2
+a²c+b²c=1. ∴cos〈BDBD1 AC61,AC〉=
BD6
1AC
=
∴AC与BD61夹角的余弦值为
6
=a2+b2+c2
+2(a²b+b²c+c²a)
=1+1+1+2³ 1 212+12 =6, ∴|AC1|=6, 即AC1的长为6.
(2)BD1=b+c-a,AC=a+b, ∴|BD1|=2,|AC|=3,BD1²AC =(b+c-a)²(a+b)
=b2-a2
+a²c+b²c=1. ∴cos〈BDBD1 AC61,AC〉=
BD6
1AC
=
∴AC与BD61夹角的余弦值为
6