平行六面体ABCD-A1B1C1D1中,AB=4,AD=3,AA1=5,角BAD=角BAA1=角DAA1=60°则AC1的长=?

向量AC'=向量AB+向量AD+向量AA'
=>
AC'^2 = (向量AB+向量AD+向量AA')^2
=
AB^2 + AD^2 + AA'^2 + 2(向量AB*向量AD+向量AA'*向量AB+向量AD*向量AA')
=
AB^2 + AD^2 + AA'^2 + 2AB*ADcos60+2AA'*ABcos45+2AD*AA'cos45
=
25+9+49+15+35√2+21√2
=
98+56√2
=>
AC' = √(98+56√2)

∵六面体ABCD-A1B1C1D1是平行六面体，
∵
AC1
=
AA1
+
AD
+
AB
∴|
AC1
| 2=（
AA1
+
AD
+
AB
）2=|
AA1
| 2+|
AB
| 2+|
AD
| 2+2
AA1
•
AD
+2
AA1
•
AB
+2
AB
•
AD
又∵∠BAD=∠A1AB=∠A1AD=60°，AD=4，AB=3，AA1=5，
∴|
AC1
| 2=16+9+25+2×5×4×cos60°+2×5×3×cos60°+2×3×4×cos60°=97
∴|
AC1
|=
97
故答案为
97

向量AC'=向量AB+向量AD+向量AA'=>AC'^2 = (向量AB+向量AD+向量AA')^2 = AB^2 + AD^2 + AA'^2 + 2(向量AB*向量AD+向量AA'*向量AB+向量AD*向量AA')=AB^2 + AD^2 + AA'^2 + 2AB*ADcos60+2AA'*ABcos60+2AD*AA'cos60=16+9...