1.六面体ABCD-A'B'C'D'中,AB=5,AD=3,AA'=7,角BAD=60度,角BAA'=角DAA'=45度,求AC'的长2.正方体ABCD-A'B'C'D'棱长为a,求A'B和B'C夹角.求证：A'B垂直于AC'

Question

1.六面体ABCD-A'B'C'D'中,AB=5,AD=3,AA'=7,角BAD=60度,角BAA'=角DAA'=45度,求AC'的长
2.正方体ABCD-A'B'C'D'棱长为a,求A'B和B'C夹角.
求证：A'B垂直于AC'

Answer 1

①.
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)
②.以下均为向量
A'B垂直AC',即证A'B*A'C=0,所以有：
A'B*A'C=(A'A+AB)(AC+CC')=A'A*CC'+AB*CC'+AB*(AC+CC')
=A'A*CC'+AB*CC'+AB*(AB+BC+CC')
=-(AA')^2+0+(AB)^2+0+0
=-a^2+a^2=0
所以垂直（因为正方体,所以有很多已知的垂直,利用他们）
因为/A'B/=/A'A+AB/=√((A'A+AB)^2)=√(A'A^2+AB^2+0)=√2*a
/B'C/=/B'C'+C'C/=√((B'C'+C'C)^2)=√(B'C'^2+C'C^2+0)=√2*a
A'B*B'C=(A'A+AB)(B'C'+C'C)=0+a^2+0+0=a^2
所以,设夹角为θ,
COSθ=COS=A'B*B'C/(/A'B/*/B'C/)=1/2
所以,夹角θ=60