已知长方体ABCD-A'B'C'D'中,AB=BC=4,CC'=2则直线BC'和平面DBB'D'所成交的余弦值为

Question

已知长方体ABCD-A'B'C'D'中,AB=BC=4,CC'=2则直线BC'和平面DBB'D'所成交的余弦值为

Answer 1

设正方形A'B'C'D'的对角线的交点为O'. 知C'O'垂直于B'D'.
又BB'垂直于平面A'B'C'D'.故BB'垂直于其内的直线C'O'.
即C'O'垂直于相交直线B'D',BB'. 故C'O'垂直于平面DBB'D'.
连接O'B, 则角O'BC' 即为BC'与平面DBB'D'所成角.
在三角形BO'C'中: BC'=根号(16+4)=2根号5.
O'C'=0.5*根号(16+16)=2根号2.
BO'=根号(4+8)=2根号3
故cos角O'BC'=[12+20-8]/[2*(2根号3)*2根号5]=24/[8根号15]=3/[根号15]
即所求角的余弦为:(根号15)/5