题目

如图,在正三棱柱ABC—A1B1C1中,AB=AA1,E是棱BB1上的点,且平面A1EC⊥平面AA1C1C.(1)试确定点E的位置;(2)若把平面A1EC与平面A1B1C1所成的锐二面角为60°时的正三棱柱称为“黄金棱柱”,请判断此棱柱是否为“黄金棱柱”,并说明理由. 答案：解法一:(1)连结AC1交A1C于点O,连结AE、EC.由于AB=AA1,三棱柱ABC—A1B1C1为正三棱柱,∴四边形AA1C1C为正方形.∴AO⊥A1C.∵平面A1EC⊥平面AA1C1C且平面A1EC∩平面AA1C1C=A1C,∴AO⊥平面A1EC.∴AO⊥OE.                                              在△AEC1中,AE=EC1,即.而AB=B1C1,∴BE=B1E,即点E为棱BB1的中点.                                          (2)延长CE交C1B1的延长线于F,连结A1F,而B1E∥C1C,∴=.∴FB1=B1C1.∴FB1=B1C1=A1B1.∴FA1⊥A1C1.                                                             ∵面AA1C1C⊥面A1B1C1,面AA1C1C∩面A1B1C1=A1C1,∴FA1⊥面AA1C1C.∴FA1⊥CA1，FA1⊥C1A1.∴∠CA1C1就是二面角CA1FC1的平面角.                                     而∠CA1C1=45°≠60°,∴此三棱柱不是“黄金棱柱”.                                              解法二:(1)如图建立空间直角坐标系,设AB=AA1=a,B1E=b,则A1(0,0,0)、B1(,,0)、C1(0,a,0)、E(a,,b)、C(0,a,a).                 ∴=(a,,b),=(0,a,a).                                      设平面A1EC的法向量为m=(x,y,z),则∴可取m=(,-1,1).                                                   显然平面A1C1CA的法向量可取为n=(1,0,0).由题意,平面A1EC⊥平面AA1C1C,∴m⊥n.∴m·n=0,即=0,b=.∴点E为BB1的中点.                                                     (2)由(1)知,m=(0,-1,1),显然平面A1B1C1的法向量可取为l=(0,0,1),∴cos〈m,l〉=.                                         ∴〈m,l〉=45°.∴平面A1EC与平面A1B1C1所成的锐二面角为45°,即此三棱柱不是“黄金棱柱”.

查看本题答案和解析