Question

在三棱柱ABC-A1B1C1中，已知AB=AC=AA1=根号5,BC=4，点A1在底面ABC的投影是线段 求向量法

Answer 1

取坐标系O﹙0,0,0﹚ A﹙1,0,0﹚,B﹙0,2,0﹚,A1﹙0,0,2﹚ [请楼主验证，正好！]
⑴ 设E﹙λ,0，﹙1-λ﹚2﹚
OE＝﹛λ,0，﹙1-λ﹚2﹜⊥BB1 [＝AA1＝﹛-1,0,2﹜]
﹛λ,0，﹙1-λ﹚2﹚﹜•﹛-1,0,2﹜＝-λ+2-2λ＝0 ∴λ＝2/3 E﹙2/3,0,2/3﹚ AE＝√5/3
⑵ BB1C1C法向量∥OE 取n1＝﹛1,0,1﹜
A1B1C法向量∥CA1×AB＝﹛-4,-2,2﹜ 取n2＝﹛2,1,-1﹜
cos﹤n1,n2﹥＝n1•n2/﹙|n2||n2|﹚＝1/﹙√2×√6﹚＝√3/6
所求夹角余弦＝√3/6

Answer 2

Answer 3

Answer 4

推荐回答是错误的！三楼答案简洁明了正确！

Answer 5

√30/10