Question

求详细过程 高二数学！

Answer 1

AB = √{(3+2)²+(4+1)²+(4-5)²} = √51
AC = √{(3-4)²+(4-5)²+(4-0)²} = 3√2
BC = √{(-2-4)²+(-1-5)²+(5-0)²} = √97
cosA = (AB²+AC²-BC²)/(2*AB*AC) = (51+18-97)/(2*√51*3√2) = -14/(3√102)
∵S△ABD=1/3S△ABC
∴AD=1/3AC = √2
BD² = AB²+AD²-2AB*ADcosA = 51+2-2*√51*√2*{-14/(3√102)} = 187/3
BD = √561/3

Answer 2

解：三角形ABD的面积是三角形ABC面积的1/3===>AD=DC/2.===>DC=2AD
设D点坐标为（x，y，z）===>向量AD=（X-3,y-4,z-4) DC=(4-x,5-y,-z)
4-X=2(X-3) 5-Y=2(y-4) -z=2(z-4)===>X= 10/3 Y=13/3 Z=8/3
所以 ｜BD|=根号【（10/3+2)^2+(13/3+1)^2+(8/3-5)^2]=根号（561）/3。===》选B.