空间三点A(1,2,3),B(2,-1,5),C(3,2,-5),求三角形ABC的面积?求三角形ABC的边AB上的高?
问题描述：
空间三点A(1,2,3),B(2,-1,5),C(3,2,-5),求三角形ABC的面积?求三角形ABC的边AB上的高?
答
|AB|=√[(1-2)^2+(2+1)^2+(3-5)^2]=√14
|AC|=√[(1-3)^2+(2-2)^2+(3+5)^2]=√68
|BC|=√[(2-3)^2+(-1-2)^2+(5+5)^2]=√110
cosA=(AB^2+AC^2-BC^2)/2AB*AC=-√14/√17
sinA=√3/√17
S=1/2AB*AC*sinA=√42 就是
答
|AB|=√[(1-2)^2+(2+1)^2+(3-5)^2]=√14
|AC|=√[(1-3)^2+(2-2)^2+(3+5)^2]=√68
|BC|=√[(2-3)^2+(-1-2)^2+(5+5)^2]=√110
cosA=(AB^2+AC^2-BC^2)/2AB*AC=-√14/√17
sinA=√3/√17
S=1/2AB*AC*sinA=√42