Question

在直角坐标系内已知三点A(5,1,-1),B（0,-4,3）C（1,-3,7）,试求（1）三角形ABC的面积,（2）三角

Answer 1

^AB = ^A - ^B = (5,5,-4),^ BC = ^B - ^C = ( -1,-1,-4)

面积= 1/2 | ^AB X ^BC| = 1/2 | (5,5,-4) X ( -1,-1,-4)| = 1/2 | (-24,24,0)| = 12 根(2 )

高AB = 面积 / 高所垂直的边

= (12根(2 ))/ | ^A - ^B| =(12根(2 ))/(根66)= (12(根(33))/33

高BC = 面积 / 高所垂直的边

= (12根(2 ))/ | ^B - ^C| =(12根(2 ))/(3根2)= 4

高AC = 面积 / 高所垂直的边

= (12根(2 ))/ | ^A - ^C| =(12根(2 ))/|(4,4,-8)|= (12根(2 ))/(16根6)= [(3/4)根(3)]/3