Question

已知三角形三顶点A(1,-2,-3),B(-1,-1,-1),C(0,0,-5),试证明△ABC是直角三角形

Answer 1

根据坐标算距离

|AB|^2=(-1-1)^2+(-1-(-2))^2+(-1-(-3))^2=9

|AC|^2=(0-1)^2+(0-(-2))^2+(-5-(-3))^2=9

|BC|^2=(0-(-1))^2+(0-(-1))^2+(-5-(-1))^2=18

∴|AB|^2+|AC|^2=|BC|^2

∴△ABC是直角三角形