半径为1的球面上有三点ABC,若A和B、A和C的球面距离为π/2,B和C的球面距离为π/3,则球心到截面ABC的距离
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![半径为1的球面上有三点ABC,若A和B、A和C的球面距离为π/2,B和C的球面距离为π/3,则球心到截面ABC的距离](/uploads/image/z/6160864-40-4.jpg?t=%E5%8D%8A%E5%BE%84%E4%B8%BA1%E7%9A%84%E7%90%83%E9%9D%A2%E4%B8%8A%E6%9C%89%E4%B8%89%E7%82%B9ABC%2C%E8%8B%A5A%E5%92%8CB%E3%80%81A%E5%92%8CC%E7%9A%84%E7%90%83%E9%9D%A2%E8%B7%9D%E7%A6%BB%E4%B8%BA%CF%80%2F2%2CB%E5%92%8CC%E7%9A%84%E7%90%83%E9%9D%A2%E8%B7%9D%E7%A6%BB%E4%B8%BA%CF%80%2F3%2C%E5%88%99%E7%90%83%E5%BF%83%E5%88%B0%E6%88%AA%E9%9D%A2ABC%E7%9A%84%E8%B7%9D%E7%A6%BB)
半径为1的球面上有三点ABC,若A和B、A和C的球面距离为π/2,B和C的球面距离为π/3,则球心到截面ABC的距离
半径为1的球面上有三点ABC,若A和B、A和C的球面距离为π/2,B和C的球面距离为π/3,则球心到截面ABC的距离
半径为1的球面上有三点ABC,若A和B、A和C的球面距离为π/2,B和C的球面距离为π/3,则球心到截面ABC的距离
令O为圆心
由题可得三角锥0-ABC,∠BOC=π/3,∠BOA=∠AOC=π/2,OA=OB=OC=BC=1,
AB=BC=√2,RT△OAB斜边上的高为OD,OD=√2/2
设O到截面ABC的距离为x,(OD^2-x^2)∶(OA^2-x^2)=1/2∶√2
则((√2/2)^2-x^2)∶(1^2-x^2)=(1/2)^2∶(√2)^2
解得,x^2=3/7
x=√21/7