数学计算问题（关于航海）某船拟由A（30ºN 50

2个回答

a·b=|a|*|b|cosθax,ay,az;bx,by,bz a·b=axbx+ayby+azbz|a|=sqrt(ax^2+ay^2+az^2)用数量积可以求出两向量的夹角的余弦引入三维坐标,原点o地心,z北极方向,x赤道上0度经线交点方向,y赤道上90E交点方向xa=6371cos30cos50,ya=6371cos30sin50,za=6371sin30xb=6371cos60cos140,yb=6371cos60sin140,zb=6371sin60化简A：6371(0.556670399 0.663413948 0.5)B：6371(-0.383022222 0.321393805 0.866025404)a·b=6371*6371*（-0.213217133+0.213217133+0.433012702）cosAOB=0.433012702角AOB=64.34109373优弧AOB=360-64.34109373=295.6589063航程=2*pi*6371*295.6589063/360=32875.7704km(-0.213217133,0.213217133,0.433012702)是AOB所在平面的法相量,而赤道的法相量极为z轴,可以取(0,0,1)夹角余弦=0.433012702,夹角也等于64.3410937390-64.34109373=25.65890628起始角度：南偏西 25.65890628度