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平面x/3+y/4+z/5=1和柱面x^2+y^2=1的交线上到平面xoy最短的点

湖州添众1 1年前 已收到1个回答 我来回答 举报

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几何法:

设柱面x^2+y^2=1交xOy平面于圆O:x^2+y^2=1(z=0)

平面x/3+y/4+z/5=1交xOy平面于直线AB:x/3+y/4=1(z=0),A(0,4,0),B(3,0,0)

过O做OC⊥AB于C,交圆O于D

cosCOB=sinABO=4/5

sinCOB=cosABO=3/5

所以D点坐标为(4/5,3/5,0)

所求点即为过D点且垂直于xOy平面的直线与平面x/3+y/4+z/5=1的交点

将D点坐标代入平面方程即得所求点坐标(4/5,3/5,35/12)

解析法: 设该点坐标为(cosa,sina,z),a∈[0,2π)

则(cosa)/3+(sina)/4+z/5=1

z=5-(25/12)((4/5)cosa+(3/5)sina)

=5-(25/12)sin(a+b)

其中b∈(0,π/2),且sinb=4/5,cosb=3/5

当a+b=π/2+2kπ时,k∈Z

z最小为35/12

此时a=π/2-b cosa=sinb=4/5,sina=cosb=3/5

故所求点坐标为(4/5,3/5,25/12)

1年前

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