习题解答 第六章几何空间的常见曲面 习题6-1 1.试分别用正等测投影及正二等测投影画出边长等于2,3,4的长方体以及正四面体 解: 习题6-2 1.分别就下列条件求球面方程: (1)一直径的两端点为A(2,-3,5)和B(4,1,-3) (2)球心在直线4=+8=22上,且过点(,-3,6和(63,-2 (3)过点(-1,2,5),且与3个坐标平面相切; 4 (4)过点(√2,√22),且包含圆 解()球心坐标C(24,3+1,523)=(8.-11,半径 R=√(3-2)2+(-1+3)2+(1-5)2=√21 所以球面方程为(x-3)2+(y+1)2+(2-1)2=21. (2)因球心在已知直线上,故它的坐标应为(4+2,-8-4t,2+t).又因点(2,-3,6)和(6,3,-2)在 球面上所以它们到球心的距离相等,即 (4+2-2)2+(-8-4t+3)2+(2+t-6)2=(4+t-6)2+(-8-4t-3)2+(2+t+2)2 解得t=-2,从而球心坐标是(00,0),且半径等于7.球面方程为x2+y2+2=49 (3)球心与点(-1,2,5)在同一卦限内,因此可设它的坐标为(-a,a,a),则球面方程为 x+a)2+(y-a)2+(2-a)2=a2￾
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