上正垂足减去垂足x+y+z=1的2倍难包 、除正球面法平面x+y+z=1的交线就意S2.试正S法S2相同应距球面上 7.证除:过圆 S=x2+y2+2+2ux+2y+2uz+d=0,(2+n2+un2-d>0) E= Ar+ By+Cz+D=0 (A2+B2+C2≠0) 的球面族垂足二表向圆:S+2AE=0(入圆参数) S=0 前再:对切参数λ说垂足S+2E=0确表向应距球面,则圆 应该相正球面上 E=0 对任.应距过圆 E=0 的球面 已过圆应相已距球的交线 x2+y2+z2+2ux+2y+2uz+d=0 x2+y2+22+2px +2gy+2t 上.平第4题过,正圆应相平面 (p-u)I+(q-u)y+(t-w)z+ 上但E=0也过正圆,试正已距平面重合,即存相数λ使包 P-=入A,q-U=AB,t-u=C, 交长p,q,t,r,代入球面垂足即包 到直6-3 1.求面就旋转曲面的垂 (1)线=1 旋转; (2.线1= 旋转 (3)抛物线 它的准线旋转; =数=1旋转 解:(1)显然端点O相旋转。上,且的垂向向量意ξ=(1,1,1).参照(3.1),二限包垂足组 (x-x)+(y-y)+(z-2)=0 相垂足组标消去参数x,y,z′难二包 试正卦求旋转曲面的垂足圆 2(x2+y2+2)-5(xy+xz+y2)+5(x+y+z)-7=0.. )TUYZTU x + y + z = 1  2 ng=￾ x 2 + y 2 + z 2 − 2x − 2y − 2 = 0, c)53;3 x + y + z = 1 %4 S2. () S1  S2 EBC53. ∗7. : :#: ( S = x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, (u 2 + v 2 + w 2 − d > 0), E = Ax + By + Cz + D = 0, (A2 + B2 + C 2 6= 0) 53oTU.PQ#: S + 2λE = 0 (λ #b ). : N<b λ, cTU S + 2λE = 0 PpPQBC53, # ( S = 0, E = 0 Bm)53. NBC:# ( S = 0, E = 0 53 x 2 + y 2 + z 2 + 2px + 2qy + 2tz + r = 0, @:#B@C5% ( x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d = 0, x 2 + y 2 + z 2 + 2px + 2qy + 2tz + r = 0 . ;| 4 a:, )#B;3 (p − u)x + (q − v)y + (t − w)z + r − d 2 = 0 . q E = 0 :)#, ()@C;3!r, tp λ = p − u = λA, q − v = λB, t − w = λC, r − d 2 = λD. %0 p, q, t, r, JK53TUt= x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + d + 2λ(Ax + By + Cz + D) = 0. ￾
6–3 1. X34s3TU: (1)
 x 2 = y 1 = z − 1 0 t
 x = y = z s; (2)
 x − 1 1 = y −3 = z 3 t z Ms; (3) uv ( y 2 = 2px, z = 0 tAws; (4)  ( x 2 = y, x + z = 0 t
 x 1 = y 2 = z 1 s. : (1) xy7& O sM, $MTQQV ξ = (1, 1, 1). bx (3.1), .G=￾TU"    (x − x 0 ) + (y − y 0 ) + (z − z 0 ) = 0, x 2 + y 2 + z 2 = x 02 + y 02 + z 02 , x 0 2 = y 0 1 = z 0 − 1 0 , TU" yZb x 0 , y0 , z0 g.= x 2 + y 2 + z 2 − 1 = 5 9 (x + y + z − 1)2 , ()FXs3TU# 2(x 2 + y 2 + z 2 ) − 5(xy + xz + yz) + 5(x + y + z) − 7 = 0. · 4 ·