求过三条平行直线x = y = z, x +1 = y = z −1,与x −1 = y +1 = z − 2的圆柱面方程.
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![求过三条平行直线x = y = z, x +1 = y = z −1,与x −1 = y +1 = z − 2的圆柱面方程.](/uploads/image/z/12974454-54-4.jpg?t=%E6%B1%82%E8%BF%87%E4%B8%89%E6%9D%A1%E5%B9%B3%E8%A1%8C%E7%9B%B4%E7%BA%BFx+%3D+y+%3D+z%2C+x+%2B1+%3D+y+%3D+z+%26%238722%3B1%2C%E4%B8%8Ex+%26%238722%3B1+%3D+y+%2B1+%3D+z+%26%238722%3B+2%E7%9A%84%E5%9C%86%E6%9F%B1%E9%9D%A2%E6%96%B9%E7%A8%8B.)
求过三条平行直线x = y = z, x +1 = y = z −1,与x −1 = y +1 = z − 2的圆柱面方程.
求过三条平行直线x = y = z, x +1 = y = z −1,与x −1 = y +1 = z − 2的圆柱面方程.
求过三条平行直线x = y = z, x +1 = y = z −1,与x −1 = y +1 = z − 2的圆柱面方程.
这3条直线的方向向量都是L=(1,1,1)
取x=y=z上点(0,0,0),过此点的法平面为x+y+z=0
此平面与x+1=y=z-1和x-1=y+1=z-2的交点分别是(-1,0,1)、(1/3,-5/3,4/3)
那么圆柱面的轴心上的点设为(a,b,c),到这3点距离相等.满足方程组:
a²+b²+c²=(a+1)²+b²+(c-1)²=(a-1/3)²+(b+5/3)²+(c-4/3)²
解得:b=a-3/5,c=a+1
所以圆柱体轴心的直线方程为a=b+3/5=c-1
易算得x=y+3/5=z-1与x+y+z=0的交点为(-2/15,-11/15,13/15)
得到圆柱体半径r=√[(-2/15)²+(-11/15)²+(13/15)²]=7√6/15
任取圆柱面上一点(x,y,z),使得它在轴心法平面与轴心的交点为(a,b,c),满足
(x-a)+(y-b)+(z-c)=0,(x-a)²+(y-b)²+(z-c)²=r²
第一式乘以(z-c)和第二式相减,有(x-a)(x-z+1)+(y-b)(y-z+8/5)=98/75
再根据b=a-3/5以及x+y+z=a+b+c=a+(a-3/5)+(a+1)=3a+2/5
消去a、b,化简,得圆柱面的方程
x²+y²+z²-xy-yz-zx+2x/5+11y/5-13z/5=0