直线的方程可化为 (x-1)/1=(y+2)/(-1)=(z-5)/3 ,
因此方向向量为 (1,-1,3),这也是所求平面的法向量,
所以平面方程为 (x-4)-(y+1)+3(z-0)=0 ,
化简得 x-y+3z-5=0 .
求过A点 (4,-1,0)且与直线3(x-1)=3(-y-2)=(z-5)垂直的平面方程
求过A点 (4,-1,0)且与直线3(x-1)=3(-y-2)=(z-5)垂直的平面方程
数学人气:709 ℃时间:2019-12-15 05:47:10
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