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51问答网 > 一直线L过点(1,2,0)且平行于平面x-2y+z-4=0,又与直线L：x-2⼀1=y-1⼀2=z-2⼀1 相交，求直线L的方程？

一直线L过点(1,2,0)且平行于平面x-2y+z-4=0,又与直线L：x-2⼀1=y-1⼀2=z-2⼀1 相交，求直线L的方程？

谢谢各位

2025-04-08 09:17:25

推荐回答（1个）

回答1：

过点 A（1，2，0）且与平面 x-2y+z-4=0 平行的平面方程为 (x-1)-2(y-2)+(z-0)=0 ，
化简得 x-2y+z+3=0 ，
联立方程 x-2y+z+3=0 与 (x-2)/1=(y-1)/2=(z-2)/1 ，可得公共点 B（9/2，6，9/2），
所以由两点式可得直线 AB 方程为 (x-1)/(9/2-1)=(y-2)/(6-2)=(z-0)/(9/2-0) ，
化简得 (x-1)/7=(y-2)/8=(z-0)/9 。

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