求过点P(1.1.1)且与两直线L1 x=y/2=z/3,L2 (x-2)/2=y-2=(z-3)
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发布时间：2024-04-16 22:52
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热心网友
时间：2024-04-19 14:30
则向量PQ=(t-1,2t-1,3t-1),
直线PQ的方程是(x-1)/(t-1)=(y-1)/(2t-1)=(z-1)/(3t-1),①
它与L2 （x-2）/2=y-2=（z-3）/4②有公共点，
由②，x=2y-2,z=4y-5,都代入①，得
(2y-3)/(t-1)=(y-1)/(2t-1)=(4y-6)/(3t-1),
∴(2y-3)(2t-1)=(y-1)(t-1),(y-1)(3t-1)=(4y-6)(2t-1),
∴4ty-2y-6t+3=ty-y-t+1,3ty-y-3t+1=8ty-4y-12t+6,
整理得3ty-y-5t+2=0,③5ty-3y-9t+5=0,④
③*5-④*3，得4y+2t-5=0,y=(5-2t)/4,
代入③*4，得(3t-1)(5-2t)+4(-5t+2)=0,
-6t^2-3t+3=0,
解得t=-1,或1/2(舍).
代入①，得所求方程为(x-1)/(-2)=(y-1)/(-3)=(z-1)/(-4),
即(x-1)/2=(y-1)/3=(z-1)/4.
