平面x-y+z+1=0的法向量a=(1,-1,1),
直线(x-1)/2=(y-2)/3=z/1的方向向量b=(2,3,1),
设所求直线的方向向量p=(m,n,1),则
ap=m-n+1=0,
bp=2m+3n+1=0,
解得m=-4/5,n=1/5,
取所求直线的方向向量为-5p=(4,-1,-5),
∴所求直线方程是(x-1)/4=y/(-1)=(z-2)/(-5).
过点A(1,0,2)且平行平面x-y+z+1=0切垂直于直线(X-1)/2=(y-2)/3=(z/1)的直线方程
过点A(1,0,2)且平行平面x-y+z+1=0切垂直于直线(X-1)/2=(y-2)/3=(z/1)的直线方程
数学人气:737 ℃时间:2019-09-09 17:23:50
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