[题目]直线 dfrac (x+1)(-1)=dfrac (y-2)(1)=dfrac (z-1)(0) 与平面 4x+y+z=1 的-|||-夹角为() ()-|||-

直线dfrac (x-1)(1)=dfrac (y-1)(0)=dfrac (z-1)(-1)-|||-__与直线dfrac (x-1)(1)=dfrac (y

求直线dfrac (x-1)(1)=dfrac (y)(1)=dfrac (z-1)(-1)-|||-__ __在平面dfrac (x-1)(1)=dfrac

过直线_(1):dfrac (x-1)(1)=dfrac (y-2)(0)=dfrac (z-3)(-1)-|||-__ __且平行于直线_(1):dfrac

直线dfrac (x-1)(1)=dfrac (y)(2)=dfrac (z+3)(-2)-|||-__ __-|||-__与平面dfrac (x-1)(1)=

直线dfrac (x-1)(2)=dfrac (y)(-1)=dfrac (z-2)(3)-|||-__与直线dfrac (x-1)(2)=dfrac (y)(

35.求过点 (-1,0,4), 且与平面 3x-4y+z-10=0 平行,又与直线 dfrac (x+1)(1)=dfrac (y-3)(1)=dfrac (

3、过直线 _(1):dfrac (x-1)(1)=dfrac (y-2)(0)=dfrac (z-3)(-1) 且平行于直-|||-线 _(2):dfrac

[题目]直线 dfrac (x-1)(2)=dfrac (y)(1)=dfrac (z+1)(-1) 与平面 x-y+z=1 的位-|||-置关系是 ()-||

13.求下列直线的夹角:-|||-(1) =dfrac {y-3)(-12)=dfrac (z-1)(3) 和 =dfrac {z-8)(-2) x=1

直线-1=dfrac (y+3)(-2)=z-4与直线-1=dfrac (y+3)(-2)=z-4的夹角为( )-1=dfrac (y+3)(-2)=z-4-1